Answer:
1) the domain is all real numbers
2) the range is
[tex]y \geqslant 3[/tex]
which graph shows a reflection across the line Y = X
Answer:
B
Step-by-step explanation:
"A" is not a reflection, it looks like a translation.
"C" is not a reflection, it is a rotation.
So, B is a reflection.
Answer:
[tex]\large \boxed{\mathrm{Graph \ C}}[/tex]
Step-by-step explanation:
The reflection is across the line y = x.
All options show reflection. Option C shows reflection across the line y = x.
In the reflection, the points on the triangle will also be reflected.
Point S is reflected across the line y=x, the reflected point is S’.
Point R is reflected across the line y=x, the reflected point is R’.
Point Q is reflected across the line y=x, the reflected point is Q’.
Find the interquartile range of the data in the dot plot below. players blob:mo-extension://5f64da0e-f444-4fa8-b754-95
Answer:
[tex]IQR=Q_{3}-Q_{1}[/tex]
Step-by-step explanation:
The inter-quartile range is a measure of dispersion of a data set.
It is the difference between the third and the first quartile.
[tex]IQR=Q_{3}-Q_{1}[/tex]
The 1st quartile (Q₁) is well defined as the mid-value amid the minimum figure and the median of the data set. The 2nd quartile (Q₂) is the median of the data. The 3rd quartile (Q₃) is the mid-value amid the median and the maximum figure of the data set.
The distribution of baby weights at birth is left-skewed because of premies (premature babies) who have particularly low birth weights. However, within a close range of gestation times, birth weights are approximately Normally distributed. For babies born at full term (37 to 39 completed weeks of gestation), for instance, the distribution of birth weight (in grams) is approximately N(3350,440).N(3350, 440).10 Low-birth-weight babies (weighing less than 2500 grams, or about 5 pounds 8 ounces) are at an increased risk of serious health problems. Among those, very-low-birth-weight babies (weighing less than 1500 grams, or about 3 pounds 4 ounces) have the highest risk of experiencing health problems.
A. What proportion of babies born full term are low-birth-weight babies?
B. What proportion of babies born full term are very-low-birth-weight babies?
Answer:
a
[tex]P(X < 2500) = 0.02668[/tex]
b
[tex]P(X < 1500) = 0.00001[/tex]
Step-by-step explanation:
From the question we are told that
The population mean is [tex]\mu = 3350[/tex]
The standard deviation is [tex]\sigma = 440[/tex]
We also told in the question that the birth weight is approximately Normally distributed
i.e [tex]X \ \~ \ N(\mu , \sigma )[/tex]
Given that Low-birth-weight babies weighing less than 2500 grams,then the proportion of babies born full term are low-birth-weight babies is mathematically represented as
[tex]P(X < 2500) = P(\frac{ X - \mu }{\sigma } < \frac{2500 - \mu}{\sigma } )[/tex]
Generally
[tex]\frac{X - \mu}{ \sigma } = Z (The \ standardized \ value \ of \ X )[/tex]
So
[tex]P(X < 2500) = P(Z < \frac{2500 - \mu}{\sigma } )[/tex]
substituting values
[tex]P(X < 2500) = P(Z < \frac{2500 - 3350}{440 } )[/tex]
[tex]P(X < 2500) = P(Z <-1.932 )[/tex]
Now from the standardized normal distribution table(These value can also be obtained from Calculator dot com) the value of
[tex]P(Z <-1.932 ) = 0.02668[/tex]
=> [tex]P(X < 2500) = 0.02668[/tex]
Given that very-low-birth-weight babies (weighing less than 1500 grams,then the proportion of babies born full term are very-low-birth-weight babies is mathematically represented as
[tex]P(X < 1500) = P(\frac{ X - \mu }{\sigma } < \frac{1500 - \mu}{\sigma } )[/tex]
[tex]P(X < 1500) = P(Z < \frac{1500 - \mu}{\sigma } )[/tex]
substituting values
[tex]P(X < 1500) = P(Z < \frac{1500 - 3350}{440 } )[/tex]
[tex]P(X < 1500) = P(Z <-4.205 )[/tex]
Now from the standardized normal distribution table(These value can also be obtained from calculator dot com) the value of
[tex]P(Z <-1.932 ) = 0.00001[/tex]
[tex]P(X < 1500) = 0.00001[/tex]
Is the quotient of two rational numbers always a rational number? Explain.
Answer:
Yes,
Step-by-step explananation
The quotient of two rational numbers is always rational, and the reason for this lies in the fact that the product of two integers is always an rational number.
The Quotient of two Rational Numbers is a Rational Number if and only if Numerator and Denominator are Multiples.
From Algebra, we know that a Rational Number is a Real Number of the form:
[tex]x = \frac{a}{b}[/tex], [tex]a, b\in \mathbb{N}[/tex], [tex]x \in \mathbb{R}[/tex] (1)
Where:
[tex]a[/tex] - Numerator.[tex]b[/tex] - Denominator.[tex]x[/tex] - Quotient.The Quotient can be an Integer or not. In the first case, all Quotients have their equivalent Rational Numbers.
Now, if we divide a Rational Number by another Rational Number, then we have the following expression:
[tex]x' = \frac{x_{1}}{x_{2}}[/tex]
If [tex]x'[/tex] is a Rational Number, then it must also an Integer and if [tex]x'[/tex] is an Integer, then [tex]x_{1}[/tex] and [tex]x_{2}[/tex] must be Multiples of each other.
The Quotient of two Rational Numbers is a Rational Number if and only if Numerator and Denominator are Multiples.
Please see this question related to Rational Numbers: https://brainly.com/question/24398433
Graph: y < 3x + 1 please help me
Answer:
Using a graphing calc.
Step-by-step explanation:
The numbers 1,2,3,4,5,6,7,8,9. How would you put them in each of a square block to create the sum on each line to make the number 15. The sum of each diagonals should also be 15.
Answer:
Here's one way:
4 9 2
3 5 7
8 1 6
Step-by-step explanation:
Determine the number that will complete the square to solve the equation after the constant term has been written on the right side. Do not solve the equation.
x2+3x−18=0
Answer:
Step-by-step explanation:
Hello, "the constant term has been written on the right side", it means that we add 18 to both sides to get.
[tex]x^2+3x-18=0\\\\x^2+3x=18\\\\\text{We can see the beginning of } (x+\dfrac{3}{2})^2 \\\\x^2+3x=(x+\dfrac{3}{2})^2-\dfrac{3^3}{2^2}=18\\\\(x+\dfrac{3}{2})^2=18+\dfrac{9}{4}=\dfrac{18*4+9}{4}=\dfrac{81}{4}[/tex]
Hope this helps.
Thank you.
Answer:
2.25.
Step-by-step explanation:
x^2 + 3x - 18 = 0
First, we need to write the constant on the right of the equation. So, we add 18 to both sides.
x^2 + 3x = 18.
Now, we find the number that will complete the square. It will be [tex](\frac{b}{2} )^2[/tex].
In this case, b = 3.
[tex](\frac{3}{2} )^2[/tex]
= (1.5)^2
= 2.25.
So, the number that will complete the square to solve the equation is 2.25, or 2 and 1/4, or 9/4.
Hope this helps!
i will rate you brainliest
Answer:
D. Factoring trinomials
Step-by-step explanation:
The factoring trinomials method is the best way to factor the expression, since it is in the standard trinomial form ax² + bx + c
In this method, you can factor the expression by finding 2 factors of c that add up to b.
The expression is not in the simplest form, and difference of squares cannot be used because there are no perfect squares. Prime factorization is also not used for factoring expressions with variables.
So, D is the right answer.
Choose the situation that represents a function.
A) The number of raisins in an oatmeal raisin cookie is a function of the diameter of the cookie.
B) The inches of rainfall is a function of the day’s average temperature.
C) The time it takes to cook a turkey is a function of the turkey’s weight.
D) The number of sit-ups a student can do in a minute is a function of the student’s age.
Answer:c
Step-by-step explanation:
Answer: The answer is C.
Hope this helps you!
Find the number c that satisfies the conclusion of the Mean Value Theorem on the given interval. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.) f(x) = x , [0, 16]
Answer:
C = 4
Step-by-step explanation:
solution:
f(x) can be differentiated on (0,16)
By mean value theorem
= f(16) = 4
= f(0) = 0
= f(b) - f(a)/b - a
= f(4) - f(0)/ f(16) - f(0)
= f'(c) = 1/2√C
= 1/2√C = 4/16
= 1/2√C = 1/4
= 4 = 2√C
= √C = 4/2
we make c the subject of the formula and also eliminate the square root
= √C = 2
= C = 2²
= C = 4
In a random sample of 380 cars driven at low altitudes, 42 of them exceeded a standard of 10 grams of particulate pollution per gallon of fuel consumed. In an independent random sample of 90 cars driven at high altitudes, 24 of them exceeded the standard. Compute the test statistic for testing if the proportion of high-altitude vehicles exceeding the standard is greater than the proportion of low-altitude vehicles exceeding the standard.
Answer:
The test statistic for testing if the proportion of high-altitude vehicles exceeding the standard is greater than the proportion of low-altitude vehicles exceeding the standard is 3.234.
Step-by-step explanation:
We are given that in a random sample of 380 cars driven at low altitudes, 42 of them exceeded a standard of 10 grams of particulate pollution per gallon of fuel consumed.
In an independent random sample of 90 cars driven at high altitudes, 24 of them exceeded the standard.
Let [tex]p_1[/tex] = population proportion of cars driven at high altitudes who exceeded a standard of 10 grams.
[tex]p_2[/tex] = population proportion of cars driven at low altitudes who exceeded a standard of 10 grams.
So, Null Hypothesis, [tex]H_0[/tex] : [tex]p_1\leq p_2[/tex] {means that the proportion of high-altitude vehicles exceeding the standard is smaller than or equal to the proportion of low-altitude vehicles exceeding the standard}
Alternate Hypothesis, [tex]H_A[/tex] : [tex]p_1>p_2[/tex] {means that the proportion of high-altitude vehicles exceeding the standard is greater than the proportion of low-altitude vehicles exceeding the standard}
The test statistics that will be used here is Two-sample z-test statistics for proportions;
T.S. = [tex]\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }[/tex] ~ N(0,1)
where, [tex]\hat p_1[/tex] = sample proportion of cars driven at high altitudes who exceeded a standard of 10 grams = [tex]\frac{24}{90}[/tex] = 0.27
[tex]\hat p_2[/tex] = sample proportion of cars driven at low altitudes who exceeded a standard of 10 grams = [tex]\frac{42}{380}[/tex] = 0.11
[tex]n_1[/tex] = sample of cars driven at high altitudes = 90
[tex]n_2[/tex] = sample of cars driven at low altitudes = 380
So, the test statistics = [tex]\frac{(0.27-0.11)-(0)}{\sqrt{\frac{0.27(1-0.27)}{90}+\frac{0.11(1-0.11)}{380} } }[/tex]
= 3.234
The value of z-test statistics is 3.234.
2 divided by ___=42 two divided by what equals 42?
i need help will rate you branliest
Answer:
d. The graph of g(x) is the graph of f(x) reflected over the x-axis.
Step-by-step explanation:
The standard transformation
g(x) = - f(x)
is a simple reflection about the x-axis.
So the answer is the last option.
Answer:
Last one
Step-by-step explanation:
The function we are interested in are g(x) and f(x).
● g(x)= (-1/x)
● f(x)= 1/x
Notice what happens when we input the same values in both functions.
● g(1) = -1/1 = -1
● f(x) = 1/1 = 1
●g(2) = -1/2 = -0.5
● f(2) = 1/2 = 0.5
Notice that we get opposite values by imputing the same number.
Wich means:
●f(x) = -g(x)
So the graph of g(x) is the graph of f(x) reflected over the x axis.
A market research firm supplies manufacturers with estimates of the retail sales of their products from samples of retail stores. Marketing managers are prone to look at the estimate and ignore sampling error. A random sample of 36 stores this year shows mean sales of 83 units of a small appliance with a standard deviation of 5 units. During the same point in time last year, a random sample of 49 stores had mean sales of 78 units with standard deviation 3 units.Required:Construct a 95 percent confidence interval for the difference in population means.
Answer:
The 95% confidence interval for the difference in population means is (−26.325175 , 36.325175)
Step-by-step explanation:
Given that :
sample size n₁ = 36
sample mean [tex]\over\ x[/tex]₁ = 83
standard deviation [tex]\sigma[/tex]₁ = 5
sample size n₂ = 49
sample mean [tex]\over\ x[/tex]₂= 78
standard deviation [tex]\sigma[/tex]₂ = 3
The objective is to construct a 95% confidence interval for the difference in the population means
Let the population means be [tex]\mu_1[/tex] and [tex]\mu_2[/tex]
The 95% confidence interval or the difference in population means can be calculated by using the formula;
[tex](\overline{x_1} - \overline{x_2}) \pm t_{\alpha /2} \ \times s_{p}[/tex]
where;
the pooled standard deviation [tex]s_{p} = \dfrac{(n_1-1)s_1^2+(n_2-1)s^2_2}{n_1+n_2-2}[/tex]
[tex]s_{p} = \dfrac{(36-1)5^2+(49-1)3^2}{36+49-2}[/tex]
[tex]s_{p} = \dfrac{(35)25+(48)9}{83}[/tex]
[tex]s_{p} = \dfrac{875+432}{83}[/tex]
[tex]s_{p} = \dfrac{1307}{83}[/tex]
[tex]s_p[/tex] = 15.75
degree of freedom = [tex]n_1 +n_2 -2[/tex]
degree of freedom = 36+49 -2
degree of freedom = 85 - 2
degree of freedom = 83
The Critical t- value 95% CI at df = 83 is
t critical = T.INV.2T(0.05, 83) = 1.9889
Therefore, for the population mean , we have:
= (83 - 78) ± (1.9889 × 15.75)
= 5 ± 31.325175
= 5 - 31.325175 , 5 + 31.325175
= (−26.325175 , 36.325175)
Commuting times for employees of a local company have a mean of 63 minutes and a standard deviation of 3 minutes. What does Chebyshev's Theorem say about the percentage of employees with commuting times between 54 minutes and 72minutes?
Answer: At-least 89% of employees with commuting times between 54 minutes and 72 minutes .
Step-by-step explanation:
Given: Commuting times for employees of a local company have a mean of 63 minutes and a standard deviation of 3 minutes.
Now, 54 minutes = (63 - 9) minutes
= (63 -3(3)) minutes
= Mean - 3 standard deviation
72 minutes = (63 + 9) minutes
=63 +3(3) minutes
= Mean + 3 standard deviation
According to Chebyshev's theorem, at least [tex]\dfrac{8}{9}[/tex] of the data lie within 3 standard deviations of the mean.
i.e. The percentage of employees with commuting times between 54 minutes and 72 minutes = [tex]\dfrac{8}{9}\times100\approx89\%[/tex]
Hence, at-least 89% of employees with commuting times between 54 minutes and 72 minutes .
PLEASE ANSWER ASAP!!
Question is in the picture as well as the answer choices
any unrelated answers will be reported
Answer:
c
Step-by-step explanation
i will rate you brainliest
Answer:
Option (2)
Step-by-step explanation:
In an arithmetic progression,
[tex]a_1,a_2,a_3.........a_{n-1},a_n[/tex]
First term of the progression,
a = [tex]a_1[/tex]
Common difference 'd' = [tex](a_2-a_1)[/tex]
Recursive formula for the sequence,
a = [tex]a_1[/tex]
[tex]a_n=a_{n-1}+d[/tex]
By applying these rules in the recursive formula,
[tex]a_1=\frac{4}{5}[/tex]
[tex]a_n=a_{n-1}+\frac{3}{2}[/tex]
Common difference 'd' = [tex]\frac{3}{2}[/tex]
Therefore, Option (2) will be the answer.
The scale on a scale drawing is 1 : 30. What should you do with each measurement on the drawing to get the actual dimensions? Provide an example of a drawing that uses this scale. Include both the original and new dimensions.
Answer:
see below
Step-by-step explanation:
For the first question, you should multiply the scale dimension by 30 to get the actual dimension. This is because the scale is 1:30 where the scale dimension is the 1 and the actual dimension is 30, so therefore, the scale dimension is 1/30th of the actual dimension, so to get the actual dimension, we can multiply the scale dimension by 30. I'm not totally sure how to attach pictures from my phone on my computer (sorry) but an example of a drawing could be two rectangles, the first (this is the scale drawing) having dimensions of 1 by 2 units and the second (this is the actual drawing) having dimensions of 30 by 60 units. I hope this helps!
An animal population is increasing at a rate of 13 51t13 51t per year (where t is measured in years). By how much does the animal population increase between the fourth and tenth years.
Answer:
ΔP = 567
Step-by-step explanation:
The increasing rate of the population is 13,51*t.
That rate by definition is:
dP/dt where P is the population therefore
dP/dt = 13,51*t
dt = 13,51*t*dt
Integrating on both sides of the equation we get:
∫dp = ∫ 13,51*t*dt
P = 13,51*t²/2 + K ( K is population for t = 0 )
Now the population in 10 years P(₁₀)
P(₁₀) = 13,51* (10)² /2 + K
P(₁₀) = 675,5 + K (1)
And P(₄) is
P(₄) = 13,51*(4)²/2 * K
P(₄) = 108,08 + K (2)
Then substracting
P(₁₀) - P(₄) = ( 675,5 + K ) - ( 108,08 + K )
ΔP = 567,42
But we don´t have fraction of animal, then
ΔP = 567
Which is a perfect square? 6 Superscript 1 6 squared 6 cubed 6 Superscript 5 What is the length of the hypotenuse, x, if (20, 21, x) is a Pythagorean triple
Answer:
Step-by-step explanation:
Hello, by definition a perfect square can be written as [tex]a^2[/tex] where a in a positive integer.
So, to answer the first question, [tex]6^2[/tex] is a perfect square.
(a,b,c) is a Pythagorean triple means the following
[tex]a^2+b^2=c^2[/tex]
Here, it means that
[tex]x^2=20^2+21^2=841=29^2 \ \ \ so\\\\x=29[/tex]
Thank you.
Answer:
Its B
Step-by-step explanation:
Which statement best describes a sequence? a.All sequences have a common difference. b.A sequence is always infinite. c.A sequence is an ordered list. d.A sequence is always arithmetic or geometric.
Answer:
C
Step-by-step explanation:
A sequence is defined as a list of numbers or objects in a special order.
They may be arithmetic or geometric or neither.
For example
0, 1, 4, 9, 16, 25, ..... ← is the sequence of square numbers.
Note it is neither arithmetic or geometric.
Please help with this
p-value problem. Suppose the director of manufacturing at a clothing factory needs to determine wheteher a new machine is producing a particulcar type of cloth according to the manufacturer s specification which indicate that the cloth should have mean breaking strength of 70 pounds and a standard deviation of 3.5 pounds. A sample of 49 pieces reveals a sample mean of 69.1 pounds. THe p value for this hypothesis testing scenario is
Answer:
The P-Value is 0.07186
Step-by-step explanation:
GIven that :
Mean = 70
standard deviation = 3.5
sample size n = 49
sample mean = 69.1
The null hypothesis and the alternative hypothesis can be computed as follows;
[tex]H_o : \mu = 70 \\ \\ H_1 : \mu \neq 70[/tex]
The standard z score formula can be expressed as follows;
[tex]\mathtt{z = \dfrac{\overline X - \mu}{\dfrac{\sigma}{\sqrt{n}}}}[/tex]
[tex]\mathtt{z = \dfrac{69.1 - 70}{\dfrac{3.5}{\sqrt{49}}}}[/tex]
[tex]\mathtt{z = \dfrac{-0.9}{\dfrac{3.5}{7}}}[/tex]
z = -1.8
Since the test is two tailed and using the Level of significance = 0.05
P- value = 2 × P( Z< - 1.8)
From normal tables
P- value = 2 × (0.03593)
The P-Value is 0.07186
if the nth term is , then the (n+1)st is: Sorry if formatting is off, check the image to see the equation better!
Answer:
5
----------
( n+1)(n+2)
Step-by-step explanation:
5
----------
n ( n+1)
Replace n with n+1
5
----------
(n+1) ( n+1+1)
5
----------
( n+1)(n+2)
We replace every 'n' with n+1 and simplify
[tex]\frac{5}{(n+1)(n+1+1)} = \frac{5}{(n+1)(n+2)}[/tex]
Please answer this correctly without making mistakes
Answer:
[tex]\large \boxed{\mathrm{4/5 \ cups}}[/tex]
Step-by-step explanation:
Subtract 1/10 from 9/10 to find out how much is left.
9/10 - 1/10
8/10 = 4/5
Answer:
4/5 cupsStep-by-step explanation:
[tex]Volume\:of \: syrup \:in \:cup\:from\:jug = \frac{9}{10}\\\\ She \:took\: \frac{1}{10} from \:the\:cup\:into\:the \:jug \\\\Volume \:of syrup\:in\:cup=?\\\\\frac{9}{10} -\frac{1}{10} \\\\= \frac{4}{5} cups[/tex]
Should I read Fruit's Basket? I need something to keep me busy.
Answer: Yes
Explanation: Yes because, sometimes you need to do stuff to get things of your head
How many numbers from 10 to 99 have a tens place exactly 3 times greater than their ones place? PLZZZ answer this question . I will be very happy whoever answers this I will give u brainliest too.
Answer:
45
Step-by-step explanation:
2 digit number starts from 10 ends at 99
between 10 and 19 there is only one number whose tens digit is more than ones digit.
that is 10
between 20 and 29 there are two numbers
20 and 21
like the same
between 30 and 39 there are 3 numbers
10–19. 1
20–29. 2
30–39. 3
40–49. 4
50–59. 5
60–69. 6
70–79. 7
80–89. 8
99–99. 9
sum of first n natural numbers is n(n+1)/2
9(9+1)/2=45
Three numbers between 10 and 99 have tens places that are precisely three times larger than their one's places.
What is Place value?The foundation of our whole number system is place value. In this approach, the value of a digit in a number is determined by where it appears in the number.
The tens digit must be three times larger than the units digit in order to meet the requirement. The units digit can have one of the following values: 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9.
Since we are seeking two-digit integers, it is not possible if the unit digit is zero for the tens digit also be zero.
In the case when the unit digit is 1, the tens digit must be 3, resulting in the number 31. Similar to the previous example, if the unit digit is 2, the tens digit must be 6, resulting in the number 62.
As we proceed, we discover the following integers meet the condition:
31, 62, 93
Therefore, there are three numbers from 10 to 99 that have a tens place exactly 3 times greater than their one's place.
Learn more about place values here:
https://brainly.com/question/27734142
#SPJ3
Please answer this correctly without making mistakes
Answer:
so to get a third you divide it by 3
first convert it to fraction
so it is 26/3
so do 26/3 divided by 3
so we do keep switch flip
26/3*1/3
so answer is 26/9 or 2 8/9
Step-by-step explanation:
Answer:
[tex]\large \boxed{\mathrm{2 \ 8/9 \ tablespoons \ of \ red \ chilies }}[/tex]
Step-by-step explanation:
8 2/3 tablespoons of red chilies is required for a recipe.
One-third of the original recipe would mean that the quantity of red chilies will be also one-third.
8 2/3 × 1/3
Convert to an improper fraction.
26/3 × 1/3
Multiply the fractions.
26/(3 × 3) = 26/9
Convert to a mixed fraction.
26/9 = 2 8/9
What is the following simplified product? Assume x greater-than-or-equal-to 0 (StartRoot 10 x Superscript 4 Baseline EndRoot minus x StarRoot 5 x squared EndRoot) (2 StartRoot 15 x Superscript 4 Baseline EndRoot + StartRoot 3 x cubed EndRoot) 10 x Superscript 4 Baseline StartRoot 6 EndRoot + x cubed StartRoot 30 x EndRoot minus 10 x Superscript 4 Baseline StartRoot 3 EndRoot + x squared StartRoot 15 x EndRoot 11 x Superscript 4 Baseline StartRoot 6 EndRoot + x cubed StartRoot 30 x EndRoot minus x Superscript 4 Baseline StartRoot 75 EndRoot + x squared StartRoot 15 EndRoot 10 x Superscript 4 Baseline StartRoot 6 EndRoot + x cubed StartRoot 30 x EndRoot minus 10 x Superscript 4 Baseline StartRoot 3 EndRoot minus x squared StartRoot 15 EndRoot 11 x Superscript 4 Baseline StartRoot 6 EndRoot + x cubed StartRoot 30 x EndRoot minus 10 x Superscript 4 Baseline StartRoot 3 EndRoot minus x cubed StartRoot 15 x EndRoot
Answer:
[tex]\bold{10x^4\sqrt{6}+x^3\sqrt{30x}-10x^4\sqrt{3}-x^3\sqrt{15x}}[/tex]
Step-by-step explanation:
To find:
Simplified product of:
[tex](\sqrt{10x^4}-x\sqrt{5x^2})(2\sqrt{15x^4}+\sqrt{3x^3})[/tex]
Solution:
First of all, let us have a look at some of the formula:
1. [tex](a+b) (c+d) = ac+bc+ad+bd[/tex]
2. [tex]a^b\times a^c =a^{b+c }[/tex]
3. [tex]\sqrt{a^{2b}} = \sqrt{a^b.a^b}=a^b[/tex]
4. [tex]\sqrt a \times \sqrt b = \sqrt{a\times b}[/tex]
Now, let us apply the above formula to solve the given expression.
[tex](\sqrt{10x^4}-x\sqrt{5x^2})(2\sqrt{15x^4}+\sqrt{3x^3})\\\\\Rightarrow(\sqrt{10x^4})(2\sqrt{15x^4})+(\sqrt{10x^4})(\sqrt{3x^3})-(x\sqrt{5x^2})(2\sqrt{15x^4})-(x\sqrt{5x^2})(\sqrt{3x^3})\\\\\Rightarrow2\sqrt{150x^8}+\sqrt{30x^7}-2x\sqrt{75x^6}-x\sqrt{15x^5}\\\\\Rightarrow\bold{10x^4\sqrt{6}+x^3\sqrt{30x}-10x^4\sqrt{3}-x^3\sqrt{15x}}[/tex]
The answer is:
[tex]\bold{10x^4\sqrt{6}+x^3\sqrt{30x}-10x^4\sqrt{3}-x^3\sqrt{15x}}[/tex]
Answer:
Its D
Step-by-step explanation:
Use the graph showing Debra's account balance to answer the question that follows. ^
About how long will it take for Debra's account balance to equal $60?
A - 6 months
B - 6 years
C - 3 months
D - 3 years
Answer: 2 years
Step-by-step explanation:
In the given graph, we have
Account Balance ($) on y-axis
Time (years) on x-axis.
To know the time taken to get a balance of $60 , we check the point corresponding to 60 at y-axis and then join it to the line of the function and stop.
Then from there we drop a line to x-axis.
We get x=2.
That is it will take 2 years to get $60 balance in Debra's account.
So the correct answer is 2 years.