Answer:
perimeter=9cm
Area=2cm^2
step by step explanation:
Firstly solve the sides that don't have figures using trigonometry
#1..sin15°=opposite/hypotenuse
sin15=opposite/4
opposite=sin15×4
=1.035, round off to 1cm
Then find the value of the base
tan15=opposite/adjacent
tan15=1/adjacent
1=tan15adjascent
1/tan15=adjacent
adjacent or base=3.7 round off to 4cm
After finding these values find the perimeter
p=side+side+side
p=4cm+4cm+1cm
p=9cm
Find the area
1/2bh
1/2×4×1
A=2cm2
If Bobby drinks 5 waters in 10 hours how many does he drink in 1 hour ?
Water drunk by Bobby in 10 hours = 5 units
So, water drink by Bobby in 1 hour
= 5/10 units
= 1/2 units
= 0.5 units
Answer:
1/2 water
Step-by-step explanation:
We can use a ratio to solve
5 waters x waters
----------- = ------------
10 hours 1 hours
Using cross products
5*1 = 10 *x
5 = 10x
Divide by 10
5/10 = x
1/2 waters =x
Assume the random variable X is normally distributed with mean μ = 50 and standard deviation σ = 7. Compute the probability
P(35 < X < 58)= ________
Answer:
Probability-Between .8574 = 85.74%
Step-by-step explanation:
Z1=-2.14 Z2=1.14
*x-1 35
*x-2 58
*µ 50
*σ 7
1. Find the greatest number of 4-digits which is exactly divisible by each of 2,3,4,5,6 and 7.
Answer:
Step-by-step explanation:
The number which is divisible by 2, 3, 4, 5 and 6 will also be divisible by LCM of 2, 3, 4, 5 and 6. Let us divide 9999 by 60 and find the remainder, Remainder after dividing 9999 by 60 is 39. Therefore, the largest number which is divisible by these numbers should be 39 less than 9999.
forty-six times y is no more than 276.
Answer:
yip that's all
Step-by-step explanation:
not more than 276 means less or equal to 276,
46 × y ≤ 276
y ≤ 6
Answer:
46y<276
Step-by-step explanation:
no more than means less than or equal to.
Each course at college X is worth either 2 or 3 credits. The members of the men's swim team are taking a total of 48 courses that are worth a total of 107 credits. How many 2-credit courses and how many 3-credit courses are being taken?
Answer:
Let the number of courses that are worth 3 credits each be x and those worth 4 credits be y. With the given information, you can write the following equations:
x + y = 48
3x + 4y = 155
You can solve the above equations by method of elimination/substitution
x + y = 48 ⇒ x = 48 - y (Now, substitution this equation into 3x + 4y = 155)
3(48 - y) + 4y = 155
144 -3y + 4y = 155
y + 144 = 155
y = 11
Now plug this solution back into x = 48 - y
x = 48 - 11 = 37
Check work (by plugging the solutions back into the 3x + 4y and see if it's equal to 155):
3(37) + 4(11) = 155
Answer: There are 37 of the 3-credit course and 11 of the 4-credit course
find the measure of d
Find the arclength of the curve r(t) = ⟨ 10sqrt(2)t , e^(10t) , e^(−10t)⟩, 0≤t≤1
Answer:
[tex]\displaystyle AL = 2sinh(10)[/tex]
General Formulas and Concepts:
Pre-Calculus
Hyperbolic FunctionsCalculus
Differentiation
DerivativesDerivative NotationDerivative Property [Multiplied Constant]: [tex]\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)[/tex]
Derivative Rule [Chain Rule]: [tex]\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)[/tex]
Basic Power Rule:
f(x) = cxⁿf’(x) = c·nxⁿ⁻¹Exponential Differentiation
Integration
IntegralsIntegration Constant CDefinite IntegralsParametric Integration
Vector Value Functions
Vector IntegrationArc Length Formula [Vector]: [tex]\displaystyle AL = \int\limits^b_a {\sqrt{[i'(t)]^2 + [j'(t)]^2 + [k'(t)]^2}} \, dt[/tex]
Step-by-step explanation:
Step 1: Define
Identify
[tex]\displaystyle \vec{r}(t) = <10\sqrt{2}t , e^{10t} , e^{-10t} >[/tex]
Interval [0, 1]
Step 2: Find Arclength
Rewrite vector value function: [tex]\displaystyle r(t) = 10\sqrt{2}t \textbf i + e^{10t} \textbf j + e^{-10t} \textbf k[/tex]Substitute in variables [Arc Length Formula - Vector]: [tex]\displaystyle AL = \int\limits^1_0 {\sqrt{\bigg[\frac{d}{dt}[10\sqrt{2}t \textbf i]\bigg]^2 + \bigg[\frac{d}{dt}[e^{10t} \textbf j]\bigg]^2 + \bigg[\frac{d}{dt}[e^{-10t} \textbf k ]\bigg]^2}} \, dt[/tex][Integrand] Differentiate [Respective Differentiation Rules]: [tex]\displaystyle AL = \int\limits^1_0 {\sqrt{[10\sqrt{2} \textbf i]^2 + [10e^{10t} \textbf j]^2 + [-10e^{-10t} \textbf k]^2}} \, dt[/tex][Integrand] Simplify: [tex]\displaystyle AL = \int\limits^1_0 {\sqrt{200 \textbf i + 100e^{20x} \textbf j + 100e^{-20x} \textbf k}} \, dt[/tex][Integral] Evaluate: [tex]\displaystyle AL = 2sinh(10)[/tex]Topic: AP Calculus BC (Calculus I + II)
Unit: Vector Value Functions
Book: College Calculus 10e
The weekly amount of money spent on maintenance and repairs by a company was observed, over a long period of time, to be approximately normally distributed with mean $440 and standard deviation $20. How much should be budgeted for weekly repairs and maintenance so that the probability the budgeted amount will be exceeded in a given week is only 0.1
Answer:
$465.6 should be budgeted.
Step-by-step explanation:
Normal Probability Distribution
Problems of normal distributions can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
Normally distributed with mean $440 and standard deviation $20.
This means that [tex]\mu = 440, \sigma = 20[/tex]
How much should be budgeted for weekly repairs and maintenance so that the probability the budgeted amount will be exceeded in a given week is only 0.1?
The 100 - 10 = 90th percentile should be budgeted, which is X when Z has a p-value of 0.9, so X when Z = 1.28. Then
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.28 = \frac{X - 440}{20}[/tex]
[tex]X - 440 = 1.28*20[/tex]
[tex]X = 465.6[/tex]
$465.6 should be budgeted.
How much does college tuition cost? That depends, of course, on where you go to college. Construct a weighted average. Using the data from "College Affordable for Most," estimate midpoints for the cost intervals. Say 46% of tuitions cost about $4,500; 21% cost about $7,500; 7% cost about $12,000; 8% cost about $18,000; 9% cost about $24,000; and 9% cost about $31,000. Compute the weighted average of college tuition charged at all colleges.
Answer:
0.127
Step-by-step explanation:
Given f(x) = 4x - 3 and g(x) = 9x + 2, solve for (f + g)(x).
[tex]\\ \sf\longmapsto (f+g)(x)[/tex]
[tex]\\ \sf\longmapsto f(x)+g(x)[/tex]
[tex]\\ \sf\longmapsto 4x-3+9x+2[/tex]
[tex]\\ \sf\longmapsto 4x+9x-3+2[/tex]
[tex]\\ \sf\longmapsto 13x-1[/tex]
Answer:
13x - 1
Step-by-step explanation:
f(x) + g(x) = 4x - 3 + 9x + 2
f(x) + g(x) = 4x+9x + 2 - 3
f(x) + g(x) = 13x - 1
After leaving an airport, a plane flies for 2 hours on a course of 60 degrees at a speed of 200 kilometers per hour. The plane then flies for 3 hours on a course of 210 degrees at a speed of 100 kilometers per hour What is the distance of the airport from the plane in kilometers? Round to the nearest tenth
Answer: 205.3
I suppose all measures of angles are done from the same axis (for example x-axis)
Step-by-step explanation:
You just have to use the theorem of Al'Kashi:
[tex]d^2=400^2+300^2-2*300*400*cos(30^o)\\\\d\approx{205.3(km)}[/tex]
whats 5 + 5 i need help my iq is 1 i need help pls pls pls pls
Answer:
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Continued education.
A 90% confidence interval is found to be (120,140). What is the margin of error?
Answer:
There is 10% error in both minimum and extreme values i.e. 120 & 140 , Error in 120 is 10% i.e. = 12, Since value can be more or less in error ∴ Error in 120 is ±12.
4. Explain how the graphs of the functions are similar and how they are different
2x+3y=1470
And
2x+3y=1593
Answer:
they are parallel lines so have the same slope. Difference would be the where they intersect the x and y axis
Step-by-step explanation:
If it takes 5 years for an animal population to double, how many years will it take until the population
triples?
9514 1404 393
Answer:
7.92 years
Step-by-step explanation:
We want to find t such that ...
3 = 2^(t/5)
where 2^(t/5) is the annual multiplier when doubling time is 5 years.
Taking logs, we have ...
log(3) = (t/5)log(2)
t = 5·log(3)/log(2) ≈ 7.92 . . . years
It will take about 7.92 years for the population to triple.
a) __m=10km 25m =___km
b) __m=__km__m=1.5 km
Example :
a) 7250m= 7km 250m = 7.250km
Please help me
Answer:
a) 10,025 m = 10km 25m = 10.025 km
b) 1,500 m = 1 km 500 m = 1.5 km
Answer:
a) 10025m = 10km 25m = 10.025km
b) 1500m = 1km 500m = 1.5km
Step-by-step explanation:
Concept:
Here, we need to know the idea of unit conversion.
Unit conversion is the conversion between different units of measurement for the same quantity.
1 km = 1000 m
Solve:
a)
10km 25m = 10×1000 + 25 = 10025 m10km 25m = 10 + 25/1000 = 10.025 kmb)
1.5km = 1 + 0.5 × 1000 = 1km 500m1.5km = 1.5 × 1000 = 1500mHope this helps!! :)
Please let me know if you have any questions
Which of the following theorems verifies that AABC - ASTU?
A. AA
B. HL
C. HA
D. LL
Answer:
AA
Step-by-step explanation:
See In Triangle ABC and Triangle STU
[tex]\because\begin{cases}\sf \angle A=\angle S=90° \\ \sf \angle B=\angle T=31°\end{cases}[/tex]
Hence
[tex]\sf \Delta ABC~\Delta STU(Angle-Angle)[/tex]
By AA similarity triangle ABC is similar to triangle SUT. Therefore, option A is the correct answer.
What are similar triangles?Two triangles are similar if the angles are the same size or the corresponding sides are in the same ratio. Either of these conditions will prove two triangles are similar.
In the given triangle ABC, ∠C=180°-90°-31°
∠C=59°
In the given triangle SUT, ∠U=180°-90°-31°
∠U=59°
Here, ∠B=∠T (Given)
∠C=∠U (Obtained using angle sum property of a triangle)
So, by AA similarity ΔABC is similar to ΔSUT.
Therefore, option A is the correct answer.
To learn more about the similar triangles visit:
https://brainly.com/question/25882965.
#SPJ7
Suppose that 22 inches of wire costs 66 cents.
At the same rate, how much (in cents) will 17 inches of wire cost?
cents
Х
?
Answer:
51 cents for 17 inches of wire
Step-by-step explanation:
22 = 66
17 = x
22x = 66 * 17
22x = 1122
x = 51 cents
or
22 inches costs 66 cents
1 inch costs 3 cents (66 / 22 = 3 cents)
17 inches costs 51 cents (17 * 3 = 51 cents)
HELP ASAP!!
The equation (blank) has no solution
Answer:
Just to recap, an equation has no solution when it results in an incorrect "equation".
For example:
Equation: x+3 = x+4
Subtract x: 3 = 4???
But clearly, 3 is not equal to 4, so this equation has NO SOLUTION.
Now onto our problem:
13y+2-2y = 10y+3-y
11y+2 = 9y+3
2y=1
y=1/2
9(3y+7)-2 = 3(-9y+9)
27y+61 = -27y+27
54y = -34
y = -34/54
32.1y+3.1+2.4y-8.2=34.5y-5.1
34.5-5.1=34.5y-5.1
5.1=5.1
infinite solutions
5(2.2y+3.4) = 5(y-2)+6y
11y+17 = 11y-10
17 = -10??
That's not true, so the option "5(2.2y+3.4) = 5(y-2)+6y" has no solution.
Let me know if this helps
An angle, Theta. is in standard position. The terminal side of the angle passes through the point (6.-5).
Find sin Theta
9514 1404 393
Answer:
sin(θ) = (-5√61)/61
Step-by-step explanation:
The distance from the origin to the given point is ...
d = √(6² +(-5)²) = √61
The sine of the angle is the ratio ...
sin(θ) = y/d = -5/√61
Rationalizing the denominator gives us ...
sin(θ) = (-5√61)/61
A tortoise is walking in the desert. It walks 7.5 meters in 3 minutes. What is its speed?
Answer:
Step-by-step explanation:
speed is calculated using formula v=d/t
m= 7.5m
t= 3 min
v=?
v= 7.5m/3min
v= 2.5m/min
Find two power series solutions of the given differential equation about the ordinary point x = 0. Compare the series solutions with the solutions of the differential equation obtained using the method of Section 4.3. Try to explain any differences between the two forms of the solution. y'' − y' = 0 y1 = 1 − x2 2! + x4 4! − x6 6! + and y2 = x − x3 3! + x5 5! − x7 7! + y1 = x and y2 = 1 + x + x2 2! + x3 3! + y1 = 1 + x2 2! + x4 4! + x6 6! + and y2 = x + x3 3! + x5 5! + x7 7! + y1 = 1 + x and y2 = x2 2! + x3 3! + x4 4! + x5 5! + y1 = 1 and y2 = x + x2 2! + x3 3! + x4 4! +
You're looking for a solution in the form
[tex]y(x) = \displaystyle \sum_{n=0}^\infty a_nx^n[/tex]
Differentiating, we get
[tex]y'(x) = \displaystyle \sum_{n=0}^\infty na_nx^{n-1} = \sum_{n=1}^\infty na_nx^{n-1} = \sum_{n=0}^\infty (n+1)a_{n+1}x^n[/tex]
[tex]y''(x) = \displaystyle \sum_{n=0}^\infty (n+1)na_{n+1}x^{n-1} = \sum_{n=1}^\infty (n+1)na_{n+1}x^{n-1} = \sum_{n=0}^\infty (n+2)(n+1)a_{n+2}x^n[/tex]
Substitute these for y' and y'' in the differential equation:
[tex]\displaystyle \sum_{n=0}^\infty (n+2)(n+1)a_{n+2}x^n - \sum_{n=0}^\infty (n+1)a_{n+1}x^n = 0[/tex]
[tex]\displaystyle \sum_{n=0}^\infty \bigg((n+2)(n+1)a_{n+2}-(n+1)a_{n+1}\bigg)x^n = 0[/tex]
Then the coefficients of y are given by the recurrence
[tex]\begin{cases}a_0=y(0)\\a_1=y'(0)\\a_{n+2}=\frac{a_{n+1}}{n+2}&\text{for }n\ge0\end{cases}[/tex]
or
[tex]a_n = \dfrac{a_{n-1}}n[/tex]
But we cannot assume that [tex]a_0[/tex] and [tex]a_1[/tex] depend on each other; we can only guarantee that the recurrence holds for n ≥ 1, so that
[tex]a_2=\dfrac{a_1}2 \\\\ a_3=\dfrac{a_2}3=\dfrac{a_1}{3\times2} \\\\ a_4=\dfrac{a_3}4=\dfrac{a_1}{4\times3\times2} \\\\ \vdots \\\\ a_n=\dfrac{a_1}{n!}[/tex]
So in the power series solution, we split off the constant term and we're left with
[tex]y(x) = a_0 + a_1 \displaystyle \sum_{n=1}^\infty \frac{x^n}{n!}[/tex]
so that the fundamental solutions are
[tex]y_1=1[/tex]
and
[tex]y_2=x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+\cdots[/tex]
Can someone help please
Step-by-step explanation:
a) The volume of the prism is
[tex]V = (n^2 - 1)×(n^2 - 1)×(5n)[/tex]
[tex]\:\:\:\:= (n^4 - 2n^2 + 1)(5n)[/tex]
[tex]\:\:\:\:=5n^5 - 10n^3 + 5n[/tex]
b) If the dimensions L of the prism are tripled, the new volume will be
[tex]V' = (3L)^3 = 27L^3 = 27V[/tex]
so it will increase by a factor of 27.
100.331 divide 99.355
Answer:
1.009823361
Step-by-step explanation:
Just divide like this:
[tex] \frac{100.331}{99.355} = 1.009853361[/tex]
This is Some next level Geometry
Step-by-step explanation:
Measure of Angle 1 is 115 due to corresponding angles postulate
Measure of Angle 2 is 65 is linear pair postulate.
Fourteen boys and 21 girls will be equally divided into groups. Find the greatest number of groups that can be created if no one is left out.
Answer:
7 groups can be made each with five people :)
Name this triangle by its sides and angles. This is a(n) ____________________ triangle.
A.obtuse, isosceles
B.right, scalene
C.obtuse, scalene
D.right, isosceles
Answer:
right scalene
Step-by-step explanation:
Since all three sides have different lengths , this is a scalene triangle
(isosceles means two sides have the same lengths and equilateral means all three sides have the same length)
We have a right angle indicated by the box in the corner
Which of the following inequalities matches the graph?
10
6
-10
Oxs-1
Ox>-1
Oys-1
Oy 2-1
Answer:
y > -1
Step-by-step explanation:
the line is going across the y axis and is everything above -1
Two complementary angles have measures of s and t. if t is less than twice s, which system of linear equations can be used to determine the measure of each angle? Please explain answer. I know s+t=90. But how do you get to t=2s-90
The required expressions are both equations 1 and 2 as shown:
[tex]s + t = 90 ......... 1[/tex]
[tex]t<2s[/tex] .... 2
Complementary angles are angles that sum up to 90 degrees. For instance, and A and B are complementary if A + B = 90.
According to the question, if two complementary angles have measures of s and t then:
[tex]s + t = 90 ......... 1[/tex]
Twice of 's' is expressed as [tex]2s[/tex]
If t is less than twice s, this can be expressed as [tex]t<2s[/tex] .... 2
The required expressions are both equations 1 and 2 as shown:
[tex]s + t = 90 ......... 1[/tex]
[tex]t<2s[/tex] .... 2
Learn more on word problems leading to simultaneous equations here: https://brainly.com/question/14294864
Follow the instructions on the image
Answer:
k=3
Step-by-step explanation:
Assuming the centre of dilation is 0,0, we can use the formula (kx,ky) to determine it.
Here,
The co-ordinates of pre-image=(0,1),(-1,-1) & (1,-1)
The co-ordinates of image=(0,3),(-3,-3) & (3,-3)
Now,
(kx,ky)=(0,3)
(k*0,k*1)=(0,3)
Equating,
k=3
You can use the other coordinates to further solidify your answer.