First integrate the indefinite integral,
[tex]\int(1-x^2)^{3/2}dx[/tex]
Let [tex]x=\sin(u)[/tex] which will make [tex]dx=\cos(u)du[/tex].
Then
[tex](1-x^2)^{3/2}=(1-\sin^2(u))^{3/2}=\cos^3(u)[/tex] which makes [tex]u=\arcsin(x)[/tex] and our integral is reshaped,
[tex]\int\cos^4(u)du[/tex]
Use reduction formula,
[tex]\int\cos^m(u)du=\frac{1}{m}\sin(u)\cos^{m-1}(u)+\frac{m-1}{m}\int\cos^{m-2}(u)du[/tex]
to get,
[tex]\int\cos^4(u)du=\frac{1}{4}\sin(u)\cos^3(u)+\frac{3}{4}\int\cos^2(u)du[/tex]
Notice that,
[tex]\cos^2(u)=\frac{1}{2}(\cos(2u)+1)[/tex]
Then integrate the obtained sum,
[tex]\frac{1}{4}\sin(u)\cos^3(u)+\frac{3}{8}\int\cos(2u)du+\frac{3}{8}\int1du[/tex]
Now introduce [tex]s=2u\implies ds=2du[/tex] and substitute and integrate to get,
[tex]\frac{3\sin(s)}{16}+\frac{1}{4}\sin(u)\cos^3(u)+\frac{3}{8}\int1du[/tex]
[tex]\frac{3\sin(s)}{16}+\frac{3u}{4}+\frac{1}{4}\sin(u)\cos^3(u)+C[/tex]
Substitute 2u back for s,
[tex]\frac{3u}{8}+\frac{1}{4}\sin(u)\cos^3(u)+\frac{3}{8}\sin(u)\cos(u)+C[/tex]
Substitute [tex]\sin^{-1}[/tex] for u and simplify with [tex]\cos(\arcsin(x))=\sqrt{1-x^2}[/tex] to get the result,
[tex]\boxed{\frac{1}{8}(x\sqrt{1-x^2}(5-2x^2)+3\arcsin(x))+C}[/tex]
Let [tex]F(x)=\frac{1}{8}(x\sqrt{1-x^2}(5-2x^2)+3\arcsin(x))+C[/tex]
Apply definite integral evaluation from b to a, [tex]F(x)\Big|_b^a[/tex],
[tex]F(x)\Big|_b^a=F(a)-F(b)=\boxed{\frac{1}{8}(a\sqrt{1-a^2}(5-2a^2)+3\arcsin(a))-\frac{1}{8}(b\sqrt{1-b^2}(5-2b^2)+3\arcsin(b))}[/tex]
Hope this helps :)
Answer:[tex]\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{3arcsin(a) + 2a(1 - a^2)^\Big{\frac{3}{2}} + 3a\sqrt{1 - a^2}}{8} - \frac{3arcsin(b) + 2b(1 - b^2)^\Big{\frac{3}{2}} + 3b\sqrt{1 - b^2}}{8}[/tex]General Formulas and Concepts:
Pre-Calculus
Trigonometric IdentitiesCalculus
Differentiation
DerivativesDerivative NotationIntegration
IntegralsDefinite/Indefinite IntegralsIntegration Constant CIntegration Rule [Reverse Power Rule]: [tex]\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C[/tex]
Integration Rule [Fundamental Theorem of Calculus 1]: [tex]\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)[/tex]
U-Substitution
Trigonometric SubstitutionReduction Formula: [tex]\displaystyle \int {cos^n(x)} \, dx = \frac{n - 1}{n}\int {cos^{n - 2}(x)} \, dx + \frac{cos^{n - 1}(x)sin(x)}{n}[/tex]
Step-by-step explanation:
Step 1: Define
Identify
[tex]\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx[/tex]
Step 2: Integrate Pt. 1
Identify variables for u-substitution (trigonometric substitution).
Set u: [tex]\displaystyle x = sin(u)[/tex][u] Differentiate [Trigonometric Differentiation]: [tex]\displaystyle dx = cos(u) \ du[/tex]Rewrite u: [tex]\displaystyle u = arcsin(x)[/tex]Step 3: Integrate Pt. 2
[Integral] Trigonometric Substitution: [tex]\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \int\limits^a_b {cos(u)[1 - sin^2(u)]^\Big{\frac{3}{2}} \, du[/tex][Integrand] Rewrite: [tex]\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \int\limits^a_b {cos(u)[cos^2(u)]^\Big{\frac{3}{2}} \, du[/tex][Integrand] Simplify: [tex]\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \int\limits^a_b {cos^4(u)} \, du[/tex][Integral] Reduction Formula: [tex]\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{4 - 1}{4}\int \limits^a_b {cos^{4 - 2}(x)} \, dx + \frac{cos^{4 - 1}(u)sin(u)}{4} \bigg| \limits^a_b[/tex][Integral] Simplify: [tex]\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{cos^3(u)sin(u)}{4} \bigg| \limits^a_b + \frac{3}{4}\int\limits^a_b {cos^2(u)} \, du[/tex][Integral] Reduction Formula: [tex]\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{cos^3(u)sin(u)}{4} \bigg|\limits^a_b + \frac{3}{4} \bigg[ \frac{2 - 1}{2}\int\limits^a_b {cos^{2 - 2}(u)} \, du + \frac{cos^{2 - 1}(u)sin(u)}{2} \bigg| \limits^a_b \bigg][/tex][Integral] Simplify: [tex]\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{cos^3(u)sin(u)}{4} \bigg| \limits^a_b + \frac{3}{4} \bigg[ \frac{1}{2}\int\limits^a_b {} \, du + \frac{cos(u)sin(u)}{2} \bigg| \limits^a_b \bigg][/tex][Integral] Reverse Power Rule: [tex]\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{cos^3(u)sin(u)}{4} \bigg| \limits^a_b + \frac{3}{4} \bigg[ \frac{1}{2}(u) \bigg| \limits^a_b + \frac{cos(u)sin(u)}{2} \bigg| \limits^a_b \bigg][/tex]Simplify: [tex]\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{cos^3(u)sin(u)}{4} \bigg| \limits^a_b + \frac{3cos(u)sin(u)}{8} \bigg| \limits^a_b + \frac{3}{8}(u) \bigg| \limits^a_b[/tex]Back-Substitute: [tex]\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{cos^3(arcsin(x))sin(arcsin(x))}{4} \bigg| \limits^a_b + \frac{3cos(arcsin(x))sin(arcsin(x))}{8} \bigg| \limits^a_b + \frac{3}{8}(arcsin(x)) \bigg| \limits^a_b[/tex]Simplify: [tex]\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{3arcsin(x)}{8} \bigg| \limits^a_b + \frac{x(1 - x^2)^\Big{\frac{3}{2}}}{4} \bigg| \limits^a_b + \frac{3x\sqrt{1 - x^2}}{8} \bigg| \limits^a_b[/tex]Rewrite: [tex]\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{3arcsin(x) + 2x(1 - x^2)^\Big{\frac{3}{2}} + 3x\sqrt{1 - x^2}}{8} \bigg| \limits^a_b[/tex]Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]: [tex]\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{3arcsin(a) + 2a(1 - a^2)^\Big{\frac{3}{2}} + 3a\sqrt{1 - a^2}}{8} - \frac{3arcsin(b) + 2b(1 - b^2)^\Big{\frac{3}{2}} + 3b\sqrt{1 - b^2}}{8}[/tex]Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Integration
Book: College Calculus 10e
help help me please!!!!!!!
9514 1404 393
Answer:
a) 3092.5 (rounded to tenths)
b) 39,600
c) ₹28,755
Step-by-step explanation:
These are all simple calculator problems. The arithmetic involved is something you learned in 2nd or 3rd grade.
__
a) Since we divide using the division algorithm, it isn't clear what "check your answer by division algorithm" is intended to mean. The result of the division (stopping at 1 decimal place) is 3092.5.
The usual method of checking a division problem is to multiply the quotient by the divisor to see if the dividend value is the result. Here, we have ...
13×3092.5 = 40202.5
This differs by from the dividend of 40203 by 0.5, which is the remainder showing in our long division. In short, the answer checks OK.
__
b) The value of each 4 is found by setting other digits to 0.
Most significant 4: 40,000
Least significant 4: 400
Difference in place value: 40,000 -400 = 39,600
__
c) The balance in the account is found by subtracting withdrawals from deposits:
₹35000 -6245 = ₹28,755
Ethan buys a video game on sale. If the video game usually costs $60, and it was on sale for 20% off, how much did Ethan pay? Round to the nearest whole dollar.
Ethan will pay $31.99 with the discount.
How? This is the answer because:
If 39.99 is 100%, and you are trying to find 20%...
1. you need to set it up as a ratio (of course, you do not need to do this, but it is easier for me to do it this way)
2. the ratio will look like this: 39.99/100% x/20%
3. all we need to do from here is to cross multiply!
4 39.99 x
---------- = ----------
100 20
-price is on the top and percent on the bottom
-you would now do 39.99 times 20
-then divide by 100
5. once you have 20% of 39.99, you need to subtract that answer from the total
6. 39.99 - 7.998 = 31.992 (you need to round to the nearest hundredth)
Hope this helps <3
Quit smoking: In a survey of 444 HIV-positive smokers, 202 reported that they had used a nicotine patch to try to quit smoking. Can you conclude that less than half of HIV-positive smokers have used a nicotine patch
Answer:
The p-value of the test is of 0.0287 < 0.05(standard significance level), which means that it can be concluded that less than half of HIV-positive smokers have used a nicotine patch.
Step-by-step explanation:
Test if less than half of HIV-positive smokers have used a nicotine patch:
At the null hypothesis, we test if the proportion is of at least half, that is:
[tex]H_0: p \geq 0.5[/tex]
At the alternative hypothesis, we test if the proportion is below 0.5, that is:
[tex]H_1: p < 0.5[/tex]
The test statistic is:
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
In which X is the sample mean, [tex]\mu[/tex] is the value tested at the null hypothesis, [tex]\sigma[/tex] is the standard deviation and n is the size of the sample.
0.5 is tested at the null hypothesis:
This means that [tex]\mu = 0.5, \sigma = \sqrt{0.5*(1-0.5)} = 0.5[/tex]
In a survey of 444 HIV-positive smokers, 202 reported that they had used a nicotine patch to try to quit smoking.
This means that [tex]n = 444, X = \frac{202}{444} = 0.455[/tex]
Value of the test statistic:
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex]z = \frac{0.455 - 0.5}{\frac{0.5}{\sqrt{444}}}[/tex]
[tex]z = -1.9[/tex]
P-value of the test and decision:
The p-value of the test is the probability of finding a sample proportion below 0.455, which is the p-value of z = -1.9.
Looking at the z-table, z = -1.9 has a p-value of 0.0287.
The p-value is of 0.0287 < 0.05(standard significance level), which means that it can be concluded that less than half of HIV-positive smokers have used a nicotine patch.
13) What is 4 1/2 subtracted from 5.33?
A. 0.43
B. 0.53
C. 0.83
D. 1.08
Given:
[tex]4\dfrac{1}{2}[/tex] subtracted from 5.33.
To find:
The value for the given statement.
Solution:
[tex]4\dfrac{1}{2}[/tex] subtracted from 5.33 can be written as:
[tex]5.33-4\dfrac{1}{2}[/tex]
On simplification, we get
[tex]=5.33-\dfrac{8+1}{2}[/tex]
[tex]=5.33-\dfrac{9}{2}[/tex]
[tex]=5.33-4.5[/tex]
[tex]=0.83[/tex]
Therefore, the correct option is C.
The beginning balance in a person's checking account was 45 dollars. After writing three checks for $6, $17
and $18 and making a deposit of $80, what was the new balance in the account?
Initial balance = $45
Withdrawal amount
= $6 + $17 + $18
= $41
So, new balance
= $45 - $41
= $4
Deposited amount = $80
So, final balance
= $4 + $80
= $84
So, the final balance is $84.
which of the following is not an asymptote of the hyperbola xy = -42? y = 0 x = 0 y = x
Given:
The equation of the hyperbola is:
[tex]xy=-42[/tex]
To find:
The the equation which is not an asymptote of the hyperbola.
Solution:
We have,
[tex]xy=-42[/tex]
It can be written as:
[tex]y=\dfrac{-42}{x}[/tex]
Equating denominator and 0, we get
[tex]x=0[/tex]
So, the vertical asymptotic is [tex]x=0[/tex].
The degree of numerator is 0 and the degree of denominator is 1.
Since the degree of numerator is greater that the degree of denominator, therefore the horizontal asymptote is [tex]y=0[/tex] and there is no oblique asymptote.
Therefore, [tex]y=x[/tex] is not an asymptote of the given hyperbola and the correct option is C.
29 and one-fifth divided by 4 and StartFraction 6 over 7 EndFraction
Answer:
6 1/85
Step-by-step explanation:
Convert any mixed numbers to fractions.
Then your initial equation becomes:
146/5÷34/7
Applying the fractions formula for division,
146/5*7/34=1022/170
Simplifying 1022/170, the answer is
6 1/85
Complete the sentence that explains why Write an Equation is a reasonable strategy for solving this problem. Because the answer may be _________ the numbers in the problem.
Answer:
4 e
Step-by-step explanation:
dz6dxrx xrrx6 xz33x4xr4x xrx
What is the value of x?
non-integer fractions
repeating decimal
integer
irrational number
Answer:
square root 3
Step-by-step explanation:
Pythagoreans theorem
a^2+b^2=c^2
C is the hypotenuse
1^2+(✓2)^2=3
✓3
A random sample of 64 students at a university showed an average age of 25 years and a sample standard deviation of 2 years. The 98% confidence interval for the true average age of all students in the university is
Answer:
24.4185<x<25.5815
Step-by-step explanation:
Given the following:
n = 64
mean x = 25
s = 2
z is the z score at 98% CI = 2.326
Get the Confidence Interval:
CI = x±z*s/√n
CI = 25±2.326*2/√64
CI = 25±2.326*2/8
CI = 25±0.5815
CI = (25-0.5815, 25+0.5815)
CI = (24.4185, 25.5815)
CI = 24.4185<x<25.5815
Hence the 98% confidence interval for the true average age of all students in the university is 24.4185<x<25.5815
this khan academy problem confuses me... (5/3)^3= can anyone help me solve it?
Answer:
4.629
Step-by-step explanation:
(5/3)³5×5×5/3×3×3125/274.629.Hope it is helpful to you
write your answer in simplest radical form
Step-by-step explanation:
5ft hight this ancle 90°so
answer is 5ft
After the booster club sold 40 hotdogs at a football game, it had $90 in profit.
After the next game, it had sold a total of 80 hotdogs and had a total of $210
profit. Which equation models the total profit, y, based on the number of
hotdogs sold, X?
Step-by-step explanation:
x = goods y = $
x Sold = 40, Y = $90
x Sold = 80, Y = $210
sum of xHotdogs = 40+80 = 120 Hotdogs
Sum of Y$ = $90 + 210 = 300
so
X = 2A & Y = 3 its mean one hotdogs can sold for one each = $2.25 and we round it to $3
So = XY = 2A + 3
sorry if i wrong
five times a number minus two is ten .find the number
Answer:
12/5 or 2.4
Step-by-step explanation:
Form the equation:
5x - 2 = 10
^ ^ ^
^ ^ "is ten" which adds a equal sign to it
^ minus two
five times a number
Solve:
5x - 2 = 10
+2 +2
-----------------
5x = 12
---- ----
5 5
x = 12/5 or 2.4
Let the number be x
Then ATQ
5x - 2 = 10
5x = 10+2
x = 12/5
Must click thanks and mark brainliest
The quadratic equation [tex]x^2+3x+50 = 0[/tex] has roots r and s. Find a quadratic question whose roots are r^2 and s^2.
According to the question, our quadratic equation is :
\begin{gathered} \bf {x}^{2} - ( {r}^{2} + {s}^{2} )x + {r}^{2} {s}^{2} = 0 \\ \bf \implies \: {x}^{2} - ( - 91)x + {(rs)}^{2} = 0 \\ \bf \implies \: {x}^{2} + 91x + {(50)}^{2} = 0 \\ \bf \implies \: {x}^{2} + 91x + 2500 = 0\end{gathered}
x
2
−(r
2
+s
2
)x+r
2
s
2
=0
⟹x
2
−(−91)x+(rs)
2
=0
⟹x
2
+91x+(50)
2
=0
⟹x
2
+91x+2500=0
Two real estate companies, Century 21 and RE/MAX, compete with one another in a local market. The manager of the Century 21 office would like to advertise that homes listed with RE/MAX average more than 10 days on the market when compared to homes listed with his company. The following data shows the sample size and average number of days on the market for the two companies along with the population standard deviations.Sample mean Sample size Population standard deviation Century 21 22 days 36 32 days RE/MAX 144 days 30 35 daysIf Population 1 is defined as RE/MAX and Population 2 is defined as Century 21, the 80% confidence interval for the difference in population means is:_________.(17.8, 26.2)(11.5, 32.7)(5.4, 38.6)(-3.0, 47.0)
Answer:
17.8 , 26.2
Step-by-step explanation:
The confidence interval is 80% for the given population. Significance level is 0.2 [ 1 - 0.8 ] . Sample size is given and sample mean is calculated with the given standard deviation. Sample mean is 36 days and population size is 144 days.
Solve the equation
P=100x-0.1x^2
Answer:
100x - 0.01x
Step-by-step explanation:
100x-0.1x^2
100x - 0.01x
Find the perimeter: a polygon with sides 2.1 cm, 2.1 cm, 1 cm, 3.3 cm, and 1 cm
The perimeter = _______ cm
A.8.3
B.10.7
C.7.7
D.9.5
Hello!
P = 2,1cm + 2,1cm + 1cm + 3,3cm + 1cm => P = 4,2cm + 1cm + 3,3cm + 1cm => P = 5,2cm + 3,3cm + 1cm => P = 8,5cm + 1cm => P = 9,5cm
Answer: D. 9,5cm
Good luck! :)
Answer:
9.5cm
Step-by-step explanation:
To find the perimeter, add up all the sides
2.1 cm+ 2.1 cm+ 1 cm+ 3.3 cm+ 1 cm
9.5cm
Find the area and perimeter of a rectangle with length measuring 14 cm and width measuring 5 more than twice the length.
Answer:
AREA: 462cm
PERIMETER: 94cm
Step-by-step explanation:
To find the width, you have to double 14 and then add 5. That would equal 33. Then to find area, multiply 33 and 14 = 462. To find perimeter, add 33+33+14+14=94
The area of the rectangle is 462 cm² and the perimeter is 94 cm.
How to determine the area and perimeterTo find the area and perimeter of a rectangle, we need the length and width of the rectangle.
Given:
Length = 14 cm
Width = 2(14) + 5
Calculating the width:
Width = 2(14) + 5
Width = 28 + 5
Width = 33 cm
Now, we can calculate the area and perimeter of the rectangle.
Area of a rectangle:
Area = Length x Width
Substituting the values:
Area = 14 cm x 33 cm
Area = 462 cm²
Perimeter of a rectangle:
Perimeter = 2(Length + Width)
Substituting the values:
Perimeter = 2(14 cm + 33 cm)
Perimeter = 2(47 cm)
Perimeter = 94 cm
Therefore, the area of the rectangle is 462 cm² and the perimeter is 94 cm.
Learn more about area and perimeter at
https://brainly.com/question/19819849
#SPJ2
Bill invested $4000 at 6%
compounded annually. Find the
accumulated amount at the end of
12 years.
Answer:
$ 8048.79Step-by-step explanation:
P = $4000t = 12 yearsr = 6% = 0.06Formula:
A = P(1 + r)^tThe total amount:
A = 4000*(1 + 0.06)^12 = 8048.79We have to find the,
Accumulated amount at end of 12 years.
The formula we use,
→ A = P(1+r)^t
It is given that,
→ P = $4000
→ t = 12 years
Then r will be,
→ 6%
→ 6/100
→ 0.06
Then the total amount is,
→ P(1+r)^t
→ 4000 × (1 + 0.06)^12
→ 8048.79
Thus, $ 8048.79 is the amount.
What is circulatry systerm
Answer: i think you meant circulatory system but the defination for it is:
The circulatory system (also called the cardiovascular system) is the body system that moves blood around the body. It consists of the heart and blood vessels. The blood carries various materials that the body needs, and takes away waste or harmful substances. Blood vessels that take blood away from the heart are arteries.
The stem-and-leaf plot above shows house sale prices over the last week in Tacoma. What was the most
expensive house sold? Give your answer in dollars
$
Answer:
the answer is 2
Step-by-step explanation:
use the function to find f(-2) f(x)=[tex]3^{x}[/tex]
Answer:
[tex] \frac{1}{9} [/tex]
Step-by-step explanation:
[tex]f( - 2) = {3}^{ - 2} [/tex]
[tex]1 \div 9 = .111[/tex]
Riley wants to make 100ml of 25% saline but only has access to 12% and 38% saline mixtures. x= 12% y=38%
Answer:
x = 50
y = 50
Step-by-step explanation:
[tex]\begin{bmatrix}x+y=100\\ 0.12x+0.38y=25\end{bmatrix}[/tex]
.12(100-y) + .38y = 25
x = 50
y = 50
A researcher surveyed 8 people to see if there is a
relationship between years of education and starting
salaries. The data points are shown on the graph.
Which best represents the equation of the trend line
shown on the graph? (Note that the graph has a break
on the x-axis.)
O y = 0.25x + 15
O y = 0.25x + 17.5
* y = 1.25x - 10
O y = 1.25x + 7.5
Answer:
[tex]y=1.25x+7.5[/tex]
Step-by-step explanation:
We can see that the trend line is the line of best fit to the data points.
The equation of a straight line is given by:
y = mx + b:
where y, x are variables, m is the slope of the line and b is the y intercept.
From the graph, we can see that the line passes through the points (10, 20) and (14, 25). Therefore the equation of the line is given by:
[tex]y-y_1=\frac{y_2-y_1}{x_2-x_1} (x-x_1)\\\\y-20=\frac{25-20}{14-10}(x-10)\\\\y-20=1.25(x -10)\\\\y-20=1.25x-12.5\\\\y=1.25x+7.5[/tex]
Find the slope of the line
A)-1/4
B)1/4
C)-4
D)4
Answer:
b) 1/4
The slope of the line is 1/4
Which point is the vertex for the graph of y = |x| + 2?
A. (0,1)
B. (0,-2)
C. (0,2)
D. (2,0)
Answer:(0,2) is the vertex for the graph of y = |x| + 2.
Step-by-step explanation: |x| = 0, y + 2 = 2
Therefore, It would be ( 0,2)
Answer:
trust the otherguy
Step-by-step explanation:
***URGENT*** PLEASE HELP ME ASAP, ITS DUE TODAY!!!
...................................
M is the midpoint of 0A. N is the midpoint of OB. Prove that AB is parallel to MN
Answer:
Construct MN.
Since M is the midpoint of OA, OM = MA
Similarly, N is the midpoint of OB.
Thus, ON = NB.
Now, in Δs OMN and OAB,
∠MON = ∠AOB (common angle)
(sides are in proportional ratio; OA = 2OM and OB = 2ON)
∴ Δs OMN and OAB are similar (2 sides are in proportion, with the included angle)
Since they are similar, then ∠OMN = ∠OAB (corresponding angles of similar triangles are equal)
But since ∠OMN = ∠OAB, then that means MN || AB (corresponding angles of two lines must be equal since they also sit relative to the transverse line, OA)
Thus, AB || MN (QED)
Hshejoffpeowhwbwbwhjskfofofoekwwoksnfnf Helppp
Answer:
Step-by-step explanation:
3. ZW ≅ WX
The admissions officer at a small college compares the scores on the Scholastic Aptitude Test (SAT) for the school's in-state and out-of-state applicants. A random sample of 10 in-state applicants results in a SAT scoring mean of 1173 with a standard deviation of 38. A random sample of 15 out-of-state applicants results in a SAT scoring mean of 1076 with a standard deviation of 57. Using this data, find the 95% confidence interval for the true mean difference between the scoring mean for in-state applicants and out-of-state applicants. Assume that the population variances are not equal and that the two populations are normally distributed. Find the margin of error to be used in constructing the confidence interva.
Answer:
jebtucky
Step-by-step explanation:
yes yee yee yee eyetegevw
