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ozzi
3 years ago
14

Which of the following are exterior angles check all that apply?

Mathematics
1 answer:
yanalaym [24]3 years ago
7 0

The exterior angles on this diagram are angle 4 and angle 2.

We can see this because an exterior angle is the angle that forms a straight line with an interior angle, so that 180 - [interior angle] = [exterior angle].

Therefore, angles 4 and 2 are the only angles that form straight lines with interior angles, so they are thus the exterior angles.

I hope this helps!

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Help me please I have no idea on how to do this. Thanks
andriy [413]
1) P=2w +2l
46=2w +14
32=2w
16= w or width

2) A=lw
A=7 x 12
A= 84

3) A= 12 x 21
A= 252
6 0
3 years ago
6 pounds of potatoes cost $11.34. What is the unit price?
AveGali [126]
Your answer would be $1.89

You take 11.34 and divide it by 6. After that you’ll be 1.89 as your unit price.

1.89 is how much for 1 pound
8 0
2 years ago
Write the equation for Chelsea's graph in
xeze [42]

Answer:

y=-4x+16.

Step-by-step explanation:

From the given graph it is clear that the y-intercept is 16.

The line passes through (-2,24) and (2,8). So, slope of line is

Slope=\dfrac{y_2-y_1}{x_2-x_1}

Slope=\dfrac{8-24}{2-(-2)}

Slope=\dfrac{-16}{4}

Slope=-4

Slope intercept form of a line is

y=mx+b

where, m is slope and b is y-intercept.

Substitute m=-4 and b=16 in the above equation.

y=(-4)x+(16)

y=-4x+16

Therefore, the required equation is y=-4x+16.

6 0
3 years ago
9 — 3x = у<br> 3х – у = -2
Phoenix [80]

Answer:

This could be wrong, but i still wanted to help you.

Step-by-step explanation:

9-3x=y

+9      +9

3x=y

÷3   ÷3

x=y/3

Im sorry if this is wrong

6 0
3 years ago
Find the nth taylor polynomial for the function, centered at c. f(x) = ln(x), n = 4, c = 5
masha68 [24]

The nth taylor polynomial for the given function is

P₄(x) = ln5 + 1/5 (x-5) - 1/25*2! (x-5)² + 2/125*3! (x-5)³ - 6/625*4! (x - 5)⁴

Given:

f(x) = ln(x)

n = 4

c = 3

nth Taylor polynomial for the function, centered at c

The Taylor series for f(x) = ln x centered at 5 is:

P_{n}(x)=f(c)+\frac{f^{'} (c)}{1!}(x-c)+  \frac{f^{''} (c)}{2!}(x-c)^{2} +\frac{f^{'''} (c)}{3!}(x-c)^{3}+.....+\frac{f^{n} (c)}{n!}(x-c)^{n}

Since, c = 5 so,

P_{4}(x)=f(5)+\frac{f^{'} (5)}{1!}(x-5)+  \frac{f^{''} (5)}{2!}(x-5)^{2} +\frac{f^{'''} (5)}{3!}(x-5)^{3}+.....+\frac{f^{n} (5)}{n!}(x-5)^{n}

Now

f(5) = ln 5

f'(x) = 1/x ⇒ f'(5) = 1/5

f''(x) = -1/x² ⇒ f''(5) = -1/5² = -1/25

f'''(x) = 2/x³  ⇒ f'''(5) = 2/5³ = 2/125

f''''(x) = -6/x⁴ ⇒ f (5) = -6/5⁴ = -6/625

So Taylor polynomial for n = 4 is:

P₄(x) = ln5 + 1/5 (x-5) - 1/25*2! (x-5)² + 2/125*3! (x-5)³ - 6/625*4! (x - 5)⁴

Hence,

The nth taylor polynomial for the given function is

P₄(x) = ln5 + 1/5 (x-5) - 1/25*2! (x-5)² + 2/125*3! (x-5)³ - 6/625*4! (x - 5)⁴

Find out more information about nth taylor polynomial here

brainly.com/question/28196765

#SPJ4

3 0
1 year ago
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