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2 years ago

13

Lol i need help asap

1 answer:

Lisa [10]2 years ago

5 0

Answer:

102 degrees

Step-by-step explanation:

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Find the vertex of the graph of the function. f(x) = 2x^2 - 8x + 9

Rashid [163]

$\bf \textit{vertex of a vertical parabola, using coefficients} \\\\ f(x)=\stackrel{\stackrel{a}{\downarrow }}{2}x^2\stackrel{\stackrel{b}{\downarrow }}{-8}x\stackrel{\stackrel{c}{\downarrow }}{+9} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right) \\\\\\ \left(-\cfrac{-8}{2(2)}~~,~~9-\cfrac{(-8)^2}{4(2)} \right)\implies \left( \cfrac{8}{4}~~,~~9-\cfrac{64}{8} \right) \\\\\\ (2~~,~~9-8)\implies (2~~,~~1)$

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3 years ago

Use the figures below to answer parts A and B

Aneli [31]

Answer: 7 is 3216.99cm²

7 part b is 24.31 rounded the answer is 24.

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2 years ago

A particle moves along a line with acceleration a (t) = -1/(t+2)2 ft/sec2. Find the distance traveled by the particle during the

amm1812

Answer:

$s(t)=(ln|7|+ln|2|)\,ft$

Step-by-step explanation:

Acceleration is second derivative of distance and are related as:

$a(t)=\frac{d^2s}{dt^2}\\\\\frac{d^2s}{dt^2}=\frac{-1}{(t+2)^2}\\$

Integrating both sides w.r.to t

$v(t)=\frac{ds}{dt}=\frac{1}{t+2} +C\\$

Using initial value

$v(0)=\frac{1}{2}\\\\\frac{1}{2}=\frac{1}{0+2} +C\\\\C=0\\\\\frac{ds}{dt}=\frac{1}{t+2}$

We have to calculate the distance covered in time interval [0,5], so:

$\int\limits^5_0 \frac{ds}{dt}=\int\limits^5_0 {\frac{1}{t+2}} \, dt\\\\s(t)=[ln|t+2|]^5_0\\\\s(t)=ln|5+2|+ln|0+2|\\\\s(t)=(ln|7|+ln|2|)\,ft$

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3 years ago

a triangle is graphed in the coordinate plane the vertices of the triangle have coordinates (-4, 1), (1, 1), and (1, -11). what

Alisiya [41]

The triangle coordinate is -11

4 0

3 years ago

Read 2 more answers

Matthew is 5 years older than twice the age of his sister. If Matthew is 13, how old is his sister? 3 4 9 31

Zepler [3.9K]

Answer:

4

Step-by-step explanation:

first take away the 5 years (13-5=8)

then divide it by 2 (8÷2=4

4 0

3 years ago