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2 years ago

Help me ASAP please.

Mathematics

1 answer:

Leni [432]2 years ago

4 0

Answer:

Step-by-step explanation:

The vertex will occur when t=-b/(2a) for the quadratic ax^2+bx+c

In this case

t=-200/(2(-16))

t=-200/-32

t=6.25

g(6.25)=-16(6.25^2)+200(6.25)

g(6.25)=625ft

So the maximum height of the rocket is 625 feet and it reaches this height at 6.25 seconds.

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Guina made a scale drawing of a window she is making for her house.

LekaFEV [45]

Answer: To find it multiply 3 cm times 3 to get 9 in

Step-by-step explanation: If 1 cm is equal to 3 inches then to get the product multiply the number by 3 then convert to inches.

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2 years ago

Which expression is equivalent to 18x-54

Arisa [49]

Answer:

18(x - 3)

Step-by-step explanation:

18x - 54 ← factor out 18 from each term

= 18(x - 3)

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2 years ago

-6( - 3y - 6) - 10(6y - 8)

AnnyKZ [126]

Answer:

-2(21y-58)

Step-by-step explanation:

1.) Factor out -2 from the expression

2.)Distribute the three through the parenthesis

3.)distribute 5

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5.)subtract -18-40

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2 years ago

Please help! Will give brainliest

s344n2d4d5 [400]

Answer:

option A

Step-by-step explanation:

f(x)+g(x) will lead to addition of the fractions

LCM of 2x and x² is 2x²

{(2x²×½x)+(2x²×4/x²)}/2x²

x+8/2x²

5 0

2 years ago

Find the values of m and b that make the following function differentiable. the piecewise function f of x equals x cubed when x

aalyn [17]

$f(x)=\begin{cases}x^3&\text{for }x\le1\\mx+b&\text{for }x>1\end{cases}$

$f$ must be continuous in order to be differentiable, so we need to have

$\displaystyle\lim_{x\to1^-}f(x)=\lim_{x\to1^+}f(x)=f(1)$

By its definition, $f(1)=1^3=1$ , and

$\displaystyle\lim_{x\to1^-}f(x)=\lim_{x\to1}x^3=1$

$\displaystyle\lim_{x\to1^+}f(x)=\lim_{x\to1}(mx+b)=m+b$

so that $\boxed{m+b=1}$ .

We want the derivative to exist at $x=1$ , which requires that we pick an appropriate value for $f'(1)$ so that $f'(x)$ is also continuous. At the moment, we know

$f'(x)=\begin{cases}3x^2&\text{for }x1\end{cases}{/tex]so we need to pick [tex]f'(1)$ such that

$\displaystyle\lim_{x\to1^-}f'(x)=\lim_{x\to1^+}f'(x)$

We have

$\displaystyle\lim_{x\to1^-}f'(x)=\lim_{x\to1}3x^2=3$

$\displaystyle\lim_{x\to1^+}f'(x)=\lim_{x\to1}m=m$

so that $\boxed{m=3}$ (which means we need to pick $f'(1)=3$ ) and so $m+b=1\implies\boxed{b=-2}$

5 0

3 years ago