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2 years ago

See screenshot...need help its due today

Mathematics

1 answer:

Pani-rosa [81]2 years ago

4 0

Answer:

Step-by-step explanation:

f=20

Im a god bro

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V=1/2at^2 solve for t

malfutka [58]

$\dfrac{1}{2}at^2=V\qquad\text{multiply both sides by 2}\\\\at^2=2V\qquad\text{divide both sides by}\ a\neq0\\\\t^2=\dfrac{2V}{a}\to\boxed{t=\sqrt{\dfrac{2V}{a}}}$

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3 years ago

Help math question derivative!

atroni [7]

Let $f(x)=\sec^{-1}x$ . Then $\sec f(x)=x$ , and differentiating both sides with respect to $x$ gives

$(\sec f(x))'=\sec f(x)\tan f(x)\,f'(x)=1$
$f'(x)=\dfrac1{\sec f(x)\tan f(x)}$

Now, when $x=\sqrt2$ , you get

$(\sec^{-1})'(\sqrt2)=f'(\sqrt2)=\dfrac1{\sec\left(\sec^{-1}\sqrt2\right)\tan\left(\sec^{-1}\sqrt2\right)}$

You have $\sec^{-1}\sqrt2=\dfrac\pi4$ , so $\sec\left(\sec^{-1}\sqrt2\right)=\sqrt2$ and $\tan\left(\sec^{-1}\sqrt2\right)=1$ . So $(\sec^{-1})'(\sqrt2)=\dfrac1{\sqrt2\times1}=\dfrac1{\sqrt2}$

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3 years ago

A store sells boxes of juice in equal size packs. Garth bought 18 boxes, Rico bought 36 boxes, and Mai bought 45 boxes. what is

Stolb23 [73]

The anwser of your problem is 2536 do you need an explanation

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3 years ago

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katrin [286]

Answer:

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Step-by-step explanation:

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3 years ago

Write the cubic polynomial function f(x)in expanded form with zeros −6,−5,and −1, given that f(0)=60

RSB [31]

Answer:

$f(x)=2x^{3}+24x^{2}+82x+60$

Step-by-step explanation:

we know that

The roots of the polynomial are the values of x when the value of the polynomial f(x) is equal to zero

The roots of the polynomial function are

x=-6 -----> (x+6)=0

x=-5 -----> (x+5)=0

x=-1 -----> (x+1)=0

The equation of the cubic polynomial is

$f(x)=a(x+6)(x+5)(x+1)$

where

a is the leading coefficient

Remember that

f(0)=60

That means ------> For x=0 the value of f(x) is equal to 60

substitute the value of x and the value of y in the function and solve for a

$60=a(0+6)(0+5)(0+1)$

$60=a(6)(5)(1)$

$60=30a$

$a=2$

$f(x)=2(x+6)(x+5)(x+1)$

Applying the distributive property

Convert to expanded form

$f(x)=2(x+6)(x+5)(x+1)\\\\f(x)=2(x+6)(x^{2}+x+5x+5)\\\\f(x)=2(x+6)(x^{2}+6x+5)\\\\f(x)=2(x^{3}+6x^{2}+5x+6x^{2}+36x+30)\\\\f(x)=2x^{3}+24x^{2}+82x+60$

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3 years ago