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3 years ago

Which situation requires the use of the cube root?

Mathematics

1 answer:

erik [133]3 years ago

4 0

Answer:

Step-by-step explanation:

Mathematically, the volume of a cube of side L is given as;

V = L^3

Now, given V and we want to find L, what we shall simply do is to find the cube root of both sides

By finding the cube roots of both sides, we can easily get the value of the side of the cube in question

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Help please asap I’m desperate!

Verizon [17]

Answer:

f(x)=x-2x+3; f(x)=-bx

x-2x=3=-bx

x+4x+3=0

(x+3) (x+1) = 0

x=-3 or x=-1

when x+-3. f(x)=b x(-3)=18

x=-1. f(x) = - b x (-1) = b

so system of equations : ( -3, -18)

; (-),b

Step-by-step explanation:

hope it helps

8 0

2 years ago

Can someone please help me?

igomit [66]

Answer:

I don't even see the whole problem?

Step-by-step explanation:

Try asking you Mom/Dad for help.

3 0

2 years ago

Type of percent when the final amount is greater than the ordinal amount

mylen [45]

Percent of increase. When it is more than the original amount, it is an increase. When it is less, it is a percent of decrease.

4 0

3 years ago

Evaluate the limit with either L'Hôpital's rule or previously learned methods.lim Sin(x)- Tan(x)/ x^3x → 0

Vsevolod [243]

Answer:

$\dfrac{-1}{6}$

Step-by-step explanation:

Given the limit of a function expressed as $\lim_{ x\to \ 0} \dfrac{sin(x)-tan(x)}{x^3}$ , to evaluate the following steps must be carried out.

Step 1: substitute x = 0 into the function

$= \dfrac{sin(0)-tan(0)}{0^3}\\= \frac{0}{0} (indeterminate)$

Step 2: Apply L'Hôpital's rule, by differentiating the numerator and denominator of the function

$= \lim_{ x\to \ 0} \dfrac{\frac{d}{dx}[ sin(x)-tan(x)]}{\frac{d}{dx} (x^3)}\\= \lim_{ x\to \ 0} \dfrac{cos(x)-sec^2(x)}{3x^2}\\$

Step 3: substitute x = 0 into the resulting function

$= \dfrac{cos(0)-sec^2(0)}{3(0)^2}\\= \frac{1-1}{0}\\= \frac{0}{0} (ind)$

Step 4: Apply L'Hôpital's rule, by differentiating the numerator and denominator of the resulting function in step 2

$= \lim_{ x\to \ 0} \dfrac{\frac{d}{dx}[ cos(x)-sec^2(x)]}{\frac{d}{dx} (3x^2)}\\= \lim_{ x\to \ 0} \dfrac{-sin(x)-2sec^2(x)tan(x)}{6x}\\$

$= \dfrac{-sin(0)-2sec^2(0)tan(0)}{6(0)}\\= \frac{0}{0} (ind)$

Step 6: Apply L'Hôpital's rule, by differentiating the numerator and denominator of the resulting function in step 4

$= \lim_{ x\to \ 0} \dfrac{\frac{d}{dx}[ -sin(x)-2sec^2(x)tan(x)]}{\frac{d}{dx} (6x)}\\= \lim_{ x\to \ 0} \dfrac{[ -cos(x)-2(sec^2(x)sec^2(x)+2sec^2(x)tan(x)tan(x)]}{6}\\\\= \lim_{ x\to \ 0} \dfrac{[ -cos(x)-2(sec^4(x)+2sec^2(x)tan^2(x)]}{6}\\$

Step 7: substitute x = 0 into the resulting function in step 6

$= \dfrac{[ -cos(0)-2(sec^4(0)+2sec^2(0)tan^2(0)]}{6}\\\\= \dfrac{-1-2(0)}{6} \\= \dfrac{-1}{6}$

<em>Hence the limit of the function </em> $\lim_{ x\to \ 0} \dfrac{sin(x)-tan(x)}{x^3} \ is \ \dfrac{-1}{6}$ .

3 0

3 years ago

A ball is thrown and it follows a parabolic path. At each horizontal distance (x) the following equation models the ball’s heigh

Nataly_w [17]

Answer D). 9 ft is correct

Height of the ball at a horizontal distance of 10 feet is 9 ft

Step-by-step explanation:

A ball is thrown and it follows a parabolic path. At each horizontal distance (x) the following equation models the ball’s height (y).

Y(x) = $-0.02x^{2} +0.8x+3$

Question asked is " What will be height (Y) of the ball when X=10 ft "

Simply putting value of X in given equation

we get,

Y(x) = $-0.02x^{2} +0.8x+3$

Y(x) = $-0.02(10)^{2} +0.8(10)+3$

Y(x) = $-0.02(100)} +8+3$

Y(x) = $-2 +8+3$

Y(x) = $9$

<em>Therefore,</em> Height of the ball at a horizontal distance of 10 feet is 9 ft

Answer D. 9 ft is correct

6 0

3 years ago