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3 years ago

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An artist used silver wire to make a square that has a perimeter of 40 inches. She then used copper wire to make the largest cir

cle that could fit in the square, like the drawing above.
How many more inches of silver wire did the artist use compared to copper wire? (remember that = 3.14)

1 answer:

Elis [28]3 years ago

6 0

Answer:

1.4 inches

Step-by-step explanation:

if I'm wrong please tell me right away I would like to help you good day.

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Find the value of x.<br> X=____

musickatia [10]

(3x+1)+(2x+4)=90
3x+1+2x+4=90
5x+5=90
5x=85
x=17

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2 years ago

Given LaTeX: f\left(x\right)=x^{^3}-3x+4f ( x ) = x 3 − 3 x + 4, determine the intervals where the function is increasing and wh

Vedmedyk [2.9K]

Answer:

Increasing: $x$ and $x>1$ .

Decreasing: $-1$

Step-by-step explanation:

We have been given a function $f(x)=x^3-3x+4$ . We are asked to determine the intervals, where the function is increasing and where it is decreasing.

First of all, we will find critical points of our given function by equating derivative of our given function to 0.

Let us find derivative of our given function.

$f'(x)=\frac{d}{dx}(x^3)-\frac{d}{dx}(3x)+\frac{d}{dx}(4)$

$f'(x)=3x^{3-1}-3+0$

$f'(x)=3x^{2}-3$

Let us equate derivative with 0 as find critical points as:

$0=3x^{2}-3$

$3x^{2}=3$

Divide both sides by 3:

$x^{2}=1$

Now we will take square-root of both sides as:

$\sqrt{x^{2}}=\pm\sqrt{1}$

$x=\pm 1$

$x=-1,1$

We know that these critical points will divide number line into three intervals. One from negative infinity to -1, 2nd -1 to 1 and 3rd 1 to positive infinity.

Now we will check one number from each interval. If derivative of the point is greater than 0, then function is increasing, if derivative of the point is less than 0, then function is decreasing.

We will check -2 from our 1st interval.

$f'(-2)=3(-2)^{2}-3=3(4)-3=12-3=9$

Since 9 is greater than 0, therefore, function is increasing on interval $(-\infty, -1) \text{ or } x$ .

Now we will check 0 for 2nd interval.

$f'(0)=3(0)^{2}-3=0-3=-3$

Since -3 is less than 0, therefore, function is decreasing on interval $(-1,1) \text{ or } -1$ .

We will check 2 from our 3rd interval.

$f'(2)=3(2)^{2}-3=3(4)-3=12-3=9$

Since 9 is greater than 0, therefore, function is increasing on interval $(1,\infty) \text{ or } x>1$ .

6 0

3 years ago

It takes Francisco 15 minutes to ride his bike 4 km, which is the distance from home to school. What is his average speed?

Zarrin [17]

<span>It takes francisco 15 minutes to ride his bicycle 4 km, which is the distance from home to school.</span>

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3 years ago

Read 2 more answers

Solve the following quadratic by<br> completing the square.<br> y = x^2 - 6x+7

kozerog [31]

Answer:

$x= 3+\sqrt{2}, \quad x= 3 -\sqrt{2}$

Step-by-step explanation:

<u>Given quadratic equation</u>:

$y = x^2 - 6x+7$

To complete the square, begin by adding and subtracting the square of half the coefficient of the term in x:

$\implies y = x^2 - 6x+\left(\dfrac{-6}{2}\right)^2-\left(\dfrac{-6}{2}\right)^2+7$

$\implies y = x^2 - 6x+\left(-3\right)^2-\left(-3\right)^2+7$

$\implies y = x^2 - 6x+9-9+7$

Factor the perfect square trinomial:

$\implies y=(x-3)^2-9+7$

$\implies y=(x-3)^2-2$

To solve the quadratic, set it to zero and solve for x:

$\implies (x-3)^2-2=0$

$\implies (x-3)^2-2+2=0+2$

$\implies (x-3)^2=2$

$\implies \sqrt{(x-3)^2}=\sqrt{2}$

$\implies x-3= \pm\sqrt{2}$

$\implies x-3+3= 3\pm\sqrt{2}$

$\implies x= 3\pm\sqrt{2}$

Therefore, the solution to the given quadratic equation is:

$x= 3+\sqrt{2}, \quad x= 3 -\sqrt{2}$

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8 months ago

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Given f(x) = -x – 1, solve for x when f(x) = -10.<br> Plz help not sure how to do it

Paladinen [302]

Answer:

-X=9

Step-by-step explanation:

f(x)=-x-1

Plug in -10

-10=-x-1

add 1 to -10

-9=-x

which means -x=-9

to check, plug everything in

-10=-9-1

-10=-10

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3 years ago