3 years ago

Need help with this problem ^^

Mathematics

1 answer:

shutvik [7]3 years ago

8 0

Answer:

8t^2-2t+13 is the answer i believe

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Quickly IT is Due an a hour PLease show work show me how to do it and No links

djyliett [7]

Answer: 4

Step-by-step explanation:

-6x = -24 divide both sides by -6 to isolate x

x = -24/-6

x = 4

Checking my work (plug in 4 for x and solve)

-6x = -24

-6(4) = -24

-24 = -24 it's correct!

6 0

3 years ago

Which rule should be applied to reflect f(x)=x^3 over the line y=x?

Anettt [7]

Answer:

A. Substitute -x for x and simplify f(-x)

Step-by-step explanation:

i just did this

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3 years ago

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Eric paid $4.34 for peppers. Peppers cost $0.35 per pound. How many pounds of peppers did Eric buy?

n200080 [17]

Eric got 12 pounds of peppers

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4 years ago

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Mato is making a leather belt inlaid with beads. On the first day there were b beads on the belt. On

kompoz [17]

Answer:

Step-by-step explanation:

On the fourth day, there were b^3 beads and 216 beads on the belt, which means b^3 = 216. Taking the cube root of both sides, we have b = 6, which means there were 6 beads on the belt on the first day.

3 0

3 years ago

Use part 1 of the fundamental theorem of calculus to find the derivative of the function. y = cos x (3 + v5)8 dv sin x

dexar [7]

It looks like

$y(x)=\displaystyle\int_{\cos x}^{\sin x}(3+v^5)^8\,\mathrm dv$

(If the limits are in the wrong order, just multiply the result by -1)

Split the integral at an arbitrary value between $\cos x$ and $\sin x$ , and write $y(x)$ as

$y(x)=\displaystyle\left\{\int_{\cos x}^c+\int_c^{\sin x}\right\}(3+v^5)^8\,\mathrm dv$

$y(x)=\displaystyle\int_c^{\sin x}(3+v^5)^8\,\mathrm dv-\int_c^{\cos x}(3+v^5)^8\,\mathrm dv$

Then by the FTC,

$\dfrac{\mathrm dy}{\mathrm dx}=(3+\sin^5x)^8\cdot\dfrac{\mathrm d\sin x}{\mathrm dx}-(3+\cos^5x)^8\cdot\dfrac{\mathrm d\cos x}{\mathrm dx}$

$\dfrac{\mathrm dy}{\mathrm dx}=\cos x(3+\sin^5x)^8+\sin x(3+\cos^5x)^8$

7 0

4 years ago