Timothy brought in brownies for his class on his birthday.

Mathematics

1 answer:

IgorC [24]3 years ago

8 0

Answer:

Incomplete problem - reword

Step-by-step explanation:

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The sum ∫2−2() ∫52()−∫−1−2() can be written as a single integral in the form ∫() determine and

kow [346]

We have

$\displaystyle \int_{-2}^2 f(x) \, dx - \int_{-2}^{-1} f(x) \, dx = \int_{-1}^2 f(x) \, dx$

so that

$\displaystyle \int_{-2}^2 f(x) \, dx + \int_2^5 f(x) \, dx - \int_{-2}^{-1} f(x) \, dx \\\\ ~~~~~~~~~~~~ = \int_{-1}^2 f(x) \, dx + \int_2^5 f(x) \, dx \\\\ ~~~~~~~~~~~~ = \boxed{\int_{-1}^5 f(x) \, dx}$

6 0

1 year ago

What value of n makes the equation true? (2x^9y^n)(4x^2y^10)=8x^11y^20

Komok [63]

Hi there! The answer is n = 10.

$(2 {x}^{9} {y}^{n} )(4 {x}^{2} {y}^{10} ) = 8 {x}^{11} {y}^{20}$

As you see at the powers of x, we need to add the exponents of the power we when multiply them.
${x}^{9} \times {x}^{2} = {x}^{9 + 2} = x {}^{11}$

The powers of y work the same way.
${y}^{n} \times {y}^{10} = {y}^{n + 10} = {y}^{20}$

Hence, n = 10, since
${y}^{10 + 10} = {y}^{20}$

8 0

4 years ago

Horizontal asymptote

vodomira [7]

Answer:

H.A.=4

Step-by-step explanation:

If the bottom equation is factored, you get (x+1)(x-4)

If the V.A. is -1 ( gotten by putting [x+1] equal to zero I'm guessing) then the H.A. can be gotten by putting (x-4) equal to zero.

Also, I graphed it.

6 0

3 years ago