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3 years ago

What is the antiderivative of 4/e^4x?

1 answer:

7 0

$\displaystyle
F=\int \dfrac{4}{e^{4x}}\, dx\\
F=4 \int e^{-4x}\, dx\\
F=4 \cdot\left(-\dfrac{1}{4}e^{-4x}\right)+C\\
F=-e^{-4x}+C$

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A compact disc cover is 14 centimeters (cm) long, 12.4cm wide, and 1 cm deep. Which of the following represents the surface area

bogdanovich [222]

Answer:

SA= 2(l x w) |+2(l x h)+2(h x w)

= 2(14 x 12.4) +2(14 x 1)+2(1x 12.4)

=2(173.6)+2(14)+2(12.4)

=347.2|+28+24.8

Surface area is: 400 cm square

Step-by-step explanation:

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2 years ago

What is the algebraic expression for the quotient of j and 8

Dmitrij [34]

The algebraic expression for the quotient of j and 8 is j/8. Fractions are quotients.

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3 years ago

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iragen [17]

Answer:

here

Step-by-step explanation:

can you give me brainliest

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2 years ago

Please help! Thanks in advance.

alukav5142 [94]

In this problem, we need to plug in the given x values for $f(x)=0$ and find a and b.

When we plug in 1, we get: $2 \times {1}^{4} - 5 \times {1}^{3} - 14 \times {1}^{2} + a \times 1 + b = 0$

Simplify:
$2 - 5 - 14 + a + b = 0$
$- 17 + a + b = 0$
$a + b = 17$

We got our first statement about the values of the variables. If we find one more we can find those 2 variables.

We have another given root: 4.

Plug it in:
$2 \times {4}^{4} - 5 \times {4}^{3} - 14 \times {4}^{2} + a \times 4 + b = 0$
$512 - 320 - 224 + 4a + b = 0$
$- 32 + 4a + b = 0$
$4a + b = 32$

Now we have our second one. We can combine them:
$a + b = 17 \\ 4a + b = 32$

I use elimination method which is easier here.

Multiply the top equation by -1: $- a - b = - 17 \\ 4a + b = 32$

Add them up:
$3a = 15$

Simplify:
$a = 5$

Now we have a, we can plug in one of those equations to find b:
$5 + b = 17$
$b = 17 - 5$
$b = 12$

So, the answers are $a=5$ and $b=12$ .

6 0

3 years ago

solve for each please i really need help if u want to help me with my test and i get an a or a b i will give u 500 dollars

Len [333]

Answer:

The relation is not a function

The domain is {1, 2, 3}

The range is {3, 4, 5}

Step-by-step explanation:

A relation of a set of ordered pairs x and y is a function if

Every x has only one value of y

x appears once in ordered pairs

<u><em>Examples:</em></u>

The relation {(1, 2), (-2, 3), (4, 5)} is a function because every x has only one value of y (x = 1 has y = 2, x = -2 has y = 3, x = 4 has y = 5)

The relation {(1, 2), (-2, 3), (1, 5)} is not a function because one x has two values of y (x = 1 has values of y = 2 and 5)

The domain is the set of values of x

The range is the set of values of y

Let us solve the question

∵ The relation = {(1, 3), (2, 3), (3, 4), (2, 5)}

∵ x = 1 has y = 3

∵ x = 2 has y = 3

∵ x = 3 has y = 4

∵ x = 2 has y = 5

→ One x appears twice in the ordered pairs

∵ x = 2 has y = 3 and 5

∴ The relation is not a function because one x has two values of y

∵ The domain is the set of values of x

∴ The domain = {1, 2, 3}

∵ The range is the set of values of y

∴ The range = {3, 4, 5}

3 0

2 years ago