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2 years ago

What is (-1)(-2)(-3) 6 -6 5 -5

Mathematics

2 answers:

hoa [83]2 years ago

3 0

-6.

Have an amazing day :)

alisha [4.7K]2 years ago

3 0

Answer:

-6

Step-by-step explanation:

Use pemdas. You go left to right

(-1)(-2)(-3)

2(-3)

-6

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(70 points)

meriva

$f(x) = -0.26x^2 + 1.2x + 5.7\\\\\\f'(x)=(-0.26x^2 + 1.2x + 5.7)'=(-0.26x^2)'+(1.2x)'+(5.7)'=\\\\=-0.26(x^2)' + 1.2(x)' + (5.7)'=-0.26\cdot(2x) + 1.2\cdot1+0=\\\\=\boxed{-0.52x+1.2}$

Answer C)

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2 years ago

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A marble is selected from a bag contalning eight marbles numbered 1 to 8. The number on the marble selected will be recorded as

pishuonlain [190]

Answer:

immma send you the link friend me on here

Step-by-step explanation:

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2 years ago

Given the definitions of f(x) and g(x) below, find the value of g(f(-1)).

mrs_skeptik [129]

Answer:

-5

Step-by-step explanation:

At first,

f(-1) = 3 x (-1)² + 5 x (-1) -7

= 3 - 5 - 7

= -9

g(f(-1)) = g(-9)

= 2 x (-9) + 13

= -18 + 13

= -5

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2 years ago

An item is priced at $13.18. If the sales tax is 4%, what does the item cost including sales tax? If necessary, round to the nea

Rina8888 [55]

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2 years ago

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family of solutions of the second-order DE y y 0. Find a solution of the second-order IVP consisting of this differential equati

Anna35 [415]

Answer:

$y = \frac{1}{2}e^x -\frac{1}{2} e^{2-x}$

Step-by-step explanation:

Given

$y=c_1e^x +c_2e^{-x$

$y(1) = 0$

$y'(1) =e$

Required

The solution

Differentiate $y=c_1e^x +c_2e^{-x$

$y' = c_1e^x - c_2e^{-x}$

Next, we solve for c1 and c2

$y(1) = 0$ implies that; x = 1 and y = 0

So, we have:

$y=c_1e^x +c_2e^{-x$

$0 = c_1 * e^1 + c_2 * e^{-1}$

$0 = c_1 e + \frac{1}{e}c_2$ --- (1)

$y'(1) =e$ implies that: x = 1 and y' = e

So, we have:

$y' = c_1e^x - c_2e^{-x}$

$e = c_1 * e^1 - c_2 * e^{-1}$

$e = c_1 e - \frac{1}{e}c_2$ --- (2)

Add (1) and (2)

$0 + e = c_1e + c_1e + \frac{1}{e}c_2 - \frac{1}{e}c_2$

$e = 2c_1e$

Divide both sided by e

$1 = 2c_1$

Divide both sides by 2

$c_1 = \frac{1}{2}$

Substitute $c_1 = \frac{1}{2}$ in $0 = c_1 e + \frac{1}{e}c_2$

$0 = \frac{1}{2} e+ \frac{1}{e}c_2$

Rewrite as:

$\frac{1}{e}c_2 = -\frac{1}{2} e$

Multiply both sides by e

$c_2 = -\frac{1}{2} e^2$

So, we have:

$y=c_1e^x +c_2e^{-x$

$y = \frac{1}{2}e^x -\frac{1}{2} e^2 * e^{-x}$

$y = \frac{1}{2}e^x -\frac{1}{2} e^{2-x}$

6 0

2 years ago