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3 years ago

The product of three integers is -40. What could the three integers be?

Mathematics

1 answer:

Gwar [14]3 years ago

6 0

Answer:

You could choose -5, 2, 4

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Can the sides of a triangle have lengths 6.2, 0.5, and 7.8? yes or no

Vitek1552 [10]

No, it's not possible for the sides of a triangle to have those lengths.

According to the triangle inequality theorem, the sum of any two sides of the triangle has to be bigger than the last side. Let's test this.

$6.2+7.8\ \textgreater \ 0.5$

This inequality satisfies the triangle inequality theorem.

$7.8+0.5\ \textgreater \ 6.2$

This also satisfies the theorem.

$6.2+0.5 \textless \ 7.8$

Uh oh. This does not satisfy the triangle inequality theorem. Thus, it is not possible for a triangle to have these side lengths.

4 0

4 years ago

5a+c=-8a, for a how do you solve this problem?

BaLLatris [955]

The answers is A= C/-13

C= 13A

8 0

3 years ago

Can someone PLEASE help me with this question?

disa [49]

9514 1404 393

Answer:

Step-by-step explanation:

The future value of the account is given by the formula ...

A = P(1 +r/12)^(12t) . . . . principal P invested at rate r for t years

Solving for t, we find ...

A/P = (1 +r/12)^(12t) . . . . . . . . . . . divide by P

log(A/P) = 12t·log(1 +r/12) . . . . . . take logs

Divide by the coefficient of t, then fill in the numbers.

t = log(A/P)/(12·log(1 +r/12)) = log(202800/93000)/(12·log(1 +.068/12))

t ≈ 11.497

It will take about 11 years for the account balance to reach the desired amount.

4 0

3 years ago

Solve the following integral. <img src="https://tex.z-dn.net/?f=%5Cint4x%5Ccos%282-3x%29dx" id="TexFormula1" title="\int

bagirrra123 [75]

Hi there!

$\boxed{-\frac{4x}{3}sin(2-3x) + \frac{4}{9}cos(2-3x) + C}$

To find the indefinite integral, we must integrate by parts.

Let "u" be the expression most easily differentiated, and "dv" the remaining expression. Take the derivative of "u" and the integral of "dv":

u = 4x

du = 4

dv = cos(2 - 3x)

v = 1/3sin(2 - 3x)

Write into the format:

∫udv = uv - ∫vdu

Thus, utilize the solved for expressions above:

4x · (-1/3sin(2 - 3x)) -∫ 4(1/3sin(2 - 3x))dx

Simplify:

-4x/3 sin(2 - 3x) - ∫ 4/3sin(2 - 3x)dx

Integrate the integral:

∫4/3(sin(2 - 3x)dx

u = 2 - 3x

du = -3dx ⇒ -1/3du = dx

-1/3∫ 4/3(sin(2 - 3x)dx ⇒ -4/9cos(2 - 3x) + C

Combine:

$-\frac{4x}{3}sin(2-3x) + \frac{4}{9}cos(2-3x) + C$

7 0

3 years ago

Read 2 more answers

HELP!!!! HELP!!! HELP!!! Simple Intrest...!

Salsk061 [2.6K]

Answer:

67796.61

Step-by-step explanation:

4 0

2 years ago