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Gemiola [76]
4 years ago
12

Solve for x. −8x+14≥60 OR −4x+50<58

Mathematics
1 answer:
Reika [66]4 years ago
6 0

Answer: The solution is,

x\leq -\frac{23}{4} or x>-2

Step-by-step explanation:

Given compound inequality,

-8x + 14 ≥ 60 or -4x + 50 < 58,

By the subtraction property of inequality,

-8x + 14 - 14 ≥ 60 - 14 or -4x + 50 - 50  < 58 -50,

-8x ≥ 46 or -4x < 8

By the division property of inequality,

-x \geq \frac{46}{8} or -x < \frac{8}{4}

Using property a > b ⇒ - a < -b,

x\leq -\frac{23}{4} or x>-2

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Answer:

272?

Step-by-step explanation:

10 x 8 = 80

12 x 16 = 192

192 + 80 = 272.

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Which is the side length of a cube with a volume of 3375 in3?
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The product of 4x and O is.
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Step-by-step explanation:

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3 years ago
3 packs of soda cost $10 less than 5 packs of soda. Write and equation that models this, and solve.
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8 0
3 years ago
Consider the function f given by f(x)=x*(e^(-x^2)) for all real numbers x.
NISA [10]

Answer:

\frac{\sqrt{\pi}}{4}

Step-by-step explanation:

You are going to integrate the following function:

g(x)=x*f(x)=x*xe^{-x^2}=x^2e^{-x^2}  (1)

furthermore, you know that:

\int_0^{\infty}e^{-x^2}=\frac{\sqrt{\pi}}{2}

lets call to this integral, the integral Io.

for a general form of I you have In:

I_n=\int_0^{\infty}x^ne^{-ax^2}dx

furthermore you use the fact that:

I_n=-\frac{\partial I_{n-2}}{\partial a}

by using this last expression in an iterative way you obtain the following:

\int_0^{\infty}x^{2s}e^{-ax^2}dx=\frac{(2s-1)!!}{2^{s+1}a^s}\sqrt{\frac{\pi}{a}} (2)

with n=2s a even number

for s=1 you have n=2, that is, the function g(x). By using the equation (2) (with a = 1) you finally obtain:

\int_0^{\infty}x^2e^{-x^2}dx=\frac{(2(1)-1)!}{2^{1+1}(1^1)}\sqrt{\pi}=\frac{\sqrt{\pi}}{4}

5 0
3 years ago
Read 2 more answers
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