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3 years ago

Which value below is included in the solution set for the inequality statement?

Mathematics

1 answer:

mixas84 [53]3 years ago

7 0

You haven't provided any value, but I can tell you the solution set for the inequality.

First of all, expand both sides:

$-3x+12 > 6x-6$

Add 3x to both sides:

$12 > 9x-6$

Add 6 to both sides:

$18 > 9x$

Which is of course equivalent to

$9x < 18$

Divide both sides by 9

$x < 2$

So, every number smaller than 2 is part of the solution of this inequality.

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Step-by-step explanation:

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2 years ago

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sergiy2304 [10]

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3 years ago

Given b(x) = [X+41, what is b(-10)? – 10 -6 6 14

OLEGan [10]

Option C: 6 is the value of $b(-10)$

Explanation:

The given expression is $b(x)=|x+4|$

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To find the value of $b(-10)$ , let us substitute $x=-10$ in the given expression.

Thus, substituting $x=-10$ in the expression, we have,

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Since, we know the absolute rule that $|-a|=a$ and the simplified expression is of the form $|-a|$ , let us apply the absolute rule in the simplified expression.

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3 years ago

Find such m, that point (-2,4) is on line y=mx-3

Andreas93 [3]

Answer:

-3.5 ( or -7/2)

Step-by-step explanation:

we can substitute the point into this function to find out the answer

that is:

4 = m*(-2) -3

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2 years ago

What is the height of an equilateral triangle if the sides are 10

4vir4ik [10]

Answer:

The height of the equilateral triangle is $5\sqrt{3}\ units$

Step-by-step explanation:

we know that

An equilateral triangle has three congruent sides, and three congruent angles that each measure 60 degrees

To find out the height of an equilateral triangle, apply the Pythagoras Theorem in the right triangle ABD

Remember that the height of an equilateral triangle bisects the base.

see the attached figure to better understand the problem

$AB^2=AD^2+BD^2$

substitute the given values

$10^2=5^2+BD^2$

Solve for BD

$100=25+BD^2$

$BD^2=100-25$

$BD^2=75$

$BD=\sqrt{75}\ units$

simplify

$BD=5\sqrt{3}\ units$ `

therefore

The height of the equilateral triangle is $5\sqrt{3}\ units$

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3 years ago