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3 years ago

How do you do Inequalities o-o help meh pleaseeeeeeeeeee

Mathematics

1 answer:

Valentin [98]3 years ago

4 0

Answer:

To solve an inequality use the following steps:

Step 1 Eliminate fractions by multiplying all terms by the least common denominator of all fractions.

Step 2 Simplify by combining like terms on each side of the inequality.

Step 3 Add or subtract quantities to obtain the unknown on one side and the numbers on the other.

Step-by-step explanation:

Up there!

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If a quantity increases by 25% and then decreases by 25%, will that result be the original quantity? Explain.

alexandr1967 [171]

If a quantity increases by 25% and then decreases by 25%, will that result be the original quantity? Explain.
Let’s try to be able to know the answer:
=> Let’s have 10 as the value
Increase by 25%
=> 10 + 25%
=> 10 * .25 = 2.5
=> 10 + 2.5 = 12.5 the new value
Now, decrease this by 25%
=> 12.5 – 25%
=> 12.5 * .25 = 3.125
=> 12.5 – 3.125 = 9.375 or 9.4
Thus, the given statement is false.

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3 years ago

If 10+9=21 than is 1+1= 2?

Andre45 [30]

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3 years ago

What appears to be the scale factor used to create this dilation?

stealth61 [152]

The answer is 0.5

Just take one point from the original shape and one from the transformed one. The just write the ordered pair out. Mine was (3,4) and (1.5, 2) Now in order for you to get from 3 to 1.5 you need to multiply 0.5 so the scale factor that was used is 0.5

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3 years ago

Read 2 more answers

Please solve -1/3e+1=-7

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Answer:

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Step-by-step explanation:

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3 years ago

A circle has the equation 2x²+12x+2y²−16y−150=0.

KonstantinChe [14]

Answer: B. The coordinates of the center are (-3,4), and the length of the radius is 10 units.

Step-by-step explanation:

The equation of a circle in the center-radius form is:

$(x-h)^{2} +(y-k)^{2}=r^{2}$ (1)

Where $(h,k)$ are the coordinates of the center and $r$ is the radius.

Now, we are given the equation of this circle as follows:

$2x^{2}+12x+2y^{2}-16y-150=0$ (2)

And we have to write it in the format of equation (1). So, let's begin by applying common factor 2 in the left side of the equation:

$2(x^{2}+6x+y^{2}-8y-75)=0$ (3)

Rearranging the equation:

$x^{2}+6x+y^{2}-8y=75$ (4)

$(x^{2}+6x)+(y^{2}-8y)=75$ (5)

Now we have to complete the square in both parenthesis, in order to have a perfect square trinomial in the form of $(a\pm b)^{2}=a^{2}\pm+2ab+b^{2}$ :