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3 years ago

Six less than a number is negative two? In a equation

Mathematics

1 answer:

Zina [86]3 years ago

3 0

Answer:

$x-6=-2$

Step-by-step explanation:

Let $x$ , be "a number".

6 less, would be $x-6$ . "Is" means equals ( = )

You then can put that together to get your equation.

<em>Hope this helps!!!</em>

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What is the sum of the given polynomials in standard form ? (X^2-3x)+(-2x^2+5x-3)

Greeley [361]

$(x^2-3x)+(-2x^2+5x-3)=x^2-3x-2x^2+5x-3\\\\=(x^2-2x^2)+(-3x+5x)-3=-x^2+2x-3$

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3 years ago

In parallelogram EFGH, the measure of angle G is (4x + 16)° and the measure of angle H is (2x − 16)°. What is the measure of ang

asambeis [7]

Answer:

m∠H = 44°

Step-by-step explanation:

Consecutive angles of a parallelogram are supplementary. So,

∠H and ∠G are supplementary

4x + 16 + 2x - 16 = 180

6x = 180

x = 30

m∠H = 2x - 16 = 2(30) - 16 = 60 - 16 = 44°

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2 years ago

1. In the Rundhalser, 200 bags of popcorn were put out. At the end of the day,

natali 33 [55]

Answer:

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Step-by-step explanation:

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2 years ago

Which equation is the inverse of 2(x - 2)2 = 8(7 + y)?

Marina86 [1]

Answer:

The inverse of $2(x - 2)^2 = 8(7 + y)$ is $y=\sqrt{(4x+28)}+2$

Step-by-step explanation:

Given equation : $2(x - 2)^2 = 8(7 + y)$

$\Rightarrow \frac{2(x-2)^2}{8}=7+y\\\Rightarrow \frac{2(x-2)^2}{8}-7=y$

Replace x with y and y with x

So, $\Rightarrow \frac{2(y-2)^2}{8}-7=x\\\Rightarrow \frac{2(y-2)^2}{8}=x+7\\\Rightarrow 2(y-2)^2=8(x+7)\\\Rightarrow (y-2)^2=4(x+7)\\\Rightarrow (y-2)=\sqrt{4(x+7)}\\\Rightarrow y=\sqrt{4(x+7)}+2\\\Rightarrow y=\sqrt{4x+28)}+2$

So, Option C is correct

Hence The inverse of $2(x - 2)^2 = 8(7 + y)$ is $y=\sqrt{(4x+28)}+2$

4 0

2 years ago

Find the TWO integers whos product is -12 and whose sum is 1<br>

Deffense [45]

Answer:

$\rm Numbers = 4 \ and \ -3.$

Step-by-step explanation:

Given :-

The sum of two numbers is 1 .

The product of the nos . is 12 .

And we need to find out the numbers. So let us take ,

First number be x

Second number be 1-x .

According to first condition :-

$\rm\implies 1st \ number * 2nd \ number= -12\\\\\rm\implies x(1-x)=-12\\\\\rm\implies x - x^2=-12\\\\\rm\implies x^2-x-12=0\\\\\rm\implies x^2-4x+3x-12=0\\\\\rm\implies x(x-4)+3(x-4)=0\\\\\rm\implies (x-4)(x+3)=0\\\\\rm\implies\boxed{\red{\rm x = 4 , -3 }}$

Hence the numbers are 4 and -3

4 0

2 years ago

Read 2 more answers