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2 years ago

What is the slope of a line perpendicular to the line whose equation is x+y=2

Mathematics

1 answer:

vodomira [7]2 years ago

5 0

Answer:

x+y-2=0

Step-by-step explanation:

thats ur anwser you are welcome:)

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Which values are solutions to the inequality below? <br><br>Check all that apply.

matrenka [14]

You have this....

√x > 8

This means that the square root of something is greater than 8.

So plug in everything for x and find the answer.

Anything 64 and below is NOT THE ANSWER.

√64 = 8, and 8 is not greater than 8, so anything 64 and below is wrong.

So the answer is...

D and F.

4 0

2 years ago

When the sum of \, 528 \, and three times a positive number is subtracted from the square of the number, the result is \, 120. F

aleksandr82 [10.1K]

Let $x$ be the unknown number. So, three times that number means $3x$ , and the square of the number is $x^2$

We have to sum 528 and three times the number, so we have $528+3x$

Then, we have to subtract this number from $x^2$ , so we have

$x^2-(3x+528)$

The result is 120, so the equation is

$x^2 - 3x - 528 = 120 \iff x^2 - 3x - 648 = 0$

This is a quadratic equation, i.e. an equation like $ax^2+bx+c=0$ . These equation can be solved - assuming they have a solution - with the following formula

$x_{1,2} = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

If you plug the values from your equation, you have

$x_{1,2} = \dfrac{3\pm\sqrt{9-4\cdot(-648)}}{2} = \dfrac{3\pm\sqrt{9+2592}}{2} = \dfrac{3\pm\sqrt{2601}}{2} = \dfrac{3\pm51}{2}$

So, the two solutions would be

$x = \dfrac{3+51}{2} = \dfrac{54}{2} = 27$

$x = \dfrac{3-51}{2} = \dfrac{-48}{2} = -24$

But we know that x is positive, so we only accept the solution $x = 27$

6 0

3 years ago

What ratio will have a trail with 7:2

asambeis [7]

Answer:

7/2

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

15 POINTS!!!!

olganol [36]

p/5 + 6 = 10

=> p/5 = 10-6

=> p/5 = 4

=> p = 5×4

=> p = 20

C. 20

4 0

2 years ago

Read 2 more answers

What is the area of a sector of a circle with a radius of 2.0 ft and a central angle of 2<img src="https://tex.z-dn.net/?f=%20%5

Zigmanuir [339]

$\bf \qquad \textit{area of a sector of a circle}
\\\\
A=\cfrac{\theta r^2}{2}\qquad 
\begin{cases}
\theta=\textit{angle in radians}\\
r=radius
\end{cases}$

plug in the provided radius and angle

6 0

3 years ago