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3 years ago

Draw two lines that are perpendicular. What type of angle is formed where the perpendicular lines intersect?

Mathematics

2 answers:

AnnyKZ [126]3 years ago

6 0

It forms a right angle. I agree with starlipika

Vladimir79 [104]3 years ago

4 0

a right angle (90 degrees)

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Will give 20 pts! pls answer

Lena [83]

It would be Felix:

5/4 = 1.25

Hope this helps!

8 0

2 years ago

Read 2 more answers

Which ordered pair is a solution of the equation? y-3=5(x-2) A.) (2,3) B.) (3,2) C.) Both D.) Neither

White raven [17]

Answer:

A is the only solution.

Step-by-step explanation:

A simple way to solve it is to plug in the x and y values

For A, we plug in 2 for x and 3 for y

3-3=5(2-2)

0=5(0)

0=0

Ordered pair A is a solution

For B, plug in 3 for x and 2 for y

2-3=5(3-2)

-1=5(1)

-1 does not equal 5

Therefore only A is a solution

5 0

3 years ago

What is the value of the expression x^2y–y+xy^2–x for x=4 and y=0.25?

stepan [7]

Answer:

0 (zero)

Step-by-step explanation:

${x}^{2} y - y + x {y}^{2} - x \\ = {4}^{2} \times 0.25 - 0.25 + 4 \times {0.25}^{2} - 4 \\ = 16 \times 0.25 - 0.25 + 4 \times 0.0625- 4 \\ = 4 - 0.25 + 0.25 - 4 \\ = 0 \\ \\ \purple {\boxed{ \bold{ \therefore \: {x}^{2} y - y + x {y}^{2} - x = 0}}}$

8 0

3 years ago

The highest point in the United states is mount McKinley also called denali

pogonyaev

Yes, it this somehow related to mathematics? Was this a question?

3 0

2 years ago

Can anyone figure this out?

Verizon [17]

$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ N(\stackrel{x_1}{-3}~,~\stackrel{y_1}{10})\qquad A(\stackrel{x_2}{6}~,~\stackrel{y_2}{3})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ NA=\sqrt{(6+3)^2+(3-10)^2}\implies NA=\sqrt{130} \\\\[-0.35em] ~\dotfill\\\\ A(\stackrel{x_2}{6}~,~\stackrel{y_2}{3})\qquad D(\stackrel{x_1}{6}~,~\stackrel{y_1}{-1}) \\\\\\ AD=\sqrt{(6-6)^2+(-1-3)^2}\implies AD=4 \\\\[-0.35em] ~\dotfill$

$\bf D(\stackrel{x_1}{6}~,~\stackrel{y_1}{-1})\qquad N(\stackrel{x_1}{-3}~,~\stackrel{y_1}{10}) \\\\\\ DN=\sqrt{(-3-6)^2+(10+1)^2}\implies DN=\sqrt{202}$

now that we know how long each one is, let's plug those in Heron's Area formula.

$\bf \qquad \textit{Heron's area formula} \\\\ A=\sqrt{s(s-a)(s-b)(s-c)}\qquad \begin{cases} s=\frac{a+b+c}{2}\\[-0.5em] \hrulefill\\ a=\sqrt{130}\\ b=4\\ c=\sqrt{202}\\[1em] s=\frac{\sqrt{130}+4+\sqrt{202}}{2}\\[1em] s\approx 14.81 \end{cases} \\\\\\ A=\sqrt{14.81(14.81-\sqrt{130})(14.81-4)(14.81-\sqrt{202})} \\\\\\ A=\sqrt{324}\implies A=18$

5 0

3 years ago