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4 years ago

When you ask the gizmo to show the right triangle what does the hypotenuse side represent

Mathematics

2 answers:

Butoxors [25]4 years ago

7 0

Answer:

c. absolute value of z

Step-by-step explanation:

storchak [24]4 years ago

5 0

It is the longest side, directly across from the right angle. You can calculate it knowing a side and an angle (which gives you both non-right angles) or more typically by using the Pythagorean Theorem, which gives the formula.

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Factor x^4+xy^3+x^3+y^3 completely showing work.

babymother [125]

Answer:

Step-by-step explanation:

Hello, please consider the following.

$x^4+xy^3+x^3+y^3\\\\=x^3*x+x*y^3+x^3+y^3\\\\=x(x^3+y^3)+x^3+y^3\\\\\boxed{=(x+1)(x^3+y^3)}\\$

Thank you

4 0

3 years ago

Which system of equations has the same solution as the one above ?

Ber [7]

Answer:

x+4y=-35

-8x-4y=28

Step-by-step explanation:

The answer to the original equations is (1, -9). You can solve that by using substitution by moving the 4y over and substituting that equation for x on the second part of the top equation. Using elimination, the second set of answers will give you the same, (1, -9)!

7 0

3 years ago

Write the equation of the line that passes througlī the points (-2, 6)

elena-s [515]

Answer:

Horizontal line (0 slope)

Step-by-step explanation:

The equation is $\frac{y2-y1}{x2-x1}$ , so you'd just insert what you know.

It would look like this: $\frac{6-6}{4-(-2)}$ .

We would solve. 6-6=0, while 4-(-2)=6.

So, we have 0/6, which gives you 0, and it would be a horizontal line, since a horizontal line is portrayed by 0 slope.

6 0

3 years ago

Approximate value of arccos(0.36)

Alchen [17]

In degrees,

arccos(0.36) = 68.89980398 ≈ 68.9°

In radians,

arccos(0.36) = 1.202528433 ≈ 1.20

7 0

3 years ago

Find the equation of the line passing<br> through the points (3,3) and (4,5).<br> y = [? ]x +[ ?]

muminat

$\bf (\stackrel{x_1}{3}~,~\stackrel{y_1}{3})\qquad (\stackrel{x_2}{4}~,~\stackrel{y_2}{5}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{5}-\stackrel{y1}{3}}}{\underset{run} {\underset{x_2}{4}-\underset{x_1}{3}}}\implies \cfrac{2}{1}\implies 2 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{3}=\stackrel{m}{2}(x-\stackrel{x_1}{3}) \\\\\\ y-3=2x-6\implies y=2x-3$

3 0

3 years ago