Name a point that is not coplanar with R, S, and T.

The area of the square adjacent to two sides of a right triangle are 29.25 square units and 13 units 30 POINTS

Otrada [13]

Answer:

$x=6.5\ units$

Step-by-step explanation:

step 1

Find the length side of the smaller square

The area of the square is equal to

$A=a^{2}$

so

$a^{2}=13$

$a=\sqrt{13}\ units$

step 2

Find the length side of the larger square

The area of the square is equal to

$A=b^{2}$

so

$b^{2}=29.25$

$b=\sqrt{29.25}\ units$

step 3

Find the value of x

Applying the Pythagoras Theorem

$x^{2} =a^{2}+b^{2}$

substitute the values

$x^{2} =13+29.25$

$x^{2} =42.25$

$x=6.5\ units$

6 0

2 years ago

A math class is having a discussion on how to determine if the expressions 4x-x+5 and 8-3x-3 are equivalent using substitution.

docker41 [41]

Okay. In my opinion, all the class has to do is simplify the expressions and compare. But they want to substitute instead.

Well then.

First, let's notice that these are linear expressions, meaning that if they are equivalent then all their values match up.

Number 1 is not a good one. Just because they're both positive doesn't mean anything; they have to be <em>the same.</em>

This also eliminates 3.

Number 2 is a good one, but it's not as reliable. If, for instance, the two expressions are <em>not </em>equivalent and you get lucky enough to pick that one value they intersect at (or have in common), then you'd be wrong when you say they are equivalent.

Number 4 makes the most sense because if both expressions are equivalent, then every value matches up. If not, then only one will. So having two values to substitute will most definitely answer the class question.

Hope this helps, let me know if I messed up! ;)

5 0

3 years ago

Read 2 more answers

Find the value of Y<br> .....

vekshin1

Answer:

$y = 6 \sqrt{3} \: units$

Step-by-step explanation:

In order to find the value of y, first we need to find the length of the perpendicular dropped from one of the vertices of the triangle to its opposite side.

By geometric mean theorem:

Length of the perpendicular

$=\sqrt{9\times 3}$

$=\sqrt{27}\: units$

Next, by Pythagoras theorem:

${y}^{2} = {9}^{2} + {( \sqrt{27} )}^{2} \\ \\ {y}^{2} = 81 + 27 \\ \\ {y}^{2} = 108 \\ \\ y = \sqrt{108} \\ \\ y = \sqrt{36 \times 3} \\ \\ y = \sqrt{ {6}^{2} \times 3} \\ \\ y = 6 \sqrt{3} \: units$

4 0

2 years ago

PLEASE HELP!!! Find the equation , in the standard form of the line passing through the points (3,-4) and (5,1)

ExtremeBDS [4]

$\bf \begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&&(~ 3 &,& -4~) 
% (c,d)
&&(~ 5 &,& 1~)
\end{array}
\\\\\\
% slope = m
slope = m\implies 
\cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{1-(-4)}{5-3}\implies \cfrac{1+4}{5-3}\implies \cfrac{5}{2}
\\\\\\
% point-slope intercept
\stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-(-4)=\cfrac{5}{2}(x-3)\implies y+4=\cfrac{5}{2}x-\cfrac{15}{2}$

$\bf y=\cfrac{5}{2}x-\cfrac{15}{2}-4\implies y=\cfrac{5}{2}x-\cfrac{23}{2}\impliedby 
\begin{array}{llll}
\textit{now let's multiply both}\\
\textit{sides by }\stackrel{LCD}{2}
\end{array}
\\\\\\
2(y)=2\left( \cfrac{5}{2}x-\cfrac{23}{2} \right)\implies 2y=5x-23\implies \stackrel{standard~form}{-5x+2y=-23}
\\\\\\
\textit{and if we multiply both sides by -1}\qquad 5x-2y=23$

side note: multiplying by the LCD of both sides is just to get rid of the denominators

5 0

2 years ago

What is 5 divide by 845

sergeinik [125]

<h3>Answer: 0.00591716</h3>

Step-by-step explanation: Use a ti-34 calculator.

5 0

2 years ago

Read 2 more answers