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OleMash [197]
3 years ago
8

Find the length of the third side of each triangle

Mathematics
1 answer:
LuckyWell [14K]3 years ago
5 0

Answer:Where is the picture?

Step-by-step explanation:

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RT and GJ are chords that intersect at point H.
Lubov Fominskaja [6]

Answer:

20

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
(b) The area of a rectangular pool is 5488 m\2 If the length of the pool is 98 m, what is its width?​
mrs_skeptik [129]

Answer:

56

Step-by-step explanation:

To find the area of a rectangle we have the foruma A=WxL.

But we already have the area and length so we can plug that in

5488=Wx98

Now its an algebreic expression.

SInce its multiplying we do the opposite, so we divide 98 on both sides.

98/98 crosses itself out so now its 5488/98. Which equals 56. So now our expression is W=56. To fact check we put the numbers 56 and 98 into the formula to see if we get 5488.

A=56x98

A=5488

7 0
2 years ago
Does anyone know the answers to this questions
steposvetlana [31]

Answer:

6c + 36 =  6(c+6)

(6*c) + (6*6) =   6(c+6)

2(3c+3) =  6c+6

(6+c)+(6+6) =  neither

3c+6+3c =  6c+6

6c+12 =  neither

correct me if I am wrong cuz i haven't done this in 2 years

Step-by-step explanation:

going across the top row then the bottom

8 0
3 years ago
The circumference of a circle is 22 meters. What is the approximate
Aleks [24]
The Diameter of the circle =7m
7 0
3 years ago
I’m not sure how to write this equation. Please check the picture to see the graph for yourself! Help will be much appreciated!
Makovka662 [10]

let's notice something, the parabola is a vertical one, so the squared variable is the x, and is opening downwards, meaning the x² will have a negative coefficient.

the distance from the vertex to the directrix/focus is the amount of "p" units, let's see in the graph, the distance from the vertex to the directrix is 2, and since the parabola is opening downwards, "p" is a negative 2, p = -2.  The vertex is of course at (0, 2).


\bf \textit{parabola vertex form with focus point distance}
\\\\
4p(y- k)=(x- h)^2
\qquad
\begin{array}{llll}
vertex\ ( h, k)\\\\  p=\textit{distance from vertex to }\\
\qquad \textit{ focus or directrix}
\end{array}
\\\\[-0.35em]
\rule{34em}{0.25pt}\\\\
\begin{cases}
h=0\\
k=2\\
p=-2
\end{cases}\implies 4(-2)(y-2)=(x-0)^2\implies -8(y-2)=x^2
\\\\\\
y-2=\cfrac{x^2}{-8}\implies \blacktriangleright y=-\cfrac{1}{8}x^2+2 \blacktriangleleft

8 0
3 years ago
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