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Jlenok [28]
3 years ago
11

Find the missing side of this right

Mathematics
1 answer:
navik [9.2K]3 years ago
3 0

Answer:

x = 13

Step-by-step explanation:

First start with the blank equation

a^{2} + b^{2} = c^{2}

Plug in the legs and hypotenuse. Remember c squared is always the hypotenuse.

5^{2} + 12^{2} = c^{2}

Then solve

25 + 144 = c^{2}

169 =c^{2}

\sqrt{169} = \sqrt{c^{2} }

c = 13

You might be interested in
I really need help ASAP!
Semmy [17]

Answer:

Because they are parallel:

Angles 1,3,5 and 7=125

180-125=55

Angles 2,4, 6 and 8=55

Hope this helps!

3 0
3 years ago
Which expressions represent the difference of exactly two expressions? Choose 3 answers: Choose 3 answers: (Choice
Ket [755]

Solution

To find out which expressions represent the difference of exactly

two expressions

a) 6(x+7) - 2

This can be simplified as

6x + 42 - 2

6x + 40

Clearly this does not represent difference of the two

expressions .


b) -2j

Clearly this represents a single expression , this does not represent

a difference of 2 expression .


c) 4f - 2g

Here we have 2 different expressions 4f and 2g

Difference of these 2 expressions are 4f - 2g

Hence , this option represent difference of exactly two

expression


d) 3xyz - 10

3xtz - 10 represnts a single expression

Here we have a single expression

Therefore it does not represent difference of exactly 2

expression .

3 0
3 years ago
Rewrite –220° in radians. Write the solution as a fraction in simplest form.
son4ous [18]

Answer:

- \frac{11\pi }{9}

Step-by-step explanation:

To convert degrees to radians

radian measure = degree measure × \frac{\pi }{180} , then

radian measure = - 220 × \frac{\pi }{180} ( divide 220 and 180 by 20 )

                           = - \frac{11\pi }{9}

7 0
2 years ago
Please help, thank you!!
Alexxx [7]

Answer:

\sf \dfrac{Area\:of\: \triangle\: AEG}{Area \: of \: quadrilateral\:EGBH}=\dfrac{1}{7}

Step-by-step explanation:

If G is the midpoint of CD, and AC is parallel to DB, then AC = DH.

Therefore, G is the midpoint of AH and ΔACE is similar to ΔDBE.

As AC : DB = 1 : 3

⇒ Area of ΔACE : Area of ΔDBE = 1² : 3² = 1 : 9

We are told that Area ΔACE = Area ΔAEG.

⇒ Area ΔACG = 2 × Area ΔACE

As AC = DH, and G is the midpoint of CD:

⇒ ΔACG ≅ ΔHDG

⇒ Area ΔHDG = 2 × Area ΔACE

Area of quadrilateral EGHB = Area of ΔDBE - Area ΔHDG

                                              = Area of ΔDBE - 2 × Area of ΔACE

Therefore:

\sf \implies \dfrac{Area\:of\: \triangle\: AEG}{Area \: of \: quadrilateral\:EGBH}

\sf \implies \dfrac{Area\:of\: \triangle\: ACE}{Area \: of \: \triangle\:DBE - 2 \times Area\:of\: \triangle ACE}

Using the ratio of Area ΔACE : Area ΔDBE =  1 : 9

\implies \sf \dfrac{1}{9-2}

\implies \sf \dfrac{1}{7}

3 0
2 years ago
Read 2 more answers
Area of a parallelogram base 15 yards height 21 2/3 yards
Paul [167]
The area would be 324 yards square. Since the height is already found, you multiply 15 by 21 2/3 yards.
Hope this helped :) 
6 0
3 years ago
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