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irakobra [83]
2 years ago
10

Ayudame por favor alguien me ayuda porfa

Mathematics
1 answer:
LUCKY_DIMON [66]2 years ago
8 0

Answer: there no question

Step-by-step explanation:

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PLEASE HELP ME as soon as possible!!
Alenkasestr [34]

Answer:

1/3 is answer.

I hope you got it. I had added solved answer too.

plz mark brainliest

8 0
3 years ago
Find the area of the parallelogram shown below.
ale4655 [162]

Answer:

<h2><em><u>120</u></em><em><u> </u></em><em><u>square</u></em><em><u> </u></em><em><u>units</u></em><em><u> </u></em></h2>

Step-by-step explanation:

<em><u>Given</u></em><em><u>, </u></em>

Base of the parallelogram = 10

Height of the parallelogram = 12

<em><u>Therefore</u></em><em><u>, </u></em>

Area of the parallelogram = <em>base</em><em> </em><em>×</em><em> </em><em>height</em><em> </em>

= 10 × 12

= 120

<em><u>Hence</u></em><em><u>,</u></em>

<em><u>Area</u></em><em><u> </u></em><em><u>of</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>parallelogram</u></em><em><u> </u></em><em><u>is</u></em><em><u> </u></em><em><u>120</u></em><em><u> </u></em><em><u>square</u></em><em><u> </u></em><em><u>units</u></em><em><u> </u></em><em><u>(</u></em><em><u>Ans</u></em><em><u>)</u></em>

5 0
3 years ago
Find the value of x in the triangle shown below.
oee [108]

Answer: x = 5

Step-by-step explanation:

Use Pythagorean's Theorem

a^2+b^2=c^2\\12^2+x^2=13^2\\144+x^2=169\\x^2=25\\x=5

8 0
2 years ago
I need help on #10
Ede4ka [16]
D = sq.root of((x2 - x1)^2 + (y2 - y1)^2)
sq.root of 13 = sq.root of((x - (-2))^2 + (1 - 3)^2)
= sq.root of((x + 2)^2 + (-2)^2)
= sq.root of (x^2 + 4x + 4 + 4)
= sq.root of (x^2 + 4x + 8)

Now if we square both sides:
x^2 + 4x + 8 = 13
x^2 + 4x - 5 = 0
(x + 5)(x - 1) = 0
x = -5 or x = 1
6 0
3 years ago
-3+9x&gt;87 or -x-9&gt;-17​
Alex_Xolod [135]

Answer:

\large\boxed{x\in(-\infty,\ 8)\ \cup\ (10,\ \infty)}

Step-by-step explanation:

-3+9x>87\qquad\text{add 3 to both sides}\\\\9x>90\qquad\text{divide both sides by 9}\\\\\boxed{x>10}\to(1)\\\\-x-9>-17\qquad\text{add 9 to both sides}\\\\-x>-8\qquad\text{change the signs}\\\\\boxed{x10\ or\ x

7 0
3 years ago
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