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Lady bird [3.3K]
4 years ago
5

Find the value of x in the triangle above Thank you to whoever helps

Mathematics
1 answer:
kow [346]4 years ago
3 0

this is an isosceles triangle two sides are the same

so there are 180° in a triangle

180-58= 122 now divide this by two because the last two angles are equal

122/2=61° =x°

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Yall someone help I still have a bunch if other test and I'm so tired ;-;
g100num [7]

Me das mas información?

6 0
3 years ago
Please Help Fast!! 15 Point and brainiest if correct
iragen [17]

If a < b, then dividing both sides by a positive number will not flip the inequality sign. The inequality flips if you divide both sides by a negative number.

So if a < b, then a/3 < b/3 as well.

<h3>Answer: Less than sign (first choice)</h3>
6 0
3 years ago
Show work, please someone help me, i'm confused still on this concept.
BigorU [14]
Answer: choice C) -15x^4y

-----------------------------------------

Explanation:

The coefficients are -3 and 5. They are the numbers to the left of the variable terms
Multiply the coefficients to get -3*5 = -15. So -15 is the coefficient in the answer

Multiply the x terms to get x^3 times x = x^(3+1) = x^4. Notice the exponents are being added

Do the same for the y terms as well: y^2 times y^(-1) = y^(2+(-1)) = y^(2-1) = y^1 = y

So we have a final coefficient of -15, the x terms simplify to x^4 and the y terms simplify to just y

Put this all together and we end up with -15x^4y which is what choice C is showing

3 0
3 years ago
This figure is made up of a triangle and a semicircle. What is the area of the figure? Use 3.14 for π . Enter your answer, as a
lana66690 [7]
Area of ∆ = 1/2 times its base × height
=1/2 × 4 × 5 UNIT²
=10 unit²
Area of semicircle = πr²/2
= (3.14)(4)²/2
=25.12 unit²

Therefore area of figure= 10+25.12 unit² = 35.12 unit²

Hope it helps!
6 0
3 years ago
Read 2 more answers
For what interval is the function f(x) = (20 + \sqrt{x}) / (\sqrt{20 + x}) continuous?
konstantin123 [22]
Try this solution:
1. according to the condition
\left \{ {{x \geq 0} \atop {20+x\ \textgreater \ 0}} \right. \ =\ \textgreater \  \ x \geq 0.
2. for more details see the attached graph.

Answer: [0;+oo)

5 0
3 years ago
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