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S_A_V [24]
3 years ago
11

One angle of an isosceles triangle measures 28°. Which other angles could be in that isosceles triangle? Choose all that apply.

Mathematics
1 answer:
Veseljchak [2.6K]3 years ago
8 0

Answer:

Answer: 76, 28, 124

Step-by-step explanation:

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cestrela7 [59]

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Step-by-step explanation: because it is an acute triangle

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I'm not really sure what to do.
lukranit [14]
I’m not sure if this is right but this looks a bit like chemistry so maybe try to get from minutes to seconds as it becomes an ‘S’ which I’m assuming is seconds? so basically try to do dimensional analysis with it and see if that works. Don’t forget to square it though, that’ll throw you off.
4 0
4 years ago
Solve for x.show your work please<br><br> - 1/2x &lt; -12
Ratling [72]

Answer:

x <24

Step-by-step explanation:

first cancel negative signs

then transfer 2 to right side

3 0
3 years ago
Read 2 more answers
Find the measure of ∠EGC. Circle A with chords EF and CD that intersect at point G, the measure of arc EC is 50 degrees, the mea
miv72 [106K]

Answer:

m∠EGC=70°

Step-by-step explanation:

we know that

The measure of the inner angle is the semi-sum of the arcs comprising it and its opposite

so

m∠EGC=(1/2)[arc EC+arc DF]

<u><em>Find the value of x</em></u>

we have

m∠EGC=(7x+7)°

arc EC=50°

arc DF=10x°

substitute and solve for x

(7x+7)°=(1/2)[50°+10x°]

14x+14=50+10x

14x-10x=50-14

4x=36

x=9

<u><em>Find the measure of angle EGC</em></u>

m∠EGC=(7x+7)°

substitute the value of x

m∠EGC=(7(9)+7)°=70°

8 0
4 years ago
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Solve for b. Put your answer in simplest radical form. Show Steps plz!<br><br><br> A = 4πbc + πb^2
sladkih [1.3K]

Answer:

b = -2c ± [√(4π²c² + πA)]/π

Step-by-step explanation:

A = 4πbc + πb^2

A = 4πbc + πb²

πb² + 4πbc - A = 0

Using the quadratic formula to solve this quadratic equation.

The quadratic formula for the quadratic equation, pb² + qb + r = 0, is given as

b = [-q ± √(q² - 4pr)] ÷ 2p

Comparing

πb² + 4πbc - A = 0 with pb² + qb + r = 0,

p = π

q = 4πc

r = -A

b = [-q ± √(q² - 4pr)] ÷ 2p

b = {-4πc ± √[(4πc)² - 4(π)(-A)]} ÷ 2π

b = {-4πc ± √[16π²c² + 4πA]} ÷ 2π

b = (-4πc/2π) ± {√[16π²c² + 4πA] ÷ 2π}

b = -2c ± [√(4π²c² + πA)]/π

Hope this Helps!!!

4 0
3 years ago
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