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3 years ago

8

Find the base of an isosceles triangle whose area is 12 cm square and one of the equal sides is 5cm.

1 answer:

balandron [24]3 years ago

4 0

B≈6cm
or
8cm

Hope that helps!!! Good luck

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(-4) (9)<br> -----------<br> (-6)

valkas [14]

6 or -36/-6 if it works

8 0

3 years ago

Find three ratios equivalent to the given ratio 8/10 5/2

valkas [14]

For 8/10:
16/20
24/30
64/80

for 5/2:
10/4
20/8
100/40

hope this helps

7 0

3 years ago

Find the value of z please help

Elza [17]

z + 93 = 180

z = 87

Answer z = 87

Here are the answer for x and y, in case you need them.

x = 2* 93 - 112

x = 186 - 112

x = 74

y = 2 * 80 - x

y = 160 - 74

y = 86

6 0

4 years ago

Read 2 more answers

Will pick Brainliest ONLY if correct answer!

yawa3891 [41]

Answer:

x^4

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

I need help pls step by step i will really appreciate pls also l will mark you the brainly if you did !

agasfer [191]

Answer:

1. 2

2.3 $\sqrt{5}$

3.8 $\sqrt{6}$

4.2.5 $\sqrt{10}$

Step-by-step explanation:

Remember that $\frac{\sqrt{a}}{\sqrt{b}}$ = $\sqrt{\frac{a}{b} }$ , and $\frac{a}{\sqrt{b}}$ = $\frac{a}{\sqrt{b}}$ * $\frac{\sqrt{b}}{\sqrt{b}}$ = $\frac{a\sqrt{b}}{b}$

1. $\frac{\sqrt{48}}{\sqrt{12} }$ = $\sqrt{\frac{48}{12} }$ = $\sqrt{4}$ =2

2. $\frac{15}{\sqrt{5} }$ = $\frac{15}{\sqrt{5} }$ * $\frac{\sqrt{5}}{\sqrt{5} }$ = $\frac{15\sqrt{5} }{5}}$ =3 $\sqrt{5}$

3. $\frac{48}{\sqrt{6}}$ = $\frac{48}{\sqrt{6}}$ * $\frac{\sqrt{6}}{\sqrt{6}}$ = $\frac{48\sqrt{6}}{6}$ =8 $\sqrt{6}$

4. $\frac{25}{\sqrt{10}}$ = $\frac{25}{\sqrt{10}}$ * $\frac{\sqrt{10}}{\sqrt{10}}$ = $\frac{25\sqrt{10}}{10}$ =2.5 $\sqrt{10}$

4 0

3 years ago