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Mathematics

1 answer:

Phantasy [73]3 years ago

3 0

Answer:

Step-by-step explanation:

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Is the perpendicular bisector of a line segment also an angle bisector?

algol13

Answer:

yes

Step-by-step explanation:

In an isosceles triangle, one perpendicular bisector is also an angle bisector.

8 0

3 years ago

On every 3rd day Ivan goes to the gym to exercise. Every 5th day Gavin goes to the gym to exercise. What is the first day Ivan a

Colt1911 [192]

You have to find the LCM (least common multiple) of both numbers.

3: 3, 6, 9, 12, 15
5: 5, 10, 15

ivan and gavin will be together at the gym on the 15th day.

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3 years ago

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Pls help it is due very soon.

WINSTONCH [101]

Answer:

Step-by-step explanation:

The Pythagorean theorem is sqrt(a^2 + b^ 2) = c, so:

sqrt(22^2 + 8^2) = c

sqrt(548) = c

23.41 = c

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2 years ago

Sqrt of 4a-4 + sqrt of a-1

Monica [59]

$\large\begin{array}{l} \textsf{Simplify the expression (I'm goint to call it E)}\\\\ \mathsf{E=\sqrt{4a-4}+\sqrt{a-1}} \end{array}$

$\large\begin{array}{l} \textsf{So E is a sum of two square roots. Notice that you can factor}\\\textsf{the 4 out on the first one:}\\\\ \mathsf{E=\sqrt{4\cdot a-4\cdot 1}+\sqrt{a-1}}\\\\ \mathsf{E=\sqrt{4\cdot (a-1)}+\sqrt{a-1}}\\\\ \mathsf{E=\sqrt{2^2\cdot (a-1)}+\sqrt{a-1}}\\\\ \mathsf{E=\sqrt{2^2}\cdot \sqrt{a-1}+\sqrt{a-1}}\\\\ \mathsf{E=2\cdot \sqrt{a-1}+\sqrt{a-1}} \end{array}$

$\large\begin{array}{l} \mathsf{E=2\cdot \sqrt{a-1}+1\cdot \sqrt{a-1}}\\\\ \mathsf{E=(2+1)\cdot \sqrt{a-1}}\\\\ \boxed{\begin{array}{l}\mathsf{E=3\cdot \sqrt{a-1}} \end{array}} \end{array}$

If you're having problems understanding this answer, try seeing it through your browser: brainly.com/question/2179170

$\large\textsf{I hope this helps. :-)}$

Tags: <em>square root sqrt simplify expression radical irrational algebra</em>

5 0

3 years ago

What are the values of m and n?

snow_lady [41]

180 - 138 = 42°
90 + 42 = 132°
180 - 132 = 42° = m
-
90 + 42 + 2n = 180
132 + 2n = 180
- 132
2n = 48
÷ 2
n = 24°
-
Hope this helps! Let me know if you have any questions :)

6 0

3 years ago

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