Ask question

Ask question

All categories

3 years ago

11

Which pair of angles are vertical angles? ∠KRC and ∠CRH ∠MRC and ∠KRM ∠ERT and ∠MRC ∠KRE and ∠ERT Lines K H, M T, and E C inters

ect at point R. Lines K H and M T are perpendicular. Line E C passes through angle K R T and angle M R H.

1 answer:

hram777 [196]3 years ago

8 0

The best and the most correct answer among the choices provided by the question is the first choice. The pair of angles that are vertical is <span>∠KRC and ∠CRH ∠MRC and ∠KRM ∠ERT. </span>I hope my answer has come to your help. God bless and have a nice day ahead!

You might be interested in

An egg carton holds 12 eggs. A breakfast buffet uses 96 eggs by

Blizzard [7]

Answer:

13 Cartoons

Step-by-step explanation:

As, in 1 cartoon = 12 eggs

x cartoon = 156

X = 156 / 12

X = 13

Therefore, 13 Cartoons were used after 10:30 am

4 0

3 years ago

Guys please help, does this look ok? :)<br>I'll give brainliest as well <3

Sidana [21]

Answer:

it looks good

Explanation:

3 0

3 years ago

Read 2 more answers

Solve the following and explain your steps. Leave your answer in base-exponent form. (3^-2*4^-5*5^0)^-3*(4^-4/3^3)*3^3 please st

Naily [24]

Answer:

$\boxed{2^{\frac{802}{27}} \cdot 3^9}$

Step-by-step explanation:

<u>I will try to give as many details as possible. </u>

First of all, I just would like to say:

$\text{Use } \LaTeX !$

Texting in Latex is much more clear and depending on the question, just writing down without it may be confusing or ambiguous. Be together with Latex! （＊＾Ｕ＾）人（≧Ｖ≦＊）/

$(3^{-2} \cdot 4^{-5} \cdot 5^0)^{-3} \cdot (4^{-\frac{4}{3^3} })\cdot 3^3$

Note that

$\boxed{a^{-b} = \dfrac{1}{a^b}, a\neq 0 }$

The denominator can't be 0 because it would be undefined.

So, we can solve the expression inside both parentheses.

$\left(\dfrac{1}{3^2} \cdot \dfrac{1}{4^5} \cdot 5^0 \right)^{-3} \cdot \left(\dfrac{1}{4^{\frac{4}{3^3} } }\right)\cdot 3^3$

Also,

$\boxed{a^{0} = 1, a\neq 0 }$

$\left(\dfrac{1}{9} \cdot \dfrac{1}{1024} \cdot 1 \right)^{-3} \cdot \left(\dfrac{1}{4^{\frac{4}{27} } }\right)\cdot 27$

Note

$\boxed{\dfrac{1}{a} \cdot \dfrac{1}{b}= \frac{1}{ab} , a, b \neq 0}$

$\left(\dfrac{1}{9216} \right)^{-3} \cdot \left(\dfrac{1}{4^{\frac{4}{27} } }\right)\cdot 27$

$\left(\dfrac{1}{9216} \right)^{-3} \cdot \left(\dfrac{27}{4^{\frac{4}{27} } }\right)$

$\left( \dfrac{1}{\left(\dfrac{1}{9216}\right)^3} \right)\cdot \left(\dfrac{27}{4^{\frac{4}{9} } }\right)$

$\left( \dfrac{1}{\left(\dfrac{1}{9216}\right)^3} \right)\cdot \left(\dfrac{27}{4^{\frac{4}{27} } }\right)$

Note

$\boxed{\dfrac{1}{\dfrac{1}{a} } = a}$

$9216^3\cdot \left(\dfrac{27}{4^{\frac{4}{9} } }\right)$

$\left(\dfrac{ 9216^3\cdot 27}{4^{\frac{4}{27} } }\right)$

Once

$9216=2^{10}\cdot 3^2 \implies 9216^3=2^{30}\cdot 3^6$

$\boxed{(a \cdot b)^n=a^n \cdot b^n}$

And

$4^{\frac{4}{27}} = 2^{\frac{8}{27}$

We have

$\left(\dfrac{ 2^{30} \cdot 3^6\cdot 27}{2^{\frac{8}{27} } }\right)$

Also, once

$\boxed{\dfrac{c^a}{c^b}=c^{a-b}}$

$2^{30-\frac{8}{27}} \cdot 3^6\cdot 27$

As

$30-\dfrac{8}{27} = \dfrac{30 \cdot 27}{27}-\dfrac{8}{27} =\dfrac{802}{27}$

$2^{30-\frac{8}{27}} \cdot 3^6\cdot 27 = 2^{\frac{802}{27}} \cdot 3^6 \cdot 3^3$

$2^{\frac{802}{27}} \cdot 3^9$

4 0

3 years ago

=−2x^2 +8 +3<br> write in vertex form

Oxana [17]

Answer:

$y=-(x+0)^2+11$

8 0

3 years ago

What system of equations represents the graph

Elodia [21]

Answer:

The answer is C because the two figures are not congruent.

4 0

4 years ago