All categories

3 years ago

2r - 12 = 12 - 10r whats the answer

Mathematics

2 answers:

Serga [27]3 years ago

7 0

Answer:

dude what is r?

Step-by-step explanation:

for example

3 + 3z - 10

if z is 2

3 + 3 × 2 - 10

Andrew [12]3 years ago

4 0

Answer:

r=2

Step-by-step explanation:

bring the numbers to the other sides

You might be interested in

What is the answer to 2-(-8)+(-3)=

Otrada [13]

The answer to 2-(-8)+(-3)= 7

7 0

3 years ago

Alejandra is continuing a dice game she started playing with her cousin yesterday. Her score at the end of today's game is 135.

IgorC [24]

Answer:

She has 95 points in the beginning of the game.

Step-by-step explanation:

I dont really know how to explain how i got the answer, but it should be correct :)

8 0

3 years ago

Read 2 more answers

Please help! Need help asap!

Sedbober [7]

Answer:

Step-by-step explanation:

The two angles in the triangle in the question are 58 degrees and 54 degrees.

Three angles in a triangle add up to 180 degrees

58 + 54 + x = 180

112 + x = 180

subtract 112 on both sides

x = 180 - 112 = 68 degrees

Only option A has a triangle that has the same angles, thus triangles are similar.

please give thanks by clicking heart button! :)

6 0

3 years ago

Given that CD¯¯¯¯¯¯¯¯ is a perpendicular bisector of AB¯¯¯¯¯¯¯¯, where D is on AB¯¯¯¯¯¯¯¯, how can you use the Pythagorean Theor

Liula [17]

Answer:

$(AD)^2 + (CD)^2 = (CA)^2$ and $(CD)^2 + (BD)^2 = (CB)^2$

Step-by-step explanation:

Given

Bisector: CD

of Line AB

Required

Apply Pythagoras Theorem

From the question, CD bisects AB and it bisects it at D.

The relationship between AB and CD is given by the attachment

Considering ACD

From the attachment, we have that:

$Hypothenuse = CA$

$Opposite = CD$

$Adjacent = AD$

By Pythagoras Theorem, we have

$(AD)^2 + (CD)^2 = (CA)^2$

Considering CBD

From the attachment, we have that:

$Hypothenuse = CB$

$Opposite = CD$

$Adjacent = BD$

By Pythagoras Theorem, we have:

$(CD)^2 + (BD)^2 = (CB)^2$

4 0

2 years ago

Read 2 more answers

7,,,,,,,,,,,,,,,,,,,,please help

Goryan [66]

Answer:

Step-by-step explanation:

The given expression is

$\frac{2^{\frac{11}{2} }\times 2^{-7}\times 2^{-5}}{2^3\times 2^{\frac{1}{2} }\times 2^{-10}}$

Recall and use

$a^m\times a^n=a^{m+n}$

$\frac{2^{\frac{11}{2} -7-5}}{ 2^{3+\frac{1}{2} -10}}$

$\frac{2^{-\frac{13}{2}}}{ 2^{-\frac{13}{2}}}$

Same bases are dividing, so we write one base and subtract the exponent.

$2^{-\frac{13}{2}+\frac{13}{2}}$

$2^0=1$

8 0

3 years ago