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3 years ago

∠1 and ∠2 are complementary angles.

Mathematics

2 answers:

Gwar [14]3 years ago

5 0

Angle two would be 180-m<1

180-52= 128

Measurement of angle two is 128 degrees

Nuetrik [128]3 years ago

5 0

Answer:

128

Step-by-step explanation:

Angles have to equal up to 180. So what I would do is subtract 52 from 180. 180-52=128.

Or Your answer would be 38.

Because there also has to be 3 angle measures. So 52+38+120=180. So the answer is between 128, or 38

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CAN U HELP ASAP ILL GIVE 15 POINTS AND PROLLY BRAINLIEST IT

gogolik [260]

Answer:

(2, 8) and (3, 11) --- > For example

Step-by-step explanation:

Plug in any number for x

I will use 2.

So if the equation is y = 3x + 2, then the new equation will be ...

y = 3(2) + 2

y = 6 + 2

y = 8 ----> One coordinate can be (2, 8)

Another number you could use is 3.

y = 3(3) + 2

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3 years ago

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Maru [420]

Answer:

He multiplied the radius by 2 instead of squaring it

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3 years ago

A three-bedroom house in North Augusta sold for $190,000. If housing prices are expected to increase 1.8% annually in that town,

Ivahew [28]

Answer:

we know that the initial price of the house is $190,000.

We also know that the expected increase is 1.8%, this means that the year zero we have no increase; the price is $190,000.

one year after we have an increase of the 1.8% (or 0.018 in decimal form), this means that the new price is:

$190,000 + $190,000*0.018 = $190,000*(1.018)

another year after, we have:

$190,000*(1.018)*(1.018) = $190,000*(1.018)^2

and etc

The function that models the price as a function of the years, represented as y, is:

P(y) = $190,000*(1.018)^y

If we want to know the price in 5 years, we need to replace y = 5 in the equation:

p(5) = $190,000*(1.018)^5 = $207,726.80

Step-by-step explanation:

hope this helps you

3 0

3 years ago

Solve 12^x^2+5x-4 = 12^2x+6

faust18 [17]

The solutions for ‘x’ are 2 and -5

<u>Step-by-step explanation:</u>

Given equation:

$12^{x^{2}+5 x-4}=12^{2 x+6}$

Since the base on both sides as ‘12’ are the same, we can write it as

$x^{2}+5 x-4=2 x+6$

$x^{2}+5 x-2 x-4-6=0$

$x^{2}+3 x-10=0$

Often, the value of x is easiest to solve by $a x^{2}+b x+c=0$ by factoring a square factor, setting each factor to zero, and then isolating each factor. Whereas sometimes the equation is too awkward or doesn't matter at all, or you just don't feel like factoring.