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Elena L [17]
3 years ago
7

The measure of angle 1 is 130°. 2 lines intersect to form 4 angles. From top left, clockwise, the angles are 1, 4, 3, 2. Which o

ther angle must also measure 130°? angle I REALLY NEED HELP WITH THIS QUESTION PLEASSSS...

Mathematics
2 answers:
NemiM [27]3 years ago
5 0

Vertical angles are the same measurement and are opposite each other.

Angle 3 is opposite angle 1 and would also be 130 degrees.

adell [148]3 years ago
4 0

Answer:

∠3

Step-by-step explanation:

See attachment.

As we can see, ∠1 and ∠3 lie opposite each other, while ∠2 and ∠4 lie opposite each other. In other words, ∠1 and ∠3 are vertical angles; by definition, then, ∠1 = ∠3.

We already know that ∠1 = 130°, so that must mean ∠3 also equals 130°.

The answer is thus ∠3.

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60 + 7.20 + 7.80 please math is so dumb ​
Alja [10]
75
60+7.20= 67.20
67.20+7.80= 75
6 0
3 years ago
Joe has an ice cream stand. last week he spent $85 for ice cream and supplies. he earned $125. what was his profit?
Elodia [21]

Answer:

40 dollars

Step-by-step explanation:

Profit is what he earned minus his costs

Profit = 125 -85

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3 0
2 years ago
What is the absolute value of | r - 8| = 5?
fomenos

Answer:

r = 13, 3

Step-by-step explanation:

| r - 8| = 5

All solutions for r by breaking the absolute value into the positive and negative components

r = 13, 3

3 0
3 years ago
How many places do you need to move the decimal to the right to write
GenaCL600 [577]

Answer:

7 places.

Step-by-step explanation:

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6 0
3 years ago
3^x= 3*2^x solve this equation​
kompoz [17]

In the equation

3^x = 3\cdot 2^x

divide both sides by 2^x to get

\dfrac{3^x}{2^x} = 3 \cdot \dfrac{2^x}{2^x} \\\\ \implies \left(\dfrac32\right)^x = 3

Take the base-3/2 logarithm of both sides:

\log_{3/2}\left(\dfrac32\right)^x = \log_{3/2}(3) \\\\ \implies x \log_{3/2}\left(\dfrac 32\right) = \log_{3/2}(3) \\\\ \implies \boxed{x = \log_{3/2}(3)}

Alternatively, you can divide both sides by 3^x:

\dfrac{3^x}{3^x} = \dfrac{3\cdot 2^x}{3^x} \\\\ \implies 1 = 3 \cdot\left(\dfrac23\right)^x \\\\ \implies \left(\dfrac23\right)^x = \dfrac13

Then take the base-2/3 logarith of both sides to get

\log_{2/3}\left(2/3\right)^x = \log_{2/3}\left(\dfrac13\right) \\\\ \implies x \log_{2/3}\left(\dfrac23\right) = \log_{2/3}\left(\dfrac13\right) \\\\ \implies x = \log_{2/3}\left(\dfrac13\right) \\\\ \implies x = \log_{2/3}\left(3^{-1}\right) \\\\ \implies \boxed{x = -\log_{2/3}(3)}

(Both answers are equivalent)

8 0
2 years ago
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