Help me! A B C D E F

What is the sum of m A and m B if angles A and B are supplementary?

jok3333 [9.3K]

Answer:

following

Step-by-step explanation:

6 0

2 years ago

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Jessica's Furniture Store manager is trying to figure out how much to charge for an armchair that just arrived. The chair was pu

VARVARA [1.3K]

Answer: The manager should sell the chair for $1740.

Step-by-step explanation:

45% of 1200 Solve this to find the amount increase

0.45 * 1200 = 540 Now add this amount to find the price.

$1200 + $540 = $1740

or you could do it this way

100% + 45% = 145%

145% of 1200 =?

1.45 * 1200 = 1740

3 0

2 years ago

Macy is making a border out of strips of wall boarder placed end to end along the 12-foot wall in her room. Each strip is foot l

slavikrds [6]

3 for each wall so the totally is 12

4 0

2 years ago

I urgently need the answer plzzzzzzz with explanation too plzzzzz

Eddi Din [679]

The pizza is fresh and the doe is fresh made

3 0

3 years ago

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A standard piece of paper is 0.05 mm thick. Let's imagine taking a piece of paper and folding the paper in half multiple times.

tatyana61 [14]

Answer:

(a) $g(n)=0.05\cdot 2^n$

(b) $g^{-1}(n)=\log_{2}20g(n)$

(c)43 times

Step-by-step explanation:

Part A

The paper's thickness = 0.05mm

When the paper is folded, its width doubles (increases by 100%).

The thickness of the paper grows exponentially and can be modeled by the function:

$g(n)=0.05(1+100\%)^n\\\\g(n)=0.05\cdot 2^n$

Part B

$g(n)=0.05\cdot 2^n\\2^n=\dfrac{g(n)}{0.05}\\ 2^n=20g(n)\\$Changing to logarithm form, we have:\\\log_{2}20g(n)=n\\$Therefore:\\g^{-1}(n)=\log_{2}20g(n)$

Part C

If the thickness of the paper, g(n)=384,472,300,000 mm

Then:

$g^{-1}(n)=\log_{2}20g(n)\\g^{-1}(n)=\log_{2}20\times 384,472,300,000\\=\dfrac{\log 20\times 384,472,300,000}{\log 2} \\g^{-1}(n)=42.8 \approx 43\\n=43$

You must fold the paper 43 times to make the folded paper have a thickness that is the same as the distance from the earth to the moon.

3 0

2 years ago