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3 years ago

PLEASE HELP AND EXPLAIN!! 30 POINTS AND WILL MARK BRAINLIEST!!!

Mathematics

1 answer:

Sunny_sXe [5.5K]3 years ago

8 0

I am prertty sure it is the third opption. if not then I think you are right about the one you already chose. Super sorry if wrong

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Tracey paid $165 for an item that was originally priced at $470. What percent of the original price did Tracey pay? Round your p

solong [7]

(165/470) x 100%= 35.11%

4 0

3 years ago

I need help with these two trigonometry problems.

Sloan [31]

Answer:

$B \approx 32.1\°; A \approx 39.0\°$

Step-by-step explanation:

Sine theorem: in any triangle, the ratio between a side and the sine of the opposite angle is constant

$\frac{sin\ \alpha}a = \frac{sin\ \beta}b =\frac{sin\ \gamma}c$

In our case, for the left triangle,

$\frac {sin\ 103\°}{11} =\frac{sin\ B} 6 \rightarrow sin\ B = \frac6{11} sin\ 103\°$

Time to grab a calculator and crunch numbers. Double check your calculator is in degrees and not in radians (plug in sin 30°, if you're getting 0.5 you're good) and you will get

$sin\ B \approx 0.53 \rightarrow B \approx 32.1\°$

Same difference with the right triangle. With the same calculations

$sin\ A = \frac{26}{41} sin 83\° \approx 0.68 \rightarrow A \approx 39.0\°$

5 0

2 years ago

What are the vertices of A'B'C if ABC is dilated by a scale factor of 2?

malfutka [58]

The vertices of A'B'C' if A(0, 10), B(8, 6), C(4, 2) is dilated by a scale factor of 2 are;

C. A'(0, 20), B'(16, 12), C'(8, 4)

<h3>Which method can be used to find the vertices of A'B'C'?</h3>

Given that triangle ABC is dilated by a scale factor of 2, we have;

A'B' = 2 × AB

A'C' = 2 × AC

B'C' = 2 × BC

Length of AB = √((8-0)^2 + (6-10)^2) = 4•√5

Length of AC = √((4-0)^2 + (2-10)^2) = 4•√5

Length of BC = √((8-4)^2 + (6-2)^2) = 4•√2

By multiplying the given coordinates by the scale factor, we have;

A' = 2 × (0, 10) = (0, 20)

B' = 2 × (8, 6) = (16, 12)

C' = 2 × (4, 2) = (8, 4)

A'B' = √((16-0)^2 + (12-20)^2) = 8•√5

A'C' = √((8-0)^2 + (4-20)^2) = 8•√5

B'C' = √((16-8)^2 + (12-4)^2) = 8•√2

Therefore when we have;

A'(0, 20), B'(16, 12), C'(8, 4), we get;

A'B' = 2 × AB

A'C' = 2 × AC
B'C' = 2 × BC

The correct option is therefore;

C. A'(0, 20), B'(16, 12), C'(8, 4)

Learn more about finding the distance between points on the coordinate plane here:

brainly.com/question/7243416

#SPJ1

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2 years ago

List these numbers in order from least to greatest -4, 6.2, 18 1/2 , -5.9, 21, -1/4 , 1.75

Anna [14]

-5.9, -4, -1/4, 1.75, 6.2, 18 1/2, 21

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3 years ago

What's a word problem for 3.05÷0.05?

KATRIN_1 [288]

<span>Number story for 3.05 divided by 0.05?</span>

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4 years ago