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3 years ago

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Will give brianlist

1 answer:

MrRa [10]3 years ago

4 0

A. 120 millimeters.

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It costs $859.32 to have a school dance (tickets $8)

djverab [1.8K]

A) 859.32/8= 107.415=108 tickets

B) 980.68/8= 122.585=123 tickets

5 0

3 years ago

Read 2 more answers

Using the relationships between

Pavlova-9 [17]

Answer:

They will always intersect at one point

Step-by-step explanation:

^^

6 0

2 years ago

Find the length BC

Nady [450]

Answer:

The length of BC = 12

Step-by-step explanation:

c^2 = a^2 + b^2

15^2 = 9^2 + b^2

225 = 81 + b^2

b^2 = 225 - 81

b^2 = 144

b = 12

The length of BC is 12

5 0

3 years ago

4.) What is the exact value of sinθ when θ lies in Quadrant II and cosθ=−513

dimaraw [331]

Answer:

Part 4) $sin(\theta)=\frac{12}{13}$

Part 10) The angle of elevation is $40.36\°$

Part 11) The angle of depression is $78.61\°$

Part 12) $arcsin(0.5)=30\°$ or $arcsin(0.5)=150\°$

Part 13) $-45\°$ or $225\°$

Step-by-step explanation:

Part 4) we have that

$cos(\theta)=-\frac{5}{13}$

The angle theta lies in Quadrant II

so

The sine of angle theta is positive

Remember that

$sin^{2}(\theta)+ cos^{2}(\theta)=1$

substitute the given value

$sin^{2}(\theta)+(-\frac{5}{13})^{2}=1$

$sin^{2}(\theta)+(\frac{25}{169})=1$

$sin^{2}(\theta)=1-(\frac{25}{169})$

$sin^{2}(\theta)=(\frac{144}{169})$

$sin(\theta)=\frac{12}{13}$

Part 10)

Let

$\theta$ ----> angle of elevation

we know that

$tan(\theta)=\frac{85}{100}$ ----> opposite side angle theta divided by adjacent side angle theta

$\theta=arctan(\frac{85}{100})=40.36\°$

Part 11)

Let

$\theta$ ----> angle of depression

we know that

$sin(\theta)=\frac{5,389-2,405}{3,044}$ ----> opposite side angle theta divided by hypotenuse

$sin(\theta)=\frac{2,984}{3,044}$

$\theta=arcsin(\frac{2,984}{3,044})=78.61\°$

Part 12) What is the exact value of arcsin(0.5)?

Remember that

$sin(30\°)=0.5$

therefore

$arcsin(0.5)$ -----> has two solutions

$arcsin(0.5)=30\°$ ----> I Quadrant

or

$arcsin(0.5)=180\°-30\°=150\°$ ----> II Quadrant

Part 13) What is the exact value of $arcsin(-\frac{\sqrt{2}}{2})$

The sine is negative

so

The angle lies in Quadrant III or Quadrant IV

Remember that

$sin(45\°)=\frac{\sqrt{2}}{2}$

therefore

$arcsin(-\frac{\sqrt{2}}{2})$ ----> has two solutions

$arcsin(-\frac{\sqrt{2}}{2})=-45\°$ ----> IV Quadrant

or

$arcsin(-\frac{\sqrt{2}}{2})=180\°+45\°=225\°$ ----> III Quadrant

5 0

3 years ago

Points A, B, C, and D form the rectangle ABCD. E, F, and H are three points of another rectangle EFGH. Where does the fourth poi

ycow [4]

Answer:

D. (7.4); scale factor 2.

I’m not sure how to explain this but I’ve done it last year.

Hope that helps.

5 0

3 years ago