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almond37 [142]
3 years ago
8

Someone help me with this

Mathematics
1 answer:
Jet001 [13]3 years ago
7 0

2a. 14,9; 2b. 15,4; 2c. 30,8; 3. 32

For 2a., you have to set it up like this: csc 48° = 20⁄x [OR sin 48° = ˣ⁄20]. Then you would have to isolate the variable by getting rid of the denominator. The cosecant function has an extra step because you will get xcsc 48° = 20. As stated, isolate the variable; this time, divide by csc 48°. This is what 20 will be divided by to get your x, whereas the other side cancels out. Then you have to round to the nearest tenth degree.

For 2b., you have to set it up like this: sec 39° = ˣ⁄12 [OR cos 39° = 12⁄x. Then you would have to isolate the variable by getting rid of the denominator. The cosine function has an extra step because you will get xcos 39° = 12. As stated, isolate the variable; this time, divide by cos 39°. This is what 12 will be divided by to get your x, whereas the other side cancels out. Then you have to round to the nearest tenth degree.

For 2c., you have to set it up like this: cot 64° = 15⁄x [OR tan 64° = ˣ⁄15]. Then you would have to isolate the variable by getting rid of the denominator. The cotangent function has an extra step because you will get xcot 64° = 15. As stated, isolate the variable; this time, divide by cot 64°. This is what 15 will be divided by to get your x, whereas the other side cancels out. Then you have to round to the nearest tenth degree.

Now, for 3., it is unique, but similar concept. In this exercise, we are solving for an angle measure, so we have to use inverse trigonometric ratios. So, we set it up like this: cot⁻¹ 1⅗ = m<x [OR tan⁻¹ ⅝ = m<x]. We simply input this into our calculator and we get 32,00538321°. When rounded to the nearest degree, we get 32°.

WARNING: If you use a graphing calculator, you have to input it uniquely because most graphing calculators do not have the inverse trigonometric ratios programmed in their systems. This is how you would write this: tan⁻¹ 1⅗⁻¹. You set 1⅗ to the negative first power, ALONG WITH the inverse tangent function, because without it, your answer will be thrown off. Since Cotangent and Tangent are multiplicative inverses of each other, that is the reason why the negative first power is applied ALONG WITH the inverse tangent function.

**NOTE: 1⅗ = 8⁄5

Take into consideration:

sin <em>θ</em> = O\H

cos <em>θ</em><em> </em>=<em> </em>A\H

tan <em>θ</em><em> </em>= O\A

sec <em>θ</em><em> </em>= H\A

csc <em>θ</em><em> </em>= H\O

cot <em>θ</em><em> </em>= A\O

I hope this helps you out alot, but if you are still in need of assistance, do not hesitate to let me know and subscribe to my channel [username: MATHEMATICS WIZARD], and as always, I am joyous to assist anyone at any time.

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Answer:

S varies partly directly as M and Q.

S=C.

S=KMQ+C.

For the first one...

speed=80,m=220,Q=30.

80=K20×30+C.

80=600K+C......(I).equation one.

For the second one....

speed=60,m=300,Q=40.

60=K300×40+C.

60=12000K+C.....(ii). equation two.

Minus eqtn(I) from eqtn(ii).

80=600K+C.

- 60=12000K+C.

K=0.01754~0.018.

Substitute K=0.018 into eqtn(I).

80=600K+C

80=600×0.018+C.

80=10.8+C.

C=80-10.8=69.2.

The relation is S=0.018MQ+69.2

when speed is 100 and mass is 250 find the volume.

100=0.018×250×Q+69.2.

100=4.5Q+69.2.

4.5Q=100-69.2

4.5Q=30.8.

Q=30.8/4.5.

Q=6.8~7litres.

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3 years ago
Jerry’s rhombus has a base of 15 inches and a height of 9 inches. Charlene’s rhombus has a base and a height that are 13 the bas
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Answer:  The area of Charlene's rhombus is nine times smaller than the area of Jerry's rhombus.

Step-by-step explanation:

I will assume that the exercise says "\frac{1}{3} times the base and height of Jerry’s rhombus".

The area of a rhombus can be calculated with the following formula:

A=b*a

Where "b" is the length of the base  and "a" is the altitude or the height.

Then, you can calculate the area using the formula shown above.

Therefore, you get:

1. Jerry's rhombus:

A_1=(15\ in)(9\ in)\\\\A_1=135\ in^2

2. Charlene's rhombus:

A_2=(\frac{1}{3}*15\ in)(\frac{1}{3}*9\ in)\\\\A_2=(5\ in)(3\ in)\\\\A_2=15\ in^2

Dividing the area calculated, you get:

\frac{135\ in^2}{15\ in^2}=9

Therefore, you can conclude that the area of Charlene's rhombus is nine times smaller than the area of Jerry's rhombus.

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3 years ago
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Answer:

Step-by-step explanation:

(7^0/7^5)^1/5

= (1 / 7^5)^ 1/5

= (7^-5)^1/5

= 7^-1

= 1/7 is the correct answer.

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3 years ago
Suppose the mean height for adult males in the U.S. is about 70 inches and the standard deviation is about 3 inches. Assume men’
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Question options :

a. They should be between 64 and 76 inches tall.

b. They should be close to the height that is 95% of the mean. That is, 66.5 inches, plus or minus 2 standard deviations.

c. They should be at or below the 95th percentile, which is 74.92 inches.

d. None of the above.

Answer: a. They should be between 64 and 76 inches tall.

Step-by-step explanation:

Given the following :

Assume men's height follow a normal curve ; and :

Mean height = 70 inches

Standard deviation= 3 inches

According to the empirical rule ;

Assuming a normal distribution with x being random variables ;

About 68% of x-values lie between -1 to 1 standard deviation of the mean. With about 95% of the x values lying between - 2 and +2 standard deviation of mean. With 99.7% falling between - 3 to 3 standard deviations from the mean.

Using the empirical rule :

95% will fall between + or - 2 standard deviation of the mean.

Lower limit = - 2(3) = - 6

Upper limit = 2(3) = 6

(-6+mean) and (+6+ mean)

(-6 + 70) and (6+70)

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