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3 years ago

Please answer these I will give you all my points!

Mathematics

1 answer:

makvit [3.9K]3 years ago

5 0

ok so put it like that and it will be correct.............................................................................(long pause)........................................................................................................... (thinks) I think that the right answer.

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Here is the next questions brainliest to the first who answer it

qaws [65]

Answer:

Look at the attachment

7 0

3 years ago

What number is 120% of 12

Rasek [7]

120% of 12 is 14.4.I got this by doing this:

12/x=100/120
(12/x)*x=(100/120)*x - we multiply both sides of the equation by x
12=0.833333333333*x - we divide both sides of the equation by (0.833333333333) to get x
12/0.833333333333=x 
14.4=x 
x=14.4

5 0

3 years ago

Find sin(a)&cos(B), tan(a)&cot(B), and sec(a)&csc(B).

Reil [10]

Answer:

Part A) $sin(\alpha)=\frac{4}{7},\ cos(\beta)=\frac{4}{7}$

Part B) $tan(\alpha)=\frac{4}{\sqrt{33}},\ tan(\beta)=\frac{4}{\sqrt{33}}$

Part C) $sec(\alpha)=\frac{7}{\sqrt{33}},\ csc(\beta)=\frac{7}{\sqrt{33}}$

Step-by-step explanation:

Part A) Find $sin(\alpha)\ and\ cos(\beta)$

we know that

If two angles are complementary, then the value of sine of one angle is equal to the cosine of the other angle

In this problem

$\alpha+\beta=90^o$ ---> by complementary angles

$sin(\alpha)=cos(\beta)$

Find the value of $sin(\alpha)$ in the right triangle of the figure

$sin(\alpha)=\frac{8}{14}$ ---> opposite side divided by the hypotenuse

simplify

$sin(\alpha)=\frac{4}{7}$

therefore

$sin(\alpha)=\frac{4}{7}$

$cos(\beta)=\frac{4}{7}$

Part B) Find $tan(\alpha)\ and\ cot(\beta)$

we know that

If two angles are complementary, then the value of tangent of one angle is equal to the cotangent of the other angle

In this problem

$\alpha+\beta=90^o$ ---> by complementary angles

$tan(\alpha)=cot(\beta)$

Find the value of the length side adjacent to the angle alpha

Applying the Pythagorean Theorem

Let

x ----> length side adjacent to angle alpha

$14^2=x^2+8^2\\x^2=14^2-8^2\\x^2=132$

$x=\sqrt{132}\ units$

simplify

$x=2\sqrt{33}\ units$

Find the value of $tan(\alpha)$ in the right triangle of the figure

$tan(\alpha)=\frac{8}{2\sqrt{33}}$ ---> opposite side divided by the adjacent side angle alpha

simplify

$tan(\alpha)=\frac{4}{\sqrt{33}}$

therefore

$tan(\alpha)=\frac{4}{\sqrt{33}}$

$tan(\beta)=\frac{4}{\sqrt{33}}$

Part C) Find $sec(\alpha)\ and\ csc(\beta)$

we know that

If two angles are complementary, then the value of secant of one angle is equal to the cosecant of the other angle

In this problem

$\alpha+\beta=90^o$ ---> by complementary angles

$sec(\alpha)=csc(\beta)$

Find the value of $sec(\alpha)$ in the right triangle of the figure

$sec(\alpha)=\frac{1}{cos(\alpha)}$

Find the value of $cos(\alpha)$

$cos(\alpha)=\frac{2\sqrt{33}}{14}$ ---> adjacent side divided by the hypotenuse

simplify

$cos(\alpha)=\frac{\sqrt{33}}{7}$

therefore

$sec(\alpha)=\frac{7}{\sqrt{33}}$

$csc(\beta)=\frac{7}{\sqrt{33}}$

6 0

3 years ago

there are about 1.5 grams of fat in 1 tabespoon of humus.how many grams of fat are in 2 1/2 cups of hummus ?

Kipish [7]

We'll first we need to know the in 1 cup there is 16 tablespoon.
so we could multiply we need 2 1/2 cup so that is 16 (1cup) + 16 (1cup) +8 (1/2 cup)= 40 tablespoons... and 1 tablespoon is 1.5 grams of fat then we multiply 40 times 1.5 .... This is equal to 60 grams fat
So there is 60 grams of fat in 2 1/5 cups of hummus.

6 0

3 years ago

Calculate the average time the car took to reach each checkpoint. Record the average time in Table D of your Student Guide. The

zhuklara [117]

The average time the car took to reach each checkpoint are:

2.07
3.16
4.11
4.92

<h3>Average time</h3>

Given:

Time interval

1 2 3 4

2.02 3.17 4.12 4.93

2.05 3.07 3.98 4.81

2.15 3.25 4.23 5.01

Hence:

First quarter checkpoint

Average time= (2.02 + 2.05 + 2.15) / 3

Average time=6.22/3

Average time= 2.07s

Second quarter checkpoint

Average time= (3.17 +3.07 + 3.25) / 3

Average time=9.49/3

Average time = 3.16 s

Third quarter check point

Average time= (4.12 + 3.98 + 4.23) / 3

Average time=12.33/3

Average time= 4.11 s

Fourth quarter check point

Average time = (4.93 + 4.81 + 5.01) / 3

Average time=14.75/3

Average time= 4.917 s

Average time=4.92s (Approximately)

Therefore the average time the car took to reach each checkpoint are: 2.07, 3.16, 4.11, 4.92.

Learn more about average time here:brainly.com/question/19136062

#SPJ1

6 0

2 years ago