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Zarrin [17]
2 years ago
11

If tan a= 7/24 find sin a and cos a.. ​

Mathematics
1 answer:
aev [14]2 years ago
5 0
Tan=sin/cos
So sin a= 7
Cos a= 24
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Tenths and hundredth
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19. No
20. No
21. Yes
23. 1.8 miles
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3 years ago
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Someone please help me <br> The top number is 18
kotegsom [21]

Answer:

x = 24

Step-by-step explanation:

Since this is a right angle, we can use the Pythagorean Theorem to find x. In the formula, a^2 + b^2 = c^2, variables a and b represent the lengths of the legs, and variable c represents the length of the hypotenuse.

a^2 + b^2 = c^2

18^2 + x^2 = 30^2

324 + x^2 = 900

x^2 = 576

x = 24

4 0
1 year ago
Explain how you would determine if these amounts are the same or
Inga [223]

Answer: they are he same as 75%

Step-by-step explanation:

onvert 3/4 to a percent. Begin by converting the fraction 3/4 into decimal. Multiply the decimal by 100 and write the result with the percentage sign: 0.75 × 100 = 75%.

6 out of 8 can be written as 6/8 and equals to 75%. Let's understand the conversion of a fraction to a percentage. To find the percent for this fraction, we have to find the number of parts that would be shaded out of 100. To convert a fraction to percent, we multiply it by 100/100.

3 0
2 years ago
Which expression is a possible leading term for the polynomial function graphed below?
Anon25 [30]

Answer:

D, 22x^9

Step-by-step explanation:

Here's a key for end behavior: Look at the leading term

x^even = x -> neg inf, f(x) -> pos inf; x -> pos inf, f(x) -> pos inf

-x^even = x -> neg inf, f(x) -> neg inf; x -> pos inf, f(x) -> neg inf

x^odd = x -> neg inf, f(x) -> neg inf; x -> pos inf, f(x) -> pos inf

-x^odd = x -> neg inf, f(x) -> pos inf; x -> pos inf, f(x) -> neg inf

6 0
2 years ago
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The position of an object along a vertical line is given by s(t) = −t3 + 3t2 + 7t + 4, where s is measured in feet and t is meas
saw5 [17]

Answer:

The maximum velocity of the object in the time interval [0, 4] is 10 ft/s.

Step-by-step explanation:

Given : The position of an object along a vertical line is given by s(t) = -t^3+3t^2+7t +4, where s is measured in feet and t is measured in seconds.

To find : What is the maximum velocity of the object in the time interval [0, 4]?

Solution :

The velocity is rate of change of distance w.r.t time.

Distance in terms of t is given by,

s(t) = -t^3+3t^2+7t +4

Derivate w.r.t. time,

v(t)=s'(t) = -3t^2+6t+7

It is a quadratic function so its maximum is at vertex of the function.

The x point of the function is given by,

x=-\frac{b}{2a}

Where, a=-3, b=6 and c=7

t=-\frac{6}{2(-3)}

t=-\frac{6}{-6}

t=1

As 1 lie between interval [0,4]

Substitute t=1 in the function,

v(t)= -3(1)^2+6(1)+7

v(t)= -3+6+7

v(1)=10

Th maximum velocity is 10 ft/s.

Therefore, the maximum velocity of the object in the time interval [0, 4] is 10 ft/s.

8 0
2 years ago
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