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Elan Coil [88]
3 years ago
11

WORK THIS CHALLENGE EXERCISE. A is known to be 6,500 feet above sea level; AB = 600 feet. The angle at A looking up at P is 20°.

The angle at B looking up at P is 35°. How far above sea level is the peak P? Find the height of the mountain peak to the nearest foot. Height above sea level = a0ft. (Hint: Draw a perpendicular from B to . Label the various angles. Compute BJ, then BP, then PQ, rounding to the nearest tenth in each step; and finally find 6,500 + PQ.)
Mathematics
2 answers:
Dmitry_Shevchenko [17]3 years ago
8 0

Answer:

6955ft

Step-by-step explanation:

adell [148]3 years ago
6 0

Answer:

6955ft

T

Step-by-step explanation:

Hope that helps

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4 0
3 years ago
Tiana can clear a football field of debris in 3 hours. Jacob can clear the same field in 2 hours. If they decide to clear the fi
Marianna [84]
You must find the relative rates of both workers...

T=f/3 and J=f/2  so if they work together to clear the field then:

ft/3+ft/2=f  make all terms have a common denominator of 6

(2/2)(ft/3)+(3/3)(ft/2)=(6/6)f

2ft/6+3ft/6=6f/6  multiply both sides by 6

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8 0
3 years ago
Consider the equation below.
Korvikt [17]

Answer:

Equation in square form:

y=3(x+5)^2-4

Extreme value:

(h,k)=(-5,-4)

Step-by-step explanation:

We are given

y=3x^2+30x+71

we can complete square

y=3(x^2+10x)+71

we can use formula

a^2+2ab+b^2=(a+b)^2

y=3(x^2+2\times x\times 5)+71

now, we can add and subtract 5^2

y=3(x^2+2\times x\times 5+5^2-5^2)+71

y=3(x^2+2\times x\times 5+5^2)-3\times 5^2+71

y=3(x+5)^2-75+71

So, we get equation as

y=3(x+5)^2-4

Extreme values:

we know that this parabola

and vertex of parabola always at extreme values

so, we can compare it with

y=a(x-h)^2+k

where

vertex=(h,k)

now, we can compare and find h and k

y=3(x+5)^2-4

we get

h=-5

k=-4

so, extreme value of this equation is

(h,k)=(-5,-4)

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X + 4 = 11
mojhsa [17]

Answer:

A. Subtraction

B. Subtraction

C. x+4=11

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D. x=7

Step-by-step explanation:

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