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Natalka [10]
3 years ago
6

Someone help me out please

Mathematics
1 answer:
Levart [38]3 years ago
3 0

Answer:

33 ft.

Step-by-step explanation:

First, you would put the x at the opposite side, because that's what the problem is asking for, the distance between the ground and the top of the ladder. Then, because it is opposite and adjacent, you use tangent to solve the problem. Equation: tan(70)=x/12, tan=opp/adj. Then, cross multiply and get x=12(tan(70)), and then multiply to get 32.97 and round up to 33 ft.

(hope this helps!!)

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What figure are you talking about

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You use a gift card to buy a pair of shorts for $8. There is $32 left on the gift card after your purchase. What was the origina
Olenka [21]

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40

Step-by-step explanation:

32+8=40

6 0
3 years ago
a map has a scale of 6 in : 36 mi. If Clayton and Clinton are 52 mi apart, then they are how far apart on the map?
cricket20 [7]
The map scale is 6 in : 36 mi

simplify it, divide 6 from both sides

6/6 = 1
36/6 = 6

1 in : 6 mi

To find the amount of inches in 52 miles, divide 52 with 6

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8.67 in should be your answer

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6 0
3 years ago
Find the extreme values of the function f(x, y) = 4x2 + 6y2 on the circle x2 + y2 = 1.
julia-pushkina [17]

Answer with Step-by-step explanation:

We are given that

f(x,y)=4x^2+6y^2

Let g(x,y)=x^2+y^2=1

We have to find the extreme values  of the given function

\nabla f(x,y)

\nabla g(x,y)=

Using Lagrange multipliers

\nabla f(x,y)=\lambda \nabla g(x,y)

f_x=\lambda g_x

8x=\lambda 2x

Possible value x=0 or \lambda=4

If x=0 then substitute the value in g(x,y)

Then, we get y=\pm 1

f_y=\lambda g_y

12y=\lambda 2y

If \lambda=4 and substitute in the equation

Then , we  get possible value of y=0

When y=0 substitute in g(x,y) then we get

x=\pm 1

Hence, function has possible  extreme values at points (0,1),(0,-1), (1,0) and (-1,0).

f(0,1)=6

f(0,-1)=6

f(1,0)=4

f(-1,0)=4

Therefore, the maximum value of f  on the circlex^2+y^2=1  is f(0,\pm1)=6 and minimum value of f(\pm1,0)=4

7 0
3 years ago
The waiting times between a subway departure schedule and the arrival of a passenger are uniformly distributed between 0 and 8 m
rusak2 [61]

Answer:

0.59375

Step-by-step explanation:

In a uniform distribution the probability that the time t is greater than any given value, X, is:

P(t\geq X)=1-\frac{X}{b-a}

In this problem, the limits of the distribution are a = 0 and b = 8 minutes.

For X =3.25 minutes:

P(t\geq 3.25)=1-\frac{3.25}{8-0} \\P(t\geq 3.25)=0.59375

The probability that a randomly selected passenger has a waiting time greater than 3.25 minutes is 0.59375.

8 0
3 years ago
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