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skad [1K]
2 years ago
11

Denis, Vera, and Yash are rock climbers. Yash is connected to Vera by a

Mathematics
2 answers:
grin007 [14]2 years ago
8 0

Answer:

33.14\°

Step-by-step explanation:

Let

Y ----> field of vision that Yash's camera would need

we know that

Applying the law of sines

\frac{sin(Y)}{25}=\frac{sin(41\°)}{30}

Solve for sin(Y)

sin(Y)=\frac{sin(41\°)}{30}(25)

Y=sin^{-1}[\frac{sin(41\°)}{30}(25)]

Y=33.14\°

Sphinxa [80]2 years ago
4 0

Answer:

The correct answer for khan Academy is also 33.14

Step-by-step explanation:

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Complete the solution of the equation. Find<br> the value of y when x equals -4.<br> x + 3y = 14
LekaFEV [45]

Answer:

y = 6

Step-by-step explanation:

-4 + 3y = 14

+4            +4

3y = 18

----   ----

3      3

y = 6

4 0
3 years ago
Find the area of the rectangle:
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10 ft and 2 because I did the math
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If fx = 2x^2 + 3 and g(x) = x^2 - 7 find (f - g) (x)
Semmy [17]

(f-g) (x) means to subtract g(x) from f(x)

F(x) = 2x^2 +3

g(x) = x^2

This becomes:

2x^2 + 3 - x^2 = x^2 +3

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2 years ago
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7 0
2 years ago
Is the given point interior , exterior, or on the circle k (x+2)2 + (y-3)2 =18 P (8,4)
Mnenie [13.5K]
One way would be to find the distance from the point to the center of the circle and compare it to the radius

for
(x-h)^2+(y-k)^2=r^2
the center is (h,k) and the radius is r

and the distance formula is
distance between (x_1,y_1) and (x_2,y_2) is
D=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}


r=radius
D=distance form (8,4) to center

if r>D, then (8,4) is inside the circle
if r=D, then (8,4) is on the circle
if r<D, then (8,4) is outside the circle


so
(x+2)^2+(y-3)^2=18
(x-(-2))^2+(y-3)^2=(\sqrt{18})^2
(x-(-2))^2+(y-3)^2=(3\sqrt{2})^2

the radius is 3\sqrt{2}
center is (-2,3)

find distance between (8,4) and (-2,3)

D=\sqrt{(8-(-2))^2+(4-3)^2}
D=\sqrt{(8+2)^2+(1)^2}
D=\sqrt{10^2+1}
D=\sqrt{100+1}
D=\sqrt{101}




r=3\sqrt{2}≈4.2
D=\sqrt{101}≈10.04

do r<D

(8,4) is outside the circle

6 0
3 years ago
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