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3 years ago

Find the slope and the y-intercept of the equation 4x + 5y = -9

Mathematics

1 answer:

lutik1710 [3]3 years ago

7 0

Answer:

In your equation 4x+5y=-9, you've got to but your equation in slope intercept form which is y=mx+b.

First we need to get the y by itself.

Subtract 4x from each side to get:

5y=-4x-9

Next we want to divide both sides by 5.

y=-4/5x-9/5

Your slope is m which in our case -4/5.

Then our y-intercept is b or -9/5.

Your answer would then be C. slope=-4/5 and y-intercept=-9/5.

Hope this helps ;)

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Prove this identity using the product-to-sum identity for sin... sin^2 x=(1-cos(2x))/(2)

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Answer:

sin²x = (1 - cos2x)/2 ⇒ proved down

Step-by-step explanation:

∵ sin²x = (sinx)(sinx) ⇒ add and subtract (cosx)(cosx)

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3 years ago

6. Two observers, 7220 feet apart, observe a balloonist flying overhead between them. Their measures of the

MaRussiya [10]

Answer:

The ballonist is at a height of 3579.91 ft above the ground at 3:30pm.

Step-by-step explanation:

Let's call:

h the height of the ballonist above the ground,

a the distance between the two observers,

$a_1$ the horizontal distance between the first observer and the ballonist

$a_2$ the horizontal distance between the second observer and the ballonist

$\alpha _1$ and $\alpha _2$ the angles of elevation meassured by each observer

S the area of the triangle formed with the observers and the ballonist

So, the area of a triangle is the length of its base times its height.

$S=a*h$ (equation 1)

but we can divide the triangle in two right triangles using the height line. So the total area will be equal to the addition of each individual area.

$S=S_1+S_2$ (equation 2)

$S_1=a_1*h$

But we can write $S_1$ in terms of $\alpha _1$ , like this:

$\tan(\alpha _1)=\frac{h}{a_1} \\a_1=\frac{h}{\tan(\alpha _1)} \\S_1=\frac{h^{2} }{\tan(\alpha _1)}$

And for $S_2$ will be the same:

$S_2=\frac{h^{2} }{\tan(\alpha _2)}$

Replacing in the equation 2:

$S=\frac{h^{2} }{\tan(\alpha _1)}+\frac{h^{2} }{\tan(\alpha _2)}\\S=h^{2}*(\frac{1 }{\tan(\alpha _1)}+\frac{1}{\tan(\alpha _2)})$

And replacing in the equation 1:

$h^{2}*(\frac{1 }{\tan(\alpha _1)}+\frac{1}{\tan(\alpha _2)})=a*h\\h=\frac{a}{(\frac{1 }{\tan(\alpha _1)}+\frac{1}{\tan(\alpha _2)})}$

So, we can replace all the known data in the last equation:

$h=\frac{a}{(\frac{1 }{\tan(\alpha _1)}+\frac{1}{\tan(\alpha _2)})}\\h=\frac{7220 ft}{(\frac{1 }{\tan(35.6)}+\frac{1}{\tan(58.2)})}\\h=3579,91 ft$

Then, the ballonist is at a height of 3579.91 ft above the ground at 3:30pm.

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