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Debora [2.8K]
3 years ago
8

Which is a correct verbal expression for

Mathematics
1 answer:
Oxana [17]3 years ago
3 0
It should be the product of 8 and n
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Which is the graph of the equation y- 1= 5(x-3)?
Helen [10]

Answer:

linked below

Step-by-step explanation:

If you use desmos, it show this which is the correct graph :)

8 0
3 years ago
Help how do yall find this easy i cannot-
IgorC [24]

Answer:

153/50 or 3.06

Step-by-step explanation:

Just convert them into improper fractions and put parenthesis around the numbers with a division symbol in the middle of the two numbers.

Ex. (3/2)*(4/3)

7 0
3 years ago
Add<br><br> -3/8 + 5/4<br><br> Simplify your answer. Type a fraction or an interger.
sdas [7]
Beautiful your answer is 7/8
7 0
3 years ago
Read 2 more answers
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.

6 0
3 years ago
208=(1+5x)<br><br><br><br>Need help, i don't get it.
Snezhnost [94]
X equals 41.4 I think

3 0
3 years ago
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