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TEA [102]
3 years ago
12

Help me please and thank you :)

Mathematics
1 answer:
Paul [167]3 years ago
4 0

Answer:

A

Step-by-step explanation:

We have the equation:

\displaystyle \frac{3}{8}y=2+x

And we want to determine whether or not this represents a direct proportion.

First, let's solve for <em>y. </em>Multiply both sides by 8/3:

\displaystyle y=\frac{16}{3}+\frac{8}{3}x

Remember that direct proportions must pass through the origin point on a graph. In other words, their <em>y-</em>intercept or constant value is zero.

Since the constant value here is not zero (it is 16/3), the equation is not direct proportion.

Our answer is A.

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Answer:

100

Step-by-step explanation:

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If 29 cos 0=21 find the value of 1/sin0-tan0..
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It should be 21 20. Sorry if I got it wrong
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Look at the Hexagon below
Mama L [17]

Answer:

Angle 9: 60°

Angle 10: 30°

Side = radius = 40sqrt(x/3)

Area = [800sqrt(3)]x

Step-by-step explanation:

Total angle in a hexagon:

(6 - 2) × 180

720

Each interior angle:

720/6 = 120

angle 9 = 120/2 = 60

Angle 10 = 60/2 = 30

sin(60) = 20sqrt(x)/r

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r = 40sqrt(x/3)

Side:

sin(30) = ½s/(40sqrt(x/3))

½s = 20sqrt(x/3)

s = 40sqrt(x/3)

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4 0
3 years ago
A company is making a new label for one of their containers. The container is a cylinder that is 9 inches tall and 5 inches in d
nydimaria [60]

Answer:

180.55 in².

Step-by-step explanation:

Data obtained from the question include the following:

Height (h) = 9 in.

Diameter (d) = 5 in

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Area of the label =..?

Next, we shall determine the radius.

Diameter (d) = 5 in

Radius (r) =.. ?

Radius (r) = Diameter (d) /2

r = d/2

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r = 2.5 in.

Next, we shall determine the area of the label that needs to be printed to go around the new container by calculating the surface area of the cylinder.

This is illustrated below:

Height (h) = 9 in.

Pi (π) = 3.14

Radius (r) = 2.5 in.

Surface Area (SA) =.?

SA = 2πrh + 2πr²

SA = (2×3.14×2.5×9) + (2×3.14×2.5²)

SA = 141.3 + 39.25

SA = 180.55 in²

The surface area of the cylinder is 180.55 in².

Therefore, the area of the label that needs to be printed to go around the new container is 180.55 in².

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