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3 years ago

Evaluate: 6 plus x squared times 2 for x = 4

Mathematics

1 answer:

Mkey [24]3 years ago

3 0

$4^{2} = 16$

16 times 2 = 32

32 plus 6 = 38

always remember PEMDAS

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Use the first and last data points to find the slope intercept equation of a trend line.

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Answer:

y = $\frac{16}{3}x-79$

Step-by-step explanation:

From the given table,

Two points are (1, 15) and (7, 47)

If the two points $(x_1,y_1)$ and $(x_2,y_2)$ are lying on a line then slope 'm' of the line will be,

m = $\frac{y_2-y_1}{x_2-x_1}$

= $\frac{47-15}{7-1}$

= $\frac{32}{6}$

= $\frac{16}{3}$

Let the equation of a line passing through (h, k) is,

y - h = m(x - k)

If the line passes through (1, 15)

y - 1 = $\frac{16}{3}(x-15)$

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y = $\frac{16}{3}x-79$

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3 years ago

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Answer:

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Step-by-step explanation:

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3 years ago

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Answer:

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Step-by-step explanation:

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2 years ago

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3 years ago

A right cylinder has a radius of r inches and height of 2r inches.

lys-0071 [83]

Answer:

Part a) The lateral area is $4r^{2} \pi \ in^{2}$

Part b) The area of the two bases together is $2r^{2} \pi\ in^{2}$

Part c) The surface area is $6r^{2} \pi\ in^{2}$

Step-by-step explanation:

we know that

The surface area of a right cylinder is equal to

$SA=LA+2B$

where

LA is the lateral area

B is the area of the base of cylinder

we have

$r=r\ in$

$h=2r\ in$

Part a) Find the lateral area

The lateral area is equal to

$LA=2\pi rh$

substitute the values

$LA=2\pi r(2r)$

$LA=4r^{2} \pi\ in^{2}$

Part b) Find the area of the two bases together

The area of the base B is equal to

$B=r^{2} \pi\ in^{2}$

the area of the two bases together is

$2B=2r^{2} \pi\ in^{2}$

Part c) Find the surface area of the cylinder

$SA=LA+2B$

we have

$LA=4r^{2} \pi\ in^{2}$

$2B=2r^{2} \pi\ in^{2}$

substitute

$SA=4r^{2} \pi+2r^{2} \pi=6r^{2} \pi\ in^{2}$

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