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4 years ago

Can someone solve this?

Mathematics

1 answer:

Elis [28]4 years ago

3 0

Answer:

7. 206.9955 in^2

8.482.1372 ft^2

9.256 ft^2

Step-by-step explanation:

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What is the distance between point G(−4, 1) and point H(−12, 1) ?

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8 units

Step-by-step explanation:

because they have the same y axis all we need to do is subtract the distance between the x axis to get the distance between the points

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Write an equation in point slope form for the line with a slope of 3 passes through (1,2)

alexgriva [62]

Answer:

$y-2=3(x-1)$

Step-by-step explanation:

The point-slope form of a line is given as:

$y-y_1=m(x-x_1)$

Where

m is the slope

$x_1$ is the x-coordinate of the point given (passing through the line)

$y_1$ is the y-coordinate of the point given (passing through the line)

Now,

Given,

Slope = m = 3

$x_1$ = 1

$y_1$ = 2

Now, we simply plug these into the formula for point-slope form of a line:

$y-y_1=m(x-x_1)\\y-2=3(x-1)$

This is the point-slope form.

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3 years ago

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3 years ago

13,406,109 pennis into dollars

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A rectangular box with a square base is to be constructed from material that costs $9 per ft2 for the bottom, $5 per ft2 for the

kramer

Answer:

$18.73ft^3$

Step-by-step explanation:

Let

Side of square base=x

Height of rectangular box=y

Area of square base=Area of top= $(side)^2=x^2$

Area of one side face= $l\times b=xy$

Cost of bottom=$9 per square ft

Cost of top=$5 square ft

Cost of sides=$4 per square ft

Total cost=$204

Volume of rectangular box= $V=lbh=x^2y$

Total cost= $9(x^2)+5x^2+4(4xy)=14x^2+16xy$

$204=14x^2+16xy$

$204-14x^2=16xy$

$y=\frac{204-14x^2}{16x}=\frac{102-7x^2}{8x}$

Substitute the values of y

$V(x)=x^2\times \frac{102-7x^2}{8x}=\frac{1}{8}(102x-7x^3)$

Differentiate w.r.t x

$V'(x)=\frac{1}{8}(102-21x^2)=0$

$V'(x)=0$

$\frac{1}{8}(102-21x^2)=0$

$102-21x^2=0$

$102=21x^2$

$x^2=\frac{102}{21}=4.85$

$x=\sqrt{4.85}=2.2$

It takes positive because side length cannot be negative.

Again differentiate w.r. t x

$V''(x)=\frac{1}{8}(-42x)$

Substitute the value

$V''(2.2)=-\frac{42}{8}(2.2)=-11.55$

Hence, the volume of box is maximum at x=2.2 ft

Substitute the value of x

$y=\frac{102-7(2.2)^2}{8(2.2)}=3.87 ft$

Greatest volume of box= $x^2y=(2.2)^2\times 3.87=18.73 ft^3$

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