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3 years ago

12

Please help me answer this :)! I will give braisnlt

1 answer:

djyliett [7]3 years ago

3 0

Answer: I don't see anything

Step-by-step explanation:

You might be interested in

X^2-8x+10=-4 quadratic equation written in standard form

ASHA 777 [7]

Rewriting the formula in standard form would be x^2-8x+14=0

3 0

3 years ago

Can someone help me on these

grandymaker [24]

Answer:

Almost every question you ask can and will be answered if asked in a right why

6 0

3 years ago

A rectangular box with a volume of 272ft^3 is to be constructed with a square base and top. The cost per square foot for the bot

ASHA 777 [7]

Answer:

The dimensions of the box is 3 ft by 3 ft by 30.22 ft.

The length of one side of the base of the given box is 3 ft.

The height of the box is 30.22 ft.

Step-by-step explanation:

Given that, a rectangular box with volume of 272 cubic ft.

Assume height of the box be h and the length of one side of the square base of the box is x.

Area of the base is = $(x\times x)$

$=x^2$

The volume of the box is = area of the base × height

$=x^2h$

Therefore,

$x^2h=272$

$\Rightarrow h=\frac{272}{x^2}$

The cost per square foot for bottom is 20 cent.

The cost to construct of the bottom of the box is

=area of the bottom ×20

$=20x^2$ cents

The cost per square foot for top is 10 cent.

The cost to construct of the top of the box is

=area of the top ×10

$=10x^2$ cents

The cost per square foot for side is 1.5 cent.

The cost to construct of the sides of the box is

=area of the side ×1.5

$=4xh\times 1.5$ cents

$=6xh$ cents

Total cost = $(20x^2+10x^2+6xh)$

$=30x^2+6xh$

Let

C $=30x^2+6xh$

Putting the value of h

$C=30x^2+6x\times \frac{272}{x^2}$

$\Rightarrow C=30x^2+\frac{1632}{x}$

Differentiating with respect to x

$C'=60x-\frac{1632}{x^2}$

Again differentiating with respect to x

$C''=60+\frac{3264}{x^3}$

Now set C'=0

$60x-\frac{1632}{x^2}=0$

$\Rightarrow 60x=\frac{1632}{x^2}$

$\Rightarrow x^3=\frac{1632}{60}$

$\Rightarrow x\approx 3$

Now $C''|_{x=3}=60+\frac{3264}{3^3}>0$

Since at x=3 , C''>0. So at x=3, C has a minimum value.

The length of one side of the base of the box is 3 ft.

The height of the box is $=\frac{272}{3^2}$

=30.22 ft.

The dimensions of the box is 3 ft by 3 ft by 30.22 ft.

7 0

3 years ago

What is the solution for -10 < x - 9?

ikadub [295]

ANSWER:

x > - 1

STEP-BY-STEP EXPLANATION:

- 10 < x - 9

x - 9 > - 10

x > - 10 + 9

x > - 1

6 0

4 years ago

If you have a cube measuring 1 unit on each side, how many of those cubes would fit into a space 2 units by 3 units by 4 units

alisha [4.7K]

Given:

Measure of a cube = 1 unit on each side.

Dimensions of a space 2 units by 3 units by 4 units.

To find:

Number of cubes that can be fit into the given space.

Solution:

The volume of cube is:

$V_1=a^3$

Where, a is the side length of cube.

$V_1=(1)^3$

$V_1=1$

So, the volume of the cube is 1 cubic units.

The volume of the cuboid is:

$V_2=l\times b\times h$

Where, l is length, w is width and h is height.

Putting $l=2,b=3,h=4$ , we get

$V_2=2\times 3\times 4$

$V_2=24$

So, the volume of the space is 24 cubic units.

We need to divide the volume of the space by the volume of the cube to find the number of cubes that can be fit into the given space.

$n=\dfrac{V_2}{V_1}$

$n=\dfrac{24}{1}$

$n=24$

Therefore, 24 cubes can be fit into the given space.

3 0

3 years ago