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jarptica [38.1K]

2 years ago

6

What is 5^(−3)×2^(−1)

1 answer:

Valentin [98]2 years ago

3 0

The answer to this question is 0.004

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Can someone translate this into a verbal expression 3(4j + 4 + j)?

dalvyx [7]

3(5j+4)
15j+12

final answer 15j+12

5 0

2 years ago

At Computer Central, the cost of renting a computer is $45 plus $35 per day. The total cost can be represented by the equation y

Thepotemich [5.8K]

Simple, since your equation is y=35x+45, it's quite simple to see what is the y-intercept.
It's written in y=mx+b form, the b being the y-intercept.

Thus, your y-intercept, is 45.

4 0

3 years ago

A woman drives a distance of 40km to work each day and her average speed for the journey is written as x km/h. If she increase h

nata0808 [166]

40/x = 40/x+5 + 2/60
-> i don’t have a sketchbook or calculator at the moment so pls figure out the final answer yourself

8 0

2 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

2 years ago

In a class of 40 students, 5 like neither Nokia nor

Vikki [24]

Answer:

No student liked only Nokia

Step-by-step explanation:

The information can be illustrated as shown on the Venn diagram.

The number of students who liked only Nokia is x.

The sum of all regions in the Venn diagram should be 40.

We add all to get:

$5 + 5 + 30 + x = 40$

$40 + x = 40$

$x = 40 - 40$

$x = 0$

Therefore no one liked only Nokia

8 0

2 years ago