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2 years ago

8

What is the standard equation of a line whose x- and y-intercept are 5 and 3, respectively?

1 answer:

MakcuM [25]2 years ago

7 0

Answer: 3x + 5y = 15

Step-by-step explanation:

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FIND THE VOLUME OF THE CONE

sukhopar [10]

Answer:

V=130.83

Step-by-step explanation:

V=pi*5^2*5/3=125pi/3=130.83

3 0

3 years ago

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PLEASE HELP ME SOLVE THIS ASAP WITH WORK!!

Vlad [161]

Answer:

6

Step-by-step explanation:

If you multiply a number from the big polygon by 2/3s, you will get the corresponding number on the small polygon. For example, 6 × 2/3 = 4
To find x, find the corresponding number on the big polygon and multiply it by 2/3s. The number on the big polygon is 9. 9 × 2/3 = 6

I hope this helps!

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3 years ago

Click an item in the list or group of pictures at the bottom of the problem and, holding the button down, drag it into the corre

IgorC [24]

Evaluate: x⁻² for x = -3

The negative sign of the exponent signifies that the number is a denominator with 1 as the numerator.

The number is in negative form, it becomes a positive number because it is squared. negative number times negative number equals positive number.

x⁻² where x = -3 ⇒⇒ -3⁻² = 1/-3² = 1/9

8 0

3 years ago

A rectangular schoolyard is to be fenced in using the wall of the school for one side and 150

arsen [322]

Answer:

Part 1) The expression is $A(x)=150x-2x^2$

Part 2) The area of the schoolyard when x=40 m is A=2,800 m^2

Part 3) The domain is all real numbers greater than zero and less than 75 meters

Step-by-step explanation:

Part 1) Write an expression for A(x)

Let

x -----> the length of the rectangular schoolyard

y ---> the width of the rectangular schoolyard

we know that

The perimeter of the fencing (using the wall of the school for one side) is

$P=2x+y$

$P=150\ m$

so

$150=2x+y$

$y=150-2x$ -----> equation A

The area of the rectangular schoolyard is

$A=xy$ ----> equation B

substitute equation A in equation B

$A=x(150-2x)$

$A=150x-2x^2$

Convert to function notation

$A(x)=150x-2x^2$

Part 2) What is the area of the schoolyard when x=40?

For x=40 m

substitute in the expression of Part 1) and solve for A

$A(40)=150(40)-2(40^2)$

$A(40)=2,800\ m^2$

Part 3) What is a reasonable domain for A(x) in this context

we know that

A represent the area of the rectangular schoolyard

x represent the length of of the rectangular schoolyard

we have

$A(x)=150x-2x^2$

This is a vertical parabola open downward

The vertex is a maximum

The x-coordinate of the vertex represent the length for the maximum area

The y-coordinate of the vertex represent the maximum area

The vertex is the point (37.5, 2812.5)

see the attached figure

therefore

The maximum area is 2,812.5 m^2

The x-intercepts are x=0 m and x=75 m

The domain for A is the interval -----> (0, 75)

All real numbers greater than zero and less than 75 meters

5 0

3 years ago

In a basketball game, the two teams scored a total of 210 points. The home teams won by 20 points. How many did the home team sc

olasank [31]

They won with a total of 115 you subtract 20 from 210 and get 190 then divided by 2 you should get 95 then add 20 then you get 115

4 0

3 years ago

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