Ask question

Ask question

All categories

3 years ago

14

The perimeter of the rectangular playing field is 460 yards. The length of the field is 5 yards less than quadruple the width. W

hat are the dimensions of the playing field?

1 answer:

Tpy6a [65]3 years ago

8 0

Answer:

multiply 460 by 5

Step-by-step explanation:

460 * 5 = 2,300

You might be interested in

You start a search for a buried object by marking the center of a field as (0, 0), with coordinates giving distances in yards. C

pychu [463]

The coordinates to the second spot can be represented by the coordinates

(-12 + 3, 8 -11)

Since we are going 3 yards to the east we would be affecting the x-axis and since we are going east we will need to add 3 yards the x axis of the first spot

11 yards south would be affecting the y- axis negatively. We would need to subtract 11 from the y- axis of the first spot.

the coordinates of the second spot would be

-12 +3 = -9

8 -11 = -3

(-9, -3)

7 0

2 years ago

What 2 numbers add to get 29 but when you subtract you get 1?

stiks02 [169]

Answer:

15 and 14

Step-by-step explanation:

15+14=29

15-14=1

3 0

1 year ago

Please someone help me out on this

valentinak56 [21]

Answer:

I think it is C

Step-by-step explanation:

im sorry if i am wrong

6 0

2 years ago

Read 2 more answers

What is the solution to sqrt 17-x=x+3? Show your work.

snow_lady [41]

Answer:

$x=1$

Step-by-step explanation:

Remember:

$(\sqrt[n]{a})^n=a\\\\(a+b)=a^2+2ab+b^2$

Given the equation $\sqrt{17-x}=x+3$ , you need to solve for the variable "x" to find its value.

You need to square both sides of the equation:

$(\sqrt{17-x})^2=(x+3)^2$

$17-x=(x+3)^2$

Simplifying, you get:

$17-x=x^2+2(x)(3)+3^2\\\\17-x=x^2+6x+9\\\\x^2+6x+9+x-17=0\\\\x^2+7x-8=0$

Factor the quadratic equation. Find two numbers whose sum be 7 and whose product be -8. These are: -1 and 8:

$(x-1)(x+8)=0$

Then:

$x_1=1\\x_2=-8$

Let's check if the first solution is correct:

$\sqrt{17-(1)}=(1)+3$

$4=4$ (It checks)

Let's check if the second solution is correct:

$\sqrt{17-(-8)}=(-8)+3$

$5\neq-5$ (It does not checks)

Therefore, the solution is:

$x=1$

3 0

3 years ago

Help on questions 11 and 15 Add and simplify if possible

Svetllana [295]

$\frac{16+x}{x^3}+\frac{7-4x}{x^3}=\frac{16+x+7-4x}{x^3}=\frac{23-3x}{x^3}\\\\======================================\\\\\frac{5}{t-1}+\frac{3}{t}=\frac{5t}{t(t-1)}+\frac{3(t-1)}{t(t-1)}=\frac{5t+3t-3}{t^2-t}=\frac{8t-3}{t^2-t}$

7 0

2 years ago

Read 2 more answers