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3 years ago

What is the solution of the system?

1 answer:

6 0

Equation 1 - equation 2: (y-y)=(10x-2)-(-7x+3)
17x-5=0
x=5/17
so answer A is correct.
double check: 10(5/17)-2=50/17-34/17=16/17. Yes.

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Can someone please help me

Vikki [24]

$16(4 shirts)= $64
5%=$64(0.05)=$3.20
Jamal's total is $67.20

4 0

2 years ago

you are organizing the schedule for a youth hockey season. If there are 6 teams and each team place each of the other teams twic

bija089 [108]

60 games

hope this helps

4 0

3 years ago

Mathematics

Vladimir [108]

Answer:

there no data

Step-by-step explanation:

Find the median of the first and third quartiles81, 69 ,90, 79, 59, 125, 95, 116, 183, 97, 92, 50, 122, 55, 9250, 55, 59, 69, 79, 81, 90, 92, 92, 95, 97, 116, 122, 125, 183.The first quartile: 50, 55, 59, 69, 79, 81, 90: median of the first quartile is 69The third quartile: 92, 95, 97, 116, 122, 125, 183: median of the third quartile is 116.The interquartile range is the difference between the median of both the first and third quartile: which is 116 - 69 = 47.There are no outliers in this set of data

4 0

1 year ago

GIVING BRAINLIEST! A rectangular garden has a width of 10 feet and a length of 14 feet. A cement walkway is added around the out

Serjik [45]

Given:

Width of a garden = 10 feet

Length of the garden = 14 feet

Area of the garden and the walkway together = 396 square feet.

To find:

The width of the walkway.

Solution:

A cement walkway is added around the outside of the garden.

Let x be the width of the walkway.

The width of the garden with walkway = 10+2x

The length of the garden with walkway = 14+2x

Area of a rectangle is

$Area=length\times width$

Area of garden and the walkway together is

$Area=(14+2x)\times (10+2x)$

$396=140+28x+20x+4x^2$

$396=140+48x+4x^2$

$396=4(35+12x+x^2)$

Divide both sides by 4.

$99=35+12x+x^2$

$0=35-99+12x+x^2$

$0=-64+12x+x^2$

Splitting the middle term, we get

$x^2+16x-4x-64=0$

$x(x+16)-4(x+16)=0$

$(x+16)(x-4)=0$

Using zero product property, we get

$(x+16)=0\text{ and }(x-4)=0$

$x=-16\text{ and }x=4$

Width of walkway cannot be negative. So, x=4.

Therefore, the width of the walkways is 4 feet.

6 0

2 years ago

Read 2 more answers

Your principal surprises you by buying your class a turtle. He brings the turtle to your class along with a sheet from the pet s

Nitella [24]

The height of the tank must be at least 1 foot, or 12 inches. We know the floor area (which is length x width) must be at least 400 inches. Therefore these minimum dimensions already tell us that the minimum volume is 400 x 12 = 4800 cubic inches. Since we have a maximum of 5000 cubic inches, the volume must be within the range of 4800 - 5000 cubic inches.

We can set the height at exactly 1 ft (or 12 inches). Then we can select length and width that multiply to 400 square inches, for example, L = 40 inches and W = 10 in. This gives us a tank of dimensions 40 x 10 x 12 = 4800 cubic inches, which fits all the criteria.

7 0

3 years ago