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3 years ago

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Mr. Williams is building a sand box for his children. It cost $228 for the sand if he builds a regular-sandbox with dimensions 9

‘ x 6‘. How much will the same cost if he decides to increase the size to 13 1/2‘ x 9‘?

1 answer:

erastovalidia [21]3 years ago

3 0

Answer:

It will cost $512.73

Step-by-step explanation:

9 in by 6 in sandbox cost $228

i.e., 9*6 = 54 in²

54 in² cost $228

1 in² will cost $228/54

≈ $4.22

If the new dimension is 13 1/2 in by 9 in

then area will be 13 1/2*9

27*9/2 = 121.5 in²

The cost of 121.5 in² will be

121.5*4.22 = $512.73

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A bag contains three red marbles, two green ones, one lavender one, two yellows, and two orange marbles. HINT [See Example 7.] H

lapo4ka [179]

Answer: 8

Step-by-step explanation:

Given: A bag contains three red marbles, two green ones, one lavender one, two yellows, and two orange marbles.

Total marbles other than green = 8

Total marbles other than green and yellow = 6

Then the number of sets of seven marbles include at least one yellow one but no green ones:-

$^{2}C_1\times^{6}C_6+ ^2C_2\times^6C_5\\\\= 2\times 1+1\times6\\\\=2+6=8$

Number of sets of seven marbles include at least one yellow one but no green ones = 8

3 0

3 years ago

Can u guys help i need to make this longer lol

Marizza181 [45]

I don’t under stand the question. What do you want me to answer

5 0

3 years ago

4/7 - x = 6/35 what is x? how do I determine the answer?

vivado [14]

4/7 - x = 6/35

You're trying to get x by itself, so you have to subtract 4/7 from both sides.

-x = 6/35 - 4/7

You need both of the fractions to the right of the equal sign to have the same denominator so that we could simplify them, so multiply 5/5 to -4/7.

-4 /7 × 5/5 = -20/35

So, our new equation is :

-x = 6/35 - 20/35

Simplify.

-x = -14/35

Divide both sides by -1.

x = 14/35

Divide by 7.

x = 2/5

~Hope I helped!~

7 0

3 years ago

Answers fast please. 1.) y=x−63x+2y=8 Use the substitution method. A.) (4, −2) B.) (14, 8) C.) (0, −6) D.) (3, −3) 2.) What is t

GrogVix [38]

<u>QUESTION 1</u>

The given system of equation is

$y=x-6...eqn1$

and

$3x+2y=8...eqn2$

Let us substitute equation (1) into equation (2) to get,

$3x+2(x-6)=8$

We expand the bracket to get,

$3x+2x-12=8$

We simplify to get.

$5x-12=8$

We group like terms to get

$5x=8+12$

$\Rightarrow 5x=20$

$\Rightarrow x=4$

We now substitute $x=4$ in to equation (1) to obtain,

$y=4-6=-2$

The correct answer is option A.

<u>QUESTION 2</u>

The given system of equations is

$y-x=9...eqn1$

and

$10+2x=-2y...eqn2$

We make y the subject in equation (2) to get,

$y=-x-5..eqn3$

We put equation (3) into equation (1) to obtain,

$-x-5-x=9$

We group like terms to get,

$-x-x=9+5$

This implies that,

$-2x=14$

We divide through by $-2$ to get,

$x=-7$

Hence the x-coordinate is $-7$

<u>QUESTION 3</u>

The given system is

$y=3x-1..eqn1$

and

$x-y=-9...eqn2$

We make $x$ the subject in equation (2) to get,

$x=y-9...eqn3$

We put equation (3) into equation (1) to obtain,

$y=3(y-9)-1$

We expand the bracket to get,

$y=3y-27-1$

Group like terms to get,

$y-3y=-27-1$

We simplify to get;

$-2y=-28$

This implies that,

$y=14$

Therefore the y-coordinate is 14.

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3 years ago