Ask question

Ask question

All categories

2 years ago

6

A student has an allowance of $20 a week. He wants to buy lunch at school for $3.50 everyday. Does this student have enough mone

y to buy lunch for the week?

1 answer:

Alex787 [66]2 years ago

8 0

Answer:

no

Step-by-step explanation:

in a 7 day week, he would need $24.50

You might be interested in

At cycle and sport, customers can rents bikes, electric scooters, kayaks, canoes, or paddle boards by the hour. Mr Thompson is r

siniylev [52]

Answer: is 18k+22p-10!

Step-by-step explanation:

I hope it helps UvU

6 0

2 years ago

Read 2 more answers

Mr. Johnson currently has a square garden. It is in his garden and into a range of 5 feet shorter than three times shorter than

Vlad [161]

Answer:

The Dimension of new garden is $25 \ feet\ \times 10\ feet.$

Step-by-step explanation:

Given:

Perimeter of new garden = 70 feet.

Let the length of the new garden be 'l'.

Also Let the width of the new garden be 'w'.

We need to find the dimension of new garden.

Now Given:

Length is 5 feet shorter than three times it width.

framing the equation we get;

$l =3w-5 \ \ \ \ equation\ 1$

Now we know that;

Perimeter of rectangle is equal to twice the sum of length and width.

framing in equation form we get;

$2(l+w)=70$

Now Diving both side by 2 using Division property of equality we get;

$\frac{2(l+w)}2=\frac{70}{2}\\\\l+w =35$

Now Substituting equation 1 in above equation we get;

$3w-5+w=35\\\\4w-5=35$

Adding both side by 5 Using Addition Property of equality we get'

$4w-5+5=35+5\\\\4w=40$

Now Diving both side by 4 using Division property of equality we get;

$\frac{4w}{4}=\frac{40}{4}\\\\w=10\ ft$

Now Substituting the value of 'w' in equation 1 we get;

$l =3w-5\\\\l =3\times10-5\\\\l = 30-5\\\\l= 25\ ft$

Hence The Dimension of new garden is $25 \ feet\ \times 10\ feet.$

3 0

2 years ago

The formula s = StartRoot StartFraction S A Over 6 EndFraction EndRoot gives the length of the side, s, of a cube with a surface

never [62]

Answer:

$2\sqrt{2}\ ft\ longer$

Step-by-step explanation:

<u>Area Of A Cube </u>

Suppose a cube with side length s, the area of one side is

$A_s=s^2$

Since the cube has 6 sides, the total area is

$A=6A_s=6s^2$

But if we have the area, we can solve the above formula for s to get

$A=6s^2$

$\displaystyle s=\sqrt{\frac{A}{6}}$

We have two different cubes with areas 1,200 square inches and 768 square inches. Let's compute their side lengths

$\displaystyle s_1=\sqrt{\frac{1,200}{6}}=\sqrt{200}$

$\displaystyle s_1=10\sqrt{2}\ ft$

$\displaystyle s_2=\sqrt{\frac{768}{6}}=\sqrt{128}$

$\displaystyle s_2=8\sqrt{2} ft$

The difference between them is

$10\sqrt{2}\ ft-8\sqrt{2}\ ft=2\sqrt{2}\ ft\approx 2.83\ ft$

The side of the cube with area 1,200 square inches is $2\sqrt{2}\ ft$ longer then the side of the cube with area 768 square inches

7 0

2 years ago

Read 2 more answers

A motorist traveled to a distant city in 2 hours and returned by another route that is 30 miles long. On the return trip, she tr

IRISSAK [1]

Answer:

180 miles

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

For each system of equations, drag the true statement about its solution set to the box under the system?

natta225 [31]

Answer:

y = 4x + 2

y = 2(2x - 1)

Zero solutions.

4x + 2 can never be equal to 4x - 2

y = 3x - 4

y = 2x + 2

One solution

3x - 4 = 2x + 2 has one solution

Step-by-step explanation:

* Lets explain how to solve the problem

- The system of equation has zero number of solution if the coefficients

of x and y are the same and the numerical terms are different

- The system of equation has infinity many solutions if the

coefficients of x and y are the same and the numerical terms

are the same

- The system of equation has one solution if at least one of the

coefficient of x and y are different

* Lets solve the problem

∵ y = 4x + 2 ⇒ (1)

∵ y = 2(2x - 1) ⇒ (2)

- Lets simplify equation (2) by multiplying the bracket by 2

∴ y = 4x - 2

- The two equations have same coefficient of y and x and different

numerical terms

∴ They have zero equation

y = 4x + 2

y = 2(2x - 1)

Zero solutions.

4x + 2 can never be equal to 4x - 2

∵ y = 3x - 4 ⇒ (1)

∵ y = 2x + 2 ⇒ (2)

- The coefficients of x and y are different, then there is one solution

- Equate equations (1) and (2)

∴ 3x - 4 = 2x + 2

- Subtract 2x from both sides

∴ x - 4 = 2

- Add 4 to both sides

∴ x = 6

- Substitute the value of x in equation (1) or (2) to find y

∴ y = 2(6) + 2

∴ y = 12 + 2 = 14

∴ y = 14

∴ The solution is (6 , 14)

y = 3x - 4

y = 2x + 2

One solution

3x - 4 = 2x + 2 has one solution

3 0

2 years ago