All categories

3 years ago

Mrs.jackson bought 7 dozen eggs if each student is given the same out of eggs how many eggs are left

1 answer:

3 0

Without knowing the number of students she has, you will not be able to answer the question correctly. There are 12 eggs in one dozen. 12 x 7 dozen = 84 eggs. You still need to know the number of students.

You might be interested in

A garden table and a bench cost S846 combined.The garden table costs $54 less than the bench.What is the cost of the bench?

Aneli [31]

Answer:

The bench costs 450

Step-by-step explanation:

Let t = cost of the table

b = cost of bench

t+b = 846

t = b-54

t+b =846

Substituting t = b-54 into the above equation

b-54 +b =846

Combine like terms

2b -54 = 846

Add 54 to each side

2b-54+54 = 846 +54

2b =900

Divide by 2

2b/2 = 900/2

b =450

The bench costs 450

4 0

2 years ago

Read 2 more answers

There were 407 boys and 438 girls who like to join the Independence Day parade. How many buses will be hired if 65 persons can b

AURORKA [14]

Answer:

Step-by-step explanation:

just add them both and divide by 65

4 0

2 years ago

Read 2 more answers

Consider triangle ABC. The legs have a length of 5 units each. What is the length of the hypotenuse of the triangle?

Korvikt [17]

Answer:

$5\sqrt{2}$ units.

Step-by-step explanation:

We have been given that two legs of triangle ABC measures 5 units each. We are asked to find the length of the hypotenuse.

We will use Pythagoras theorem to solve for the hypotenuse of triangle ABC.

$\text{Hypotenuse}^2=\text{Leg}^2+\text{Leg}^2$

Substituting our given values in above equation we will get,

$\text{Hypotenuse}^2=5^2+5^2$

$\text{Hypotenuse}^2=25+25$

$\text{Hypotenuse}^2=50$

Upon taking square root of both sides of our equation we will get,

$\text{Hypotenuse}=\sqrt{50}$

$\text{Hypotenuse}=\sqrt{25*2}$

$\text{Hypotenuse}=5\sqrt{2}$

Therefore, the length of hypotenuse of triangle ABC is $5\sqrt{2}$ units.

6 0

3 years ago

Read 2 more answers

What is the answer to K/2+1-1k=-2k

Mazyrski [523]

$\dfrac{k}{2}+1-1k=-2k\\\\\dfrac{1}{2}k+1-k=-2k\\\\\left(\dfrac{1}{2}k-k\right)+1=-2k\\\\-\dfrac{1}{2}k+1=-2k\qquad\text{subtract 1 from both sides}\\\\-\dfrac{1}{2}k=-2k-1\qquad\text{add 2k to both sides}\\\\1\dfrac{1}{2}k=-1\qquad\text{convert the mixed number to improper fraction}\\\\\dfrac{1\cdot2+1}{2}k=-1\\\\\dfrac{3}{2}k=-1\qqua\text{multiply both sides by}\ \dfrac{2}{3}\\\\\boxed{k=-\dfrac{2}{3}}$

4 0

2 years ago

pleaseee help context is down below $3 is the entrance fee while $.50 is for each game played that's enough evidence to solve C(

BabaBlast [244]

C= Cost
g= # of times games played

C = 0.50g + 3

C(0) = 0.50(0) + 3
C(0) = 0 + 3
C(0) = 3

This means that no games have been played yet and the entrance fee is $3

5 0

2 years ago

Read 2 more answers