All categories

3 years ago

It takes 3/4 of an hour to bake a cake. How long will it take to bake 6 cakes on after the other?

Mathematics

2 answers:

storchak [24]3 years ago

5 0

Answer:

4 1/2 hours

Step-by-step explanation:

To find how long, multiply the length of time in hours by the number of cakes. 1 cake takes 3/4 of a an hour so 6 cakes will take

3/4 * 6 = 18/4 = 4 2/4

4 2/4 simplifies to 4 1/2 hours.

almond37 [142]3 years ago

5 0

Answer:

4 1/2 because you need to multiply 3/4 by 6 and you can also add 3/4 to itself 6 times which will also give you 4 1/2

Step-by-step explanation:

I used a calculator and then checked it by doing it on paper

You might be interested in

How did the Punic Wars change the population of the cities of Rome?

PtichkaEL [24]

People moved to the cities to escape war. Some people had left the countryside to work in the arms industry and some had left for Rome looking for subsistence. Rome became the most populous city in Europe and West Asia.

6 0

3 years ago

Tammy says the quotient of 793 ÷ 6 is 132 r1. Use multiplication to check Tammy's answer. × 6 + 1 = Tammy's answer is .

Troyanec [42]

To find the answer, we should multiply 132 by 6.

132*6=792

792+1=793

Tammy’s answer is correct.

7 0

3 years ago

Need an answer for this question

Natasha_Volkova [10]

Answer: f(-2)=6.
Explanation: Since -2 is less than 1 you plug it into the top equation for -(-2)+4 = 2+4 =6

5 0

3 years ago

Suppose y varies directly with x, and y = 8 when x = –6. What direct variation equation relates x and y? What is the value of y

Fantom [35]

Y=12 given -6+4=-2 just add 4 to x

6 0

4 years ago

Read 2 more answers

Consider the function below. f(x)= x^3 + 2x^2 - x - 2 plot the x and y intercepts of the function

dimulka [17.4K]

In the Figure below is shown the graph of this function. We have the following function:

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

The $y-intercept$ occurs when $x=0$ , so:

$f(0)=(0)^3+2(0)^2-(0)-2=-2$

Therefore, the $y-intercept$ is the given by the point:

$\boxed{(0,-2)}$

From the figure we have three $x-intercepts$ :

$\boxed{P_{1}(-2,0)} \\ \boxed{P_{2}(-1,0)} \\ \boxed{P_{3}(1,0)}$

So, the $x-intercepts$ occur when $y=0$ . Thus, proving this:

$f(x)=x^3+2x^2-x-2 \\ \\ For \ P_{1}:\\ If \ x=-2, \ y=(-2)^3+2(-2)^2-(-2)-2=0 \\ \\ For \ P_{2}:\\ If \ x=-1, \ y=(-1)^3+2(-1)^2-(-1)-2=0 \\ \\ For \ P_{3}:\\ If \ x=1, \ y=(1)^3+2(1)^2-(1)-2=0$

7 0

3 years ago

Read 2 more answers