Ask question

Ask question

All categories

3 years ago

6

Tara baked 6 1/2 dozen cookies she sold 3 2/6 dozen of cookies she made how many dozens of cookies does Tara have remaining

1 answer:

zaharov [31]3 years ago

4 0

Convert the mixed numbers into improper fractions.
6 1/2= 13/2
3 2/6= 20/6
Now, find a common denominator. In this case, the common denominator is 6.
13/2*3/3=39/6
20/6 stays as 20/6
Subtract both fractions.
39/6-20/6=19/6
19/6 can be written as 3 1/6.
Tara has 3 1/6 dozen of cookies remaining.

You might be interested in

Find the volume of the pyramid to the nearest hundredth please.

denis23 [38]

Answer:

234

Step-by-step explanation:

8 0

3 years ago

PLEASEEEE WILL GIVE BRAINLIEST TO FIRST RIGHT ANSWER!!!!!!!!!!

Zielflug [23.3K]

Answer:

i think x = -7, y = -6

Step-by-step explanation:

good luck

5 0

3 years ago

Higher order thinking. Write an addition and subtraction equation to match the problem then solve. Li have 14 crackers. Joe has

elena55 [62]

Answer:

8+x=14 x=6, 14-8=x x=6

Step-by-step explanation:

x is the answer and what you're solving. The two equations I wrote are the simplest.

:)

4 0

3 years ago

Read 2 more answers

Suppose the doormat has a design on it that divide it into squares measuring 1/10 meter by 1/10 meter how many squares is the do

amm1812

There are 10 squares and each square is a meter

3 0

3 years ago

Pls help asap, I put attachments down below, I will also give brainliest

lesya [120]

Answer:

Question 1;

∴ The y values are 1.49 , 2.98 , 4.47 and 5.96

Question 2;

∴ The y values are 2.00 , 2.8 , 3.6 and 4.4

Question 3;

∴ The y values are -9 , -3 , 3 and 9

Question 4;

∴ y values are $\frac{1}{2}$ , $\frac{3}{4}$ , 1 and $\frac{5}{4}$

Step-by-step explanation:

Question 1;

Given data,

Apples cost $\$1.49$ per pound.

$x$ = weight in pound

$y$ = Cost in $\$$

Here the relation between $x$ and $y$ comes;

$y=x\times 1.49$ (equation 1)

Let, Plug all x value in equation 1;

when $x=1$ then $y=1\times 1.49=1.49$ ,

when $x=2$ then $y=2\times 1.49=2.98$ ,

when $x=3$ then $y=3\times 1.49=4.47$ ,

when $x=4$ then $y=4\times 1.49=5.96$ ,

∴ The y values are 1.49 , 2.98 , 4.47 and 5.96

Question 2;

From question,

Initial fees of cab is $\$$ 2.00 and Passengers are charged $\$$ 0.80 for every minute.

Let,

$x$ = Time in minutes

$y$ = Amount in $\$$

Here the relation between x and y comes;

$y=x\times 0.80 +2.00$ (equation 2)

Let,By plug all x value in equation 2;

When $x=0$ then $y=0\times 0.80+2.00=2.00$ ,

When $x=1$ then $y=1\times 0.80+2.00=2.8$ ,

When $x=2$ then $y=2\times 0.80+2.00=3.6$ ,

When $x=3$ then $y=3\times 0.80+2.00=4.4$ ,

∴ The y values are 2.00 , 2.8 , 3.6 and 4.4

Question 3;

Given relation between $x$ and $y$ ;

$y=6\times x +3$

Plug all the values of x in above equation,

When $x=-2$ then $y=(6\times -2) +3=-9$ ,

When $x=-1$ then $y=(6\times --1) +3=-3$ ,

When $x=0$ then $y=(6\times 0) +3=3$ ,

When $x=1$ then $y=(6\times 1) +3=9$ ,

∴ The y values are -9 , -3 , 3 and 9

Question 4;

Given relation between $x$ and $y$ ;

$y=(\frac{1}{4} \times x)+1$

Plug all the values of x in above equation,

When $x=-2$ then $y=(\frac{1}{4} \times -2)+1=\frac{1}{2}$ ,

When $x=-1$ then $y=(\frac{1}{4} \times -1)+1=\frac{3}{4}$ ,

When $x=0$ then $y=(\frac{1}{4} \times 0)+1=1$ ,

When $x=1$ then $y=(\frac{1}{4} \times 1)+1=\frac{5}{4}$ ,

∴ y values are $\frac{1}{2}$ , $\frac{3}{4}$ , 1 and $\frac{5}{4}$

7 0

4 years ago