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Nadusha1986 [10]
3 years ago
15

James baked 5 1/4 dozen peanut butter cookies and 4 5/6 dozen oatmeal cookies for a bake sale what was the number of dozen cooki

es he baked
Mathematics
1 answer:
Doss [256]3 years ago
3 0

Answer:

10\frac{1}{12} dozen

Step-by-step explanation:

\frac{1}{4} × 3 = \frac{3}{12}

\frac{5}{6} × 2 = \frac{10}{12}

\frac{3}{12}  + \frac{10}{12}  = \frac{13}{12}

\frac{13}{12} is equal to 1\frac{1}{12}

5 + 4 = 9 + 1\frac{1}{12} = 10\frac{1}{12} dozen


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ABC IS A TRINAGLE <br>WORK OUT ANGLE X <br>GIVE YOUR ANSWER CORRECT TO 3 S.F​
kati45 [8]

Answer:

2 {2 =  | \leqslant 3 > 3 |2 >  \times 2 \frac{?}{ \tan( \sec(\pi\% log_{?}( \beta ) ) ) } | | }^{?}

7=9/8ths

5 0
2 years ago
If the balance at the end of eight years on an investment of $630 that has been invested at a rate of 9% is $1,083.60, how much
Contact [7]
The interest is $453.60.

The total amount includes principal + interest.  The principal was 630 and the total amount was 1083.60.
1083.60 - 630 = 453.60.
3 0
3 years ago
Give work for equations following:<br><br> -x+2y&lt;-4<br> 2x+3y&gt;6<br><br> thanks sooooo much!!!!
Schach [20]

Answer: See below

Step-by-step explanation:

A) -x+2y < -4

-x+2y-2y < -4-2y

-x < -4-2y

Multiply both sides by -1 and reverse the inequality:

\left(-x\right)\left(-1\right) > -4\left(-1\right)-2y\left(-1\right)

x > 4+2y

B) 2x+3y > 6

2x+3y-3y > 6-3y

2x > 6-3y

\frac{2x}{2} > \frac{6}{2}-\frac{3y}{2}

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3 0
2 years ago
Rita had $9.00. She spent $3.67 on a magazine.
Ket [755]
She has $5.33 left


$9.00
-$3.67
————
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8 0
3 years ago
Read 2 more answers
Someone know how to do this?: 6x2+18x=0
Crazy boy [7]
The answer to the question is: Yes.

Explanation:  It can be solved by following exactly the same procedures
that solved the other two questions that you posted within the past 15 minutes.

Since I'm here already, here is the solution to this one:

6x² + 18x = 0

Divide each side of the equation by 6 :

x² + 3x = 0

Factor the left side:

x(x+ 3) = 0

This statement is true if x=0 or if x=3.

The solution to this one is even the same as the solution to the last one you posted.
You really ought to take a break, go back, review the solutions you've been given,
and try to solve a few on your own.
4 0
3 years ago
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