Someone help me with this plz

Julie has three boxes of pens. The diagram shows expressions for the number of pens in each box. Look at these equations.

vovangra [49]

Answer:

a = c + 16

Step-by-step explanation:

a = b + 12

b = c + 4

Solve for b in the first equation.

a = b + 12

a - 12 = b

Put b as a - 12 in the second equation and solve for a.

a - 12 = c + 4

a = c + 4 + 12

a = c + 16

4 0

4 years ago

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what is the answer for this question the volume of a cube depends on the length of its sides. this can be written in function no

AVprozaik [17]

It means that when you have s=4, or the length of one of the sides of the cube is 4, then the volume is 64 cubic units

8 0

4 years ago

Tom baked 14 cookies with 2 scoops of flour with 11 scoops of flour how many cookies can tom bake

vodomira [7]

Answer:77

Step-by-step explanation:

4 0

3 years ago

Please help answer question above.

aliya0001 [1]

That's so easy sorry for saying that The answer is 333.8

8 0

3 years ago

Will mark Brainlest please answer. find the value of a,b. <br>,p,q from the equal order pairs

NARA [144]

Step-by-step explanation:

<h3>Question-1:</h3>

by order pair we obtain:

$\displaystyle \begin{cases} \displaystyle 3p = 2p - 1 \dots \dots i\\2q - p = 1 \dots \dots ii\end{cases}$

cancel 2p from the i equation to get a certain value of p:

$\displaystyle \begin{cases} \displaystyle p = - 1 \\2q - p = 1 \end{cases}$

now substitute the value of p to the second equation:

$\displaystyle \begin{cases} \displaystyle p = - 1 \\2q - ( - 1) = 1 \end{cases}$

simplify parentheses:

$\displaystyle \begin{cases} \displaystyle p = - 1 \\2q + 1= 1 \end{cases}$

cancel 1 from both sides:

$\displaystyle \begin{cases} \displaystyle p = - 1 \\2q = 0\end{cases}$

divide both sides by 2:

$\displaystyle \begin{cases} \displaystyle p = - 1 \\q = 0\end{cases}$

<h3>question-2:</h3>

by order pair we obtain:

$\displaystyle \begin{cases} \displaystyle 2x - y= 3 \dots \dots i\\3y= x + y \dots \dots ii\end{cases}$

cancel out y from the second equation:

$\displaystyle \begin{cases} \displaystyle 2x - y= 3 \dots \dots i\\ x = 2y \dots \dots ii\end{cases}$

substitute the value of x to the first equation:

$\displaystyle \begin{cases} \displaystyle 2.2y-y= 3 \\ x = 2y \end{cases}$

simplify:

$\displaystyle \begin{cases} \displaystyle 3y= 3 \\ x = 2y \end{cases}$

divide both sides by 3:

$\displaystyle \begin{cases} \displaystyle y= 1 \\ x = 2y \end{cases}$

substitute the value of y to the second equation which yields:

$\displaystyle \begin{cases} \displaystyle y= 1 \\ x = 2 \end{cases}$

<h3>Question-3:</h3>

by order pair we obtain;

$\displaystyle \begin{cases} \displaystyle 2p + q = 2 \dots \dots i\\3q + 2p = 3 \dots \dots ii\end{cases}$

rearrange:

$\displaystyle \begin{cases} \displaystyle 2p + q = 2 \\2p + 3q= 3 \end{cases}$

subtract and simplify

$\displaystyle \begin{array}{ccc} \displaystyle 2p + q = 2 \\2p + 3q= 3 \\ \hline - 2q = - 1 \\ q = \dfrac{1}{2} \end{array}$

substitute the value of q to the first equation:

$\displaystyle 2.p+ \frac{1}{2} = 2$

make q the subject of the equation:

$\displaystyle p = \frac{3}{4}$

hence,

$\displaystyle q = \frac{1}{2} \\ p = \frac{3}{4}$

3 0

3 years ago

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