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aleksley [76]
4 years ago
13

e There are nine water bottles in Devin’s refrigerator. He adds three full boxes of water bottles to the refrigerator. Then he a

dds two more boxes that each have 1 fewer bottle than a full box. When he is done, there are 67 bottles in the refrigerator. Write and solve an equation to find the number of bottles in a full box
Mathematics
1 answer:
Taya2010 [7]4 years ago
7 0

To solve this problem, we can write an equation letting the variable x represent the number of bottles in a box of water (this means that the expression x-1 represents the boxes that have one fewer bottle than a full box).

This allows us to create the following equation:

9 original bottles + 3x (3 full boxes) + 2(x-1) (two boxes with one fewer bottle) = 67 total bottles

OR

9 + 3x + 2(x-1) = 67

Now, we can begin to solve this equation by using the distributive property to get rid of the parentheses on the left side of the equation.

9 + 3x + 2x - 2 = 67

Next, we can combine like terms on the left side of the equation (meaning add the variable terms together and subtract the constant terms).

7 + 5x = 67

Now, we should subtract 7 from both sides of the equation to get the variable term alone on the left side of the equation.

5x = 60

Next, we should divide both sides by 5 to isolate the variable x on the left side of the equation.

x = 12

Therefore, your answer is that there are 12 bottles in a full box.

Hope this helps!

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There is a box of chocolates and one person easts one-fourth of the pieces and another person eats one-half
11111nata11111 [884]

Answer:

16 - Original number of chocolates in the box.

Step-by-step explanation:

C = 6 / (3/8) =6 x (8/3) =48 / 3 = 16

6 0
3 years ago
405.5 divided by 12.5. Step by step
lyudmila [28]

Answer:

= 32.440

Step-by-step explanation:

Change the divisor 12.5 to a whole number by moving the decimal point 1 places to the right. Then move the decimal point in the dividend the same, 1 places to the right.

We then have the equations:

4055 ÷ 125 = 32.440

and therefore:

405.5 ÷ 12.5 = 32.440

Both calculated to 3 decimal places.

   

3 0
3 years ago
Solve this system of equations using the addition method.
damaskus [11]
Eqt 1 = 4a-5b=7
eqt 2 = 4a+5b=17
eqt1+eqt2= 8a=24
a=3  then 4*3-5b=7 
so b=1
6 0
3 years ago
Use a matrix to find the coordinates of the endpoints or vertices of the image of each figure under the given reflection.
hjlf
The vertices of the original quadrilateral can be written in matrix form using the vertex matrix. The vertex matrix is
     \begin{pmatrix}-5&-1&-3&-7\\ 4&-1&-6&-3\end{pmatrix}

To find the coordinates of the endpoints or vertices of the image of the given coordinate points reflected about the y-axis, we just need to multiply the transformation matrix by the vertex matrix. The transformation matrix for this particular problem is 
     \begin{pmatrix}-1&0\\ 0&1\end{pmatrix}

Multiplying the two matrices, we have
     \begin{pmatrix}-1&0\\ 0&1\end{pmatrix}\begin{pmatrix}-5&-1&-3&-7\\ 4&-1&-6&-3\end{pmatrix}=\begin{pmatrix}5&1&3&7\\ 4&-1&-6&-3\end{pmatrix}

Therefore, the coordinates of the endpoints or vertices of the image are (5,4), (1,-1), (3, -6) and (7, -3).
     
7 0
3 years ago
Find the HCF of 2^120-1 and 2^100-1
zlopas [31]
There may be more brilliant solution than the following, but here are my thoughts.

We make use of Euclid's algorithm to help us out.
Consider finding the hcf of A=2^(n+x)-1 and B=2^(n)-1.

If we repeated subtract B from A until the difference C is less than B (smaller number), the hcf between A and B is the same as the hcf between B and C.

For example, we would subtract 2^x times B from A, or
C=A-2^xB=2^(n+x)-2^x(2^n-1)=2^(n+x)-2^(n+x)+2^n-1=2^n-1
By the Euclidean algorithm, 
hcf(A,B)=hcf(B,C)=hcf(2^n-1,2^x-1)
If n is a multiple of x, then by repetition, we will end up with
hcf(A,B)=hcf(2^x-1,2^x-1)=2^x-1

For the given example, n=100, x=20, so
HCF(2^120-1, 2^100-1)=2^(120-100)-1=2^20-1=1048575
(since n=6x, a multiple of x).

4 0
3 years ago
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