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3 years ago

2. Write a problem that matches: Bob has 20 pogs. He keeps 4 for himself and then equally divides the rest among his 4 brothers.

Mathematics

2 answers:

dlinn [17]3 years ago

7 0

Answer:

20 minus 4 equals 16. 16 divided by 4 is 4

Bob gives 4 stickers to each of his brothers

I just noticed that you said PEMDAS sorry :|

madreJ [45]3 years ago

4 0

Bob gives his brother 4

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Ryan wants to buy a new guitar that costs $325.00. He already has $75 in his account and makes $20 per day. How many days will h

san4es73 [151]

Answer:I pretty sure it will be 12.5
I subtracted the total cost for the guitar by the money he already had and got 250
Then I did 250/20 and got 12.5
250 is the money that he still needs for the guitar

Hope it helps

5 0

2 years ago

Are the ratios 2:3 and 8:12 equivalent? Justify you answer.

Lunna [17]

Answer:Yes

Step-by-step explanation:You are multiplying the ratios by 4

3 0

3 years ago

Simplifly the expression 64^7/12

Evgesh-ka [11]

Answer:

$8\sqrt{2}$

Step-by-step explanation:

we have

$64^{\frac{7}{12}}$

we know that

$64=2^{6}$

substitute

$64^{\frac{7}{12}}=(2^{6})^{\frac{7}{12}}\\ \\=(2)^{\frac{6*7}{12}} \\ \\=(2)^{\frac{7}{2}} \\ \\=\sqrt{2^{7}}\\ \\=2^{3}\sqrt{2} \\ \\=8\sqrt{2}$

7 0

3 years ago

(12x + y + z = 26

mel-nik [20]

Option D. D has the matrix of constants [[12], [11], [4]].

Step-by-step explanation:

Step 1:

With the given equations, we can form matrices to represent them.

The coefficients of x, y, and z form a matrix of order 3 ×3, the variables x, y, and z form a matrix of order 1 ×3 and the constants form a matrix of order 1 ×3.

Step 2:

The linear system A is represented as

$\left[\begin{array}{ccc}12&1&1\\1&-11&0\\1&-1&4\end{array}\right]$ $\left[\begin{array}{ccc}x\\y\\z\end{array}\right]$ $= \left[\begin{array}{ccc}26\\17\\23\end{array}\right]$ .

Step 3:

The linear system B is represented as

$\left[\begin{array}{ccc}4&1&1\\1&-11&0\\1&-1&12\end{array}\right]$ $\left[\begin{array}{ccc}x\\y\\z\end{array}\right]$ $= \left[\begin{array}{ccc}23\\17\\26\end{array}\right]$ .

Step 4:

The linear system C is represented as

$\left[\begin{array}{ccc}1&1&1\\1&-1&0\\1&-1&1\end{array}\right]$ $\left[\begin{array}{ccc}x\\y\\z\end{array}\right]$ $= \left[\begin{array}{ccc}4\\11\\12\end{array}\right]$ .

Step 5:

The linear system D is represented as

$\left[\begin{array}{ccc}1&1&1\\1&-1&0\\1&-1&1\end{array}\right]$ $\left[\begin{array}{ccc}x\\y\\z\end{array}\right]$ $= \left[\begin{array}{ccc}12\\11\\4\end{array}\right]$ .

Step 6:

Of the four options, the linear system D has the matrix of constants [[12], [11], [4]]. So the answer is option D. D.

4 0

3 years ago

Is there a multiple of 3? And why not or why yes?

rodikova [14]

Yes there is because u can multiply it by different numbers except for 1 and 0

6 0

3 years ago

Read 2 more answers