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Vanyuwa [196]
3 years ago
15

5×[(12() 3)-(15-9)=150

Mathematics
1 answer:
coldgirl [10]3 years ago
3 0
X=13/14
I don’t know if you meant for the parentheses to be flipped so I flipped them normally (between 12 and 3)
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This answer ASAP.......
babymother [125]
Volume = 9 1/3 * 4 *  6 1/2 
=  28/3 * 4 * 13/2
=  728/3 =  242 2//3  in^3


6 0
3 years ago
What is 37/10 as a mixed number?
Dafna11 [192]
First, we need to find how many times 10 can fit into 37.

10*3 = 30

3 is the whole number. 

37-30 = 7

7 remaining, so its the numerator.

3  7/10

Answer:  3  \frac{7}{10}
3 0
3 years ago
Read 2 more answers
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Anna11 [10]

Answer:

x=\frac{50}{7} =7.1

Step-by-step explanation:

\frac{5}{7} =\frac{x}{10}

Cross multiply the proportion:

5*10=x*7

Solve for x:

7x=50\\ x=\frac{50}{7}

6 0
3 years ago
Read 2 more answers
Ms.Kitts works at a music store. Last week she sold 6 more than 3 times the number of CDs that she sold this week. Ms.Kitts sold
Murrr4er [49]

Answer:  \left \{ {{l=3t+6} \atop {l+t=110}} \right.

Step-by-step explanation:

Let be "l" the number of CDs Ms.Kitts sold last week, and "t" the number of CDs she sold this week.

You need to remember that:

1. The word "times" indicates a multiplication.

2. The word "more" indicates addition.

Then, the sentence "Last week she sold 6 more than 3 times the number of CDs that she sold this week" can be represented with equation:

l=3t+6

According to the information given in the exercise, the sum of the number of CDs Ms. Kitts sold last week and this week is: 110 CDs.

Since a sum is the result an Addition, you can write the following equation:

l+t=110

Therefore, the System of equations that can be used to find the number of CDs she sold last week and the number of CDs she sold this week, is:

\left \{ {{l=3t+6} \atop {l+t=110}} \right.

7 0
3 years ago
PLEASE HELP QWQ AsAp with these 4 questions
den301095 [7]

Answer:

Step-by-step explanation:

I can't believe I'm doing this for 5 points, but ok!

For the first 3, we are going to multiply to find the value of that 3 x 3 matrix by picking up the first 2 columns and plopping them down at the end and then multiplying through using the rules for multiplying matrices:

\left[\begin{array}{ccccc}7&4&6&7&4\\-4&8&9&-4&8\\1&8&7&1&8\end{array}\right]  and from there find the sum of the products of the main axes minus the sum of the products of the minor axes, as follows (I'm not going to state the process in the next 2 problems, so make sure you follow it here. This is called the determinate. The determinate is what you get when you evaluate or find the value of a matrix. Just so you know):

(7*8*7)+(4*9*1)+(6*-4*8)-[(1*8*6)+(8*9*7)+(7*-4*4)] which gives us:

392 + 36 - 192 - [48 + 504 - 112] which simplifies to

236 - 440 which is -204

On to the second one:

\left[\begin{array}{ccccc}-8&-4&-1&-8&-4\\1&7&-3&1&7\\8&9&9&8&9\end{array}\right] and multiplying gives us

(-8*7*9)+(-4*-3*8)+(-1*1*9)-[(8*7*-1)+(9*-3*-8)+(9*1*-4)] which gives us:

-504 + 96 - 9 - [-56 + 216 - 36] which simplifies to

-417 - 124 which is -541, choice c.

Now for the third one:

\left[\begin{array}{ccccc}-2&-2&-5&-2&-2\\2&7&-3&2&7\\8&9&9&8&9\end{array}\right] and multiplying gives us

(-2*7*9)+(-2*-3*8)+(-5*2*9)-[(8*7*-5)+(9*-3*-2)+(9*2*-2)] which gives us:

-126+48-90-[-280+54-36] which simplifies to

-168 - (-262) which is 94, choice c again.

Now for the last one. I'll show you the set up for the matrix equation; I solved it using the inverse matrix. So I'll also show you the inverse and how I found it.

\left[\begin{array}{cc}-4&-5&\\-6&-8\\\end{array}\right] \left[\begin{array}{c}x\\y\\\end{array}\right] = \left[\begin{array}{c}-5\\-2\\\end{array}\right] and I found the inverse of the 2 x 2 matrix on the left.

Find the inverse by:

* finding the determinate

* putting the determinate under a 1

* multiply that by the "mixed up matrix (you'll see...)

First things first, the determinate:

|A| = (-4*-8) - (-6*-5) which simplifies to

|A| = 32 - 30 so

|A| = 2; now put that under a 1 and multiply it by the mixed up matrix. The mixed up matrix is shown in the next step:

\frac{1}{2}\left[\begin{array}{cc}-8&5\\6&-4\end{array}\right]  (to get the mixed up matrix, swap the positions of the numbers on the main axis and then change the signs of the numbers on the minor axis). Now we multiply in the 1/2 to get the inverse:

\left[\begin{array}{cc}-4&\frac{5}{2}\\3&-2\\\end{array}\right] Multiply that inverse by both sides of the equation. This inverse "undoes" the matrix that's already there (like dividing the matrix that's already there by itself) which leaves us with just the matrix of x and y. Multiply the inverse matrix by the solution matrix:

\left[\begin{array}{c}x&y\end{array}\right] =\left[\begin{array}{cc}-4&\frac{5}{2} \\3&-2\end{array}\right] *\left[\begin{array}{c}-5&-2\\\end{array}\right] and that right side multiplies out to

x = 20 - 5 which is

x = 15 and

y = -15 + 4 which is

y = -11

(It works, I checked it)

7 0
3 years ago
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