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2 years ago

This is the question

Mathematics

1 answer:

pishuonlain [190]2 years ago

4 0

Answer:

1st blank: Variable

2nd blank: one

Step-by-step explanation:

It should be:

To solve an equation, you must apply the variable to one side of the equation.

This makes sense to me, but I'm not 100% positive.

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Caleb bought groceries and paid \$1.60$1.60dollar sign, 1, point, 60 in sales tax. The sales tax rate is 2.5\%2.5%2, point, 5, p

saveliy_v [14]

Answer:

Price of Caleb's groceries before tax = $64

Step-by-step explanation:

Let the price of groceries before tax be =$ $x$

Sales tax charged = $1.60

Sales tax rate =2.5%

Sales tax charged in terms of $x$ will be = 2.5% of the Original price of grocery = $2.5\% of x = 0.025\times x=0.025x$

So, we have,

$0.025x=1.60$

Dividing both sides by $0.025$

$\frac{0.025x}{0.025}=\frac{1.60}{0.025}$

∴ $x=64$

∴ Price of Caleb's groceries before tax = $64

5 0

3 years ago

What is -5(7x + 2) =200

blondinia [14]

Your answer is going to be X=-6

3 0

4 years ago

Read 2 more answers

Find the inverse of the given matrix, if it exists.Aequals=left bracket Start 3 By 3 Matrix 1st Row 1st Column 1 2nd Column 0 3

BabaBlast [244]

Answer:

$A^{-1}=\left[ \begin{array}{ccc} \frac{1}{9} & \frac{4}{27} & - \frac{2}{27} \\\\ \frac{8}{9} & \frac{5}{27} & \frac{11}{27} \\\\ - \frac{4}{9} & \frac{2}{27} & - \frac{1}{27} \end{array} \right]$

Step-by-step explanation:

We want to find the inverse of $A=\left[ \begin{array}{ccc} 1 & 0 & -2 \\\\ 4 & 1 & 3 \\\\ -4 & 2 & 3 \end{array} \right]$

To find the inverse matrix, augment it with the identity matrix and perform row operations trying to make the identity matrix to the left. Then to the right will be inverse matrix.

So, augment the matrix with identity matrix:

$\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 4&1&3&0&1&0 \\\\ -4&2&3&0&0&1\end{array}\right]$

Subtract row 1 multiplied by 4 from row 2

$\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ -4&2&3&0&0&1\end{array}\right]$

Add row 1 multiplied by 4 to row 3

$\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&2&-5&4&0&1\end{array}\right]$

Subtract row 2 multiplied by 2 from row 3

$\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&0&-27&12&-2&1\end{array}\right]$

Divide row 3 by −27

$\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]$

Add row 3 multiplied by 2 to row 1

$\left[ \begin{array}{ccc|ccc}1&0&0&\frac{1}{9}&\frac{4}{27}&- \frac{2}{27} \\\\ 0&1&11&-4&1&0 \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]$

Subtract row 3 multiplied by 11 from row 2

$\left[ \begin{array}{ccc|ccc}1&0&0&\frac{1}{9}&\frac{4}{27}&- \frac{2}{27} \\\\ 0&1&0&\frac{8}{9}&\frac{5}{27}&\frac{11}{27} \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]$

As can be seen, we have obtained the identity matrix to the left. So, we are done.

6 0

3 years ago

What is the answer to 1 2/3 X 2 1/3

worty [1.4K]

10/9 or 1 1/9

I suggest using an app called Photomath for equations

4 0

3 years ago

Two distinct number cubes, one red and one blue, are rolled together. Each number cube has sides numbered 1 through 6.

algol [13]

Answer:

7/18

Step-by-step explanation:

First we need to find what values of sum we can have.

The minimum value for a cube is 1, so the minimum value for the sum is 2.

The maximum value for a cube is 6, so the maximum value for the sum is 12.

Now, we find the multiples of 6 and the multiples of 4 between 2 and 12:

4, 6, 8, 12.

To have a sum of 4, we can have the pairs:

(1,3), (2,2), (3,1)

To have a sum of 6, we can have the pairs:

(1,5), (2,4), (3,3), (4,2), (5,1)

To have a sum of 8, we can have the pairs:

(2,6), (3,5), (4,4), (5,3), (6,2)

To have a sum of 12, we can have the pairs:

(6,6)

So we have 14 pairs among the 36 (6 possibilities of each cube, so 6*6=36) total possibilities of pairs, so the probability is P = 14/36 = 7/18

5 0

3 years ago