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shtirl [24]
2 years ago
10

16 + 1/2x = 21 ...................../

Mathematics
2 answers:
Ludmilka [50]2 years ago
7 0

Answer:

there is your answer

Step-by-step explanation:

-Dominant- [34]2 years ago
6 0

Answer: x=10

Step-by-step explanation:

Subtract 16 from both sides:

16+ \frac{1}{2}-16 =21 - 16

\frac{1}{2}x=5

Multiply both sides by 2:

2*\frac{1}{2}x=5*2

x=10

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Oliga [24]

Answer:

where B is the base area and h is the height. The base of the prism is a rectangle. The length of the rectangle is 9 cm and the width is 7 cm

Step-by-step explanation:

3 0
4 years ago
Pls help
Vanyuwa [196]

Answer:

24 cubic units

Step-by-step explanation:

The first prism has 20 cubic units and all you have to do is add 4 more cubic units and you will get 24 cubic units.

5 0
3 years ago
Aria studied the traffic trends in India. She found that the number of cars on the road increases by 10% each year. If there wer
denpristay [2]

Answer:

8,800,000cars

Step-by-step explanation:

Given the initial amount of cars = 80,000,000

If there is 10 % increase in the first year;

Increment = 10% of 80,000,000

Increment = 0.1 * 80,000,000

Increment = 8,000,000

Amount of car in year 2 = 80million + 8million = 88million

Year3;

Amount of car in year 3 = 88,000,000+(10%of88million)

Amount of car in year 3 = 88,000,000+8,800,000

Amount of car in year 3 =96,800,000

Differences in the number of cars = 96,800,000 - 88,000,000

Differences in the number of cars = 8,800,000cars

<em>Hence there are 8,800,000cars in years 3 than year 2</em>

7 0
3 years ago
Please help!!<br> Write a matrix representing the system of equations
frozen [14]

Answer:

(4, -1, 3)

Step-by-step explanation:

We have the system of equations:

\left\{        \begin{array}{ll}            x+2y+z =5 \\    2x-y+2z=15\\3x+y-z=8        \end{array}    \right.

We can convert this to a matrix. In order to convert a triple system of equations to matrix, we can use the following format:

\begin{bmatrix}x_1& y_1& z_1&c_1\\x_2 & y_2 & z_2&c_2\\x_3&y_2&z_3&c_3 \end{bmatrix}

Importantly, make sure the coefficients of each variable align vertically, and that each equation aligns horizontally.

In order to solve this matrix and the system, we will have to convert this to the reduced row-echelon form, namely:

\begin{bmatrix}1 & 0& 0&x\\0 & 1 & 0&y\\0&0&1&z \end{bmatrix}

Where the (x, y, z) is our solution set.

Reducing:

With our system, we will have the following matrix:

\begin{bmatrix}1 & 2& 1&5\\2 & -1 & 2&15\\3&1&-1&8 \end{bmatrix}

What we should begin by doing is too see how we can change each row to the reduced-form.

Notice that R₁ and R₂ are rather similar. In fact, we can cancel out the 1s in R₂. To do so, we can add R₂ to -2(R₁). This gives us:

\begin{bmatrix}1 & 2& 1&5\\2+(-2) & -1+(-4) & 2+(-2)&15+(-10) \\3&1&-1&8 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\0 & -5 & 0&5 \\3&1&-1&8 \end{bmatrix}

Now, we can multiply R₂ by -1/5. This yields:

\begin{bmatrix}1 & 2& 1&5\\ -\frac{1}{5}(0) & -\frac{1}{5}(-5) & -\frac{1}{5}(0)& -\frac{1}{5}(5) \\3&1&-1&8 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\3&1&-1&8 \end{bmatrix}

From here, we can eliminate the 3 in R₃ by adding it to -3(R₁). This yields:

\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\3+(-3)&1+(-6)&-1+(-3)&8+(-15) \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0&-5&-4&-7 \end{bmatrix}

We can eliminate the -5 in R₃ by adding 5(R₂). This yields:

\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0+(0)&-5+(5)&-4+(0)&-7+(-5) \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0&0&-4&-12 \end{bmatrix}

We can now reduce R₃ by multiply it by -1/4:

\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\ -\frac{1}{4}(0)&-\frac{1}{4}(0)&-\frac{1}{4}(-4)&-\frac{1}{4}(-12) \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}

Finally, we just have to reduce R₁. Let's eliminate the 2 first. We can do that by adding -2(R₂). So:

\begin{bmatrix}1+(0) & 2+(-2)& 1+(0)&5+(-(-2))\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 0& 1&7\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}

And finally, we can eliminate the second 1 by adding -(R₃):

\begin{bmatrix}1 +(0)& 0+(0)& 1+(-1)&7+(-3)\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 0& 0&4\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}

Therefore, our solution set is (4, -1, 3)

And we're done!

3 0
3 years ago
Which inequality has a solution that is the set of all real numbers?
kati45 [8]

Answer:

12<14

x<12

14>12

14>x

Step-by-step explanation:

I dont know if it is correct but I think it is that way try

6 0
3 years ago
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