1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Digiron [165]
3 years ago
13

Find the value of x...

Mathematics
1 answer:
Hoochie [10]3 years ago
6 0

Answer:

your answer is 22.67

MAY BE

Step-by-step explanation:

hey, could uh please add me on SC..

xMOHITx its my SC

You might be interested in
What does 7/3 x 19/6 equal
adelina 88 [10]

Answer:

7.38

cheak it i dont trust myself

5 0
3 years ago
Read 2 more answers
Yuri descended 450 feet down a
skelet666 [1.2K]
The change is 150 because you would divide 450/3
8 0
3 years ago
Multiply. Write your answer as a fraction in simplest form.<br> 5/2*3/10
kherson [118]
(5/2)(3/10)=(5*3)/(2*10)=15/20=3/4

Or

(5/2)(3/10)= (5)(1/2)(3)(1/10)=(5)(3)(1/2)(1/10) =(5*3)/(2*10)=15/20=3/4

Or

(5/2)(3/10)=(5)(0.5)(3)(0.1)=(2.5)(3)(0.1)=(7.5)(0.1)=.75=3/4

All of these work, but the first one is simplest
7 0
3 years ago
What is 4x4= with this problem
barxatty [35]
Hewo

Answer:

16

Explanation:

4*4 is 16
5 0
3 years ago
Read 2 more answers
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
Other questions:
  • Melania walked her dog 2 1/5 miles Cathy walked her dog 1 3/4 times as far as Melania. How many more miles did Cathy walk her do
    14·1 answer
  • Which of the following is a true statement about weight loss?
    11·2 answers
  • The product of two positive numbers is 245. determine the two numbers if one number is 5 times the other.
    10·1 answer
  • If f(1) =160 and f(n+1 )=-2 f(n) what is f(4)
    11·1 answer
  • A 16 foot pole is supported by two wires that extend from the top of the pole to points that are each 12 feet from the base of t
    14·1 answer
  • Do you not want to buy a car. Her parents loan her $5000 for five years at 5% simple interest how much will Jonah pay in interes
    6·1 answer
  • Write the equation of a line that is perpendicular to the given line and that passes through the given point.
    8·1 answer
  • Calculate the product of 8/15, 6/5, and 1/3.<br>B. 16:15<br>D. 16/75<br>A. 48<br>0115​
    5·1 answer
  • Are the polynomial functions graphed below even or odd?
    5·1 answer
  • Write a linear equation in Slope-Intercept Form that is parallel to a line containing the points: (2,7) and (4,11)
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!