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
shepuryov [24]
3 years ago
6

(10+2-8)2-5 Evaluate if necessary.

Mathematics
1 answer:
bixtya [17]3 years ago
4 0
Answer is 3. You have to do what’s in the parentheses first: 10+2-8=4. Then multiply 4 • 2 = 8 and then finally subtract: 8-5=3
You might be interested in
Evaluate the expression. Identify the property used in each step.
borishaifa [10]

Answer:

3/4 1/2    =3/5

Step-by-step explanation:

6 0
3 years ago
Each classmates contributes $2 for charity Write an expression for the amount of money raised by your class
Alecsey [184]
$2*S=A
S=Student
A=Total amount
3 0
3 years ago
Read 2 more answers
Please help me please.
shtirl [24]

Answer:

The first one is 79

The second one is 26

I hope this helped! :D

Step-by-step explanation:

For the first one you just plug in 23 into x so it would be 3(23) + 10 = x

For the second one you just plug in 88 into g so it would be 88 = 3x + 10

6 0
3 years ago
How to work it out<br> And who drove farther and how many miles farther
OlgaM077 [116]
Easy, just divide
kelly=440mi/8hr=220mi/4hr=110mi/2hr=55mi/1hr
alberto=468mi/9hr=156mi/3hr=52mi/1hr

55>52

kelly drove 55mi in 1 hour
55-52=3
she drove 3 miles more  in 1 hour
5 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
Other questions:
  • The decimal 0.3(repeating) represents 1/3. what type of number best describes 0.9(repeating), which is 3 times 0.9(repeating)? e
    8·2 answers
  • The height of a batted ball is modeled by the function h=-0.01x^2+1.22x+3, where x is the horizontal distance in feet from the p
    6·1 answer
  • 51 is the product of diego’s score and 3<br><br> Translate into an equation
    6·2 answers
  • It took Amir 2 hours to hike 5 miles. On the first part of the hike, Amir averaged 3 miles per hour. For the second part of the
    15·1 answer
  • I need the answer ill mark brainliest
    7·1 answer
  • Mathhh helpp me plzzz
    8·1 answer
  • I'm having a math test tomorrow, what the best way to study?
    7·2 answers
  • Just help me again for 2 times
    15·1 answer
  • Please help. Not sure I know what I am doing
    13·1 answer
  • NEED HELP ASAP The salaries of Nathan and Trisha are in the ratio of 5:9. Trisha's salary is £16800 more than Nathan's. Find Nat
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!