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True [87]
2 years ago
12

A number is multiplied by 6, then 8 is subtracted from the product. The result is

Mathematics
1 answer:
MrMuchimi2 years ago
6 0

Answer:

3 round to the nearest ten is 0

Step-by-step explanation:

let n be the number

(n × 6) - 8 = 10

6n= 10 + 8

6n = 18

6n ÷ 6 = 18 ÷ 6

n = 3

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I really need help someone please
Olegator [25]
The factorization of A is y = (x - 8)(x + 7).
The factorization of B is y = (x + 1)(x - 4)(x - 5)

In order to find these, you must first find where each graph crosses the x-axis. In the first problem it does so at 8 and -7. In order to find the correct parenthesis for those, you need to write it out as a statement and then solve for 0. 

x = 8 ---> subtract 8 from both sides
x - 8 = 0
This means we use (x -  8) in our factorization. 

You then need to repeat the process until you have all the pieces. In the second problem, there will be 3 instead of 2 since it crosses the axis 3 times. 
3 0
3 years ago
What are the measures of x y and z
Svetlanka [38]
Hahah what you learning...............
7 0
3 years ago
140 degrees decreased by 65%
BaLLatris [955]
Hey there! 
<span>Here are the steps involved: 
</span>
1.Change 65% into a decimal. Move the decimal two places to the left and you get 0.65.

2. Multiply 0.65 by 50.
<span>140 x 0.65 = 91.

3. Since you are decreasing 140 by 65%, subtract 140 by 91. 
140-91= 49


</span><span>(If you feel that my answer has helped you, please consider rating it and giving it a thank you! Also, feel free to make the best answer, the brainliest answer!)
</span>
<span>Thank you! :D</span>


7 0
3 years ago
PLEASE HELP ME me me me​
kykrilka [37]

Answer:

Andre has the correct answer. When simplified his answer is equivalent to the original equation.

7 0
3 years ago
Read 2 more answers
. Use the quadratic formula to solve each quadratic real equation. Round
Liono4ka [1.6K]

Answer:

A. No real solution

B. 5 and -1.5

C. 5.5

Step-by-step explanation:

The quadratic formula is:

\begin{array}{*{20}c} {\frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}} \end{array}, with a being the x² term, b being the x term, and c being the constant.

Let's solve for a.

\begin{array}{*{20}c} {\frac{{ 5 \pm \sqrt {5^2 - 4\cdot1\cdot11} }}{{2\cdot1}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 5 \pm \sqrt {25 - 44} }}{{2}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 5 \pm \sqrt {-19} }}{{2}}} \end{array}

We can't take the square root of a negative number, so A has no real solution.

Let's do B now.

\begin{array}{*{20}c} {\frac{{ 7 \pm \sqrt {7^2 - 4\cdot-2\cdot15} }}{{2\cdot-2}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 7 \pm \sqrt {49 + 120} }}{{-4}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 7 \pm \sqrt {169} }}{{-4}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 7 \pm 13 }}{{-4}}} \end{array}

\frac{7+13}{4} = 5\\\frac{7-13}{4}=-1.5

So B has two solutions of 5 and -1.5.

Now to C!

\begin{array}{*{20}c} {\frac{{ -(-44) \pm \sqrt {-44^2 - 4\cdot4\cdot121} }}{{2\cdot4}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 44 \pm \sqrt {1936 - 1936} }}{{8}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 44 \pm 0}}{{8}}} \end{array}

\frac{44}{8} = 5.5

So c has one solution: 5.5

Hope this helped (and I'm sorry I'm late!)

4 0
3 years ago
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