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deff fn [24]
3 years ago
12

Please help me with this! ASAP

Mathematics
1 answer:
kobusy [5.1K]3 years ago
4 0

The question gives you 4 different numbers that could be x, so you have to replace each x in each problem and solve. If the two answers are equivalent, that is your answer. For the first one, 6 times 2 is 12, minus 2 is 10. 5 times 2 is 10, plus 3 is 13. 10 is not equal to 13, so that is not the answer. For 5, 6 times 5 is 30, minus 2 is 28. 5 times 5 is 25, plus 3 is 28, so therefore the answer is x=5

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-6x - 10 y=-12<br> -12x - 20 y=-24
EleoNora [17]

Answer:

minus x minus y and you get your answer

7 0
2 years ago
if the lesser of two consecutive even integers is seven more than half the greater, what are the two integers?
beks73 [17]

Answer: 16 and 18

Step-by-step explanation:

Let the lesser of two consecutive even integers is x.

Hence, the bigger of two consecutive even integers is (x+2).

\displaystyle\\x-\frac{x+2}{2}=7 \\

Multiply both parts of the equation by 2:

x(2)-(x+2)=7(2)\\2x-x-2=14\\x-2=14\\x-2+2=14+2\\x=16\\x+2=16+2\\x+2=18

7 0
1 year ago
Help me please please
lilavasa [31]

Answer triangle RST is congruent to the triangle LMK

If you move the triangle vertically you can find the congruency of two sides and one angle

5 0
2 years ago
What number is ten less than 658?
Karo-lina-s [1.5K]

Answer:

658-10=648

Step-by-step explanation:

subtract 10 from 658.

6 0
2 years ago
Find the other endpoint of the line segment with the given endpoint and midpoint. Endpoint: (9,8) Midpoint (0,9)​
shepuryov [24]

Answer:

(-9, 10)

Step-by-step explanation:

The location of the midpoint of a line with endpoint at (x_1,y_1) and (x_2,y_2) is given as (x, y). The location of x and y are:

x = \frac{x_1+x_2}{2},y=\frac{y_1+y_2}{2}

Given the endpoint (9,8) and Midpoint (0,9), the location of the other endpoint can be gotten from:

0=\frac{9+x_2}{2}\\ \\9+x_2=0\\\\x_2=-9\\\\Also,9=\frac{8+y_2}{2}\\ \\8+y_2=18\\\\y_2=18-8\\\\y_2=10

Hence the endpoint is at (x2, y2) which is at (-9, 10)

5 0
3 years ago
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