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omeli [17]
3 years ago
9

Evaluate the expression. ( – 5+ – 9)× – 3÷ – 2

Mathematics
2 answers:
gayaneshka [121]3 years ago
6 0

Answer:

-21.

Step-by-step explanation:

( – 5+ – 9)× – 3÷ – 2

Using order of operations we work out the parentheses first:

= (-5 - 9) * -3 / 2

= -14 * -3 / -2  Now the multiplication followed by the division:

= 42 / -2

= -21.

Ronch [10]3 years ago
3 0

Answer:

-21

Step-by-step explanation:

( - 5+ - 9) \times - 3 \div - 2

Solve the brackets first.

( - 14) \times - 3 \div - 2

Division comes next.

( - 14) \times - 3/-2

Cancel the negative signs.

- 14 \times 3/2

Multiply the terms.

-42/2

=-21

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Step-by-step explanation:

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5 0
2 years ago
13 + what = 23.6? I need help i will give u brainlest if you help plz
Bas_tet [7]

Answer:

10.6

Step-by-step explanation:

13+10.6= is 23.6

Hope this helps

3 0
3 years ago
Read 2 more answers
What is the equation of a line that passes through (1,5) and (3,11)? Show work please.
jekas [21]
First, find the gradient/slope:
Use slope formula:
m=(y2-y1)/(x2-x1)
=(11-5)/(3-1)
=6/2
=3
Then use the line equation formula:
y=mx+c
You can substitute (1,5) if you like, also must substitute the slope as well!
5=3x1+c
c=2
Then find the full equation, which gives you the answer:
y=3x+2
3 0
3 years ago
Help w this also! please<3
jolli1 [7]

Answer:

equation for line:  y = 5/3x -7/3

Step-by-step explanation:

The equation for a linear line is y = mx + c. m is the gradient of the line, also known as slope. So, m = 5/3.

Next we need to find c. Since we know a point the line must intersect, we can sub this into our line equation to get an answer:

(2,1) x=2,y=1

1 = 5/3*2 + c

1 = 10/3 + c

1 - 10/3 = c

3/3 - 10/3 = c

-7/3 = c

The final eqaution of the line is: y = 5/3x -7/3

7 0
2 years ago
Lines a and b are parallel. The slope of line b is 1/3. What is the slope of line a?
Viefleur [7K]

Answer:

1/3

Step-by-step explanation:

Parallel lines have the same slope.

Therefore, if the slope of line b is 1/3, the slope of line a is also 1/3

8 0
2 years ago
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