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swat32
3 years ago
8

3

Mathematics
2 answers:
Gnoma [55]3 years ago
6 0

Answer:

-35 - 5

Step-by-step explanation:

2 + 12 = 14, which is positive

(-3) * (-8) = 24, which is positive

10 - (-18) = 10 + 18 = 28, which is positive

-35 - 5 = -40, which is negative

frozen [14]3 years ago
3 0

Answer:

-35-5

Step-by-step explanation:

2+12=14

The result is a positive value.

-3 \times -8=24

The result is a positive value.

10-(-8)\\10+8=18

The result is a positive value.

-35-5=-40

The result is a negative value.

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A circular disc has a diameter of 24 centimeters. What is the area of the disc?
Degger [83]

D) 144(π) cm2

See the image I have shared

4 0
3 years ago
Melissa and Joey enjoyed dinner at a restaurant. They paid the waiter $100.50 plus a 20% tip. How much did each person pay if th
babymother [125]
Okay, first we have to find the total amount paid:
10050 + 0.2(10050)
12060
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8 0
3 years ago
U-5=-8 is tell solution of equation ​
ivanzaharov [21]

Answer:

u = -3

Step-by-step explanation:

Solve for u:

u - 5 = -8

Hint: | Isolate terms with u to the left hand side.

Add 5 to both sides:

u + (5 - 5) = 5 - 8

Hint: | Look for the difference of two identical terms.

5 - 5 = 0:

u = 5 - 8

Hint: | Evaluate 5 - 8.

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6 0
3 years ago
Read 2 more answers
Please help! What is the function for the graph
Evgen [1.6K]

When x = 0 the value of the function is 2 so its either A or B.

When x = 1, y = 8 so substituting these values in A we get:-

8 = 2 * 8^1 = 16 so

its Not A. Try B:-

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7 0
3 years ago
4(2x-3)=0.2(x+5)÷5.72
rusak2 [61]

Answer:

x=1.528534

Explanation:

Simplify both sides of equation

4(2x-3)=0.2(x+5)/5.72

(4)(2x)+(4)(−3)=0.034965x+0.174825(Distribute)

8x+−12=0.034965x+0.174825

8x−12=0.034965x+0.174825

Step 2: Subtract 0.034965x from both sides.

8x−12−0.034965x=0.034965x+0.174825−0.034965x

7.965035x−12=0.174825

Step 3: Add 12 to both sides.

7.965035x−12+12=0.174825+12

7.965035x=12.174825

Step 4: Divide both sides by 7.965035.

7.965035x/7.965035=12.174825/7.965035

x=1.528534

4 0
3 years ago
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