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Reil [10]
3 years ago
13

20% of Desiree's math test problems consisted of word problems. 30% of Everett's math test problems consisted of word problems.

Which statement must be true?
Mathematics
1 answer:
Semmy [17]3 years ago
3 0
The first statement must be true (20% of Desiree's math..)
The first statement can be true while the second statement is false, but the first statement cannot be false while the second statement is true.
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3,14 x (200 - 175)2 - 150 Round your answer off to the nearest whole .​
RideAnS [48]

Answer:

7

Step-by-step explanation:

I assume that 3,14 is 3.14

=3.14 x (200-175)2 - 150

=3.14 x (25)2 - 150

=3.14 x 50 - 150

=157-150

=7

7 0
2 years ago
Find the distance CD rounded to the nearest tenth<br> C=(10,-1) and D=(-6,3)
Fynjy0 [20]

Answer:

CD = 16.5

Step-by-step explanation:

To find the distance between two points, use this formula: L = \sqrt{(x_{2} -x_{1})^{2}+(y_{2} -y_{1})^{2} }

point C can be info set 1: (10, -1)    x₁ = 10   y₁ = -1

point D can be info set 2: (-6, 3)    x₂ = -6  y₂ = 3

Substitute the information into the formula

L = \sqrt{(x_{2} -x_{1})^{2}+(y_{2} -y_{1})^{2} }

CD = \sqrt{(-6 -10)^{2}+(3 -(-1))^{2} }    Simplify inside each bracket

CD = \sqrt{(-16)^{2}+(4)^{2} }    Square the numbers

CD = \sqrt{256+16 }     Add inside the root

CD = \sqrt{272}    Enter into calculator

CD = 16.5  Rounded to the nearest tenth, the first decimal

The distance CD is 16.5.

4 0
3 years ago
The difference of c and 2 is greater than or equal to -22
Charra [1.4K]
You can write the question as a inequality.

c - 2 ≥ -22
add 2 on both sides...

c ≥ -20

Your answer is c ≥ -20.
7 0
3 years ago
-1(t+8)=-7.1 solve for t
Darina [25.2K]

Answer:

t = -0.9

Step-by-step explanation:

-1(t+8) = -7.1

(t+8) = 7.1

t = -0.9

7 0
3 years ago
Read 2 more answers
At 5pm the total rainfall is 3cm. At 11 pm the total rainfall is 13cm.What is the mean hourly rainfall?
krok68 [10]

Answer:

2.16 cm

Step-by-step explanation:

5 through 11 = 6 hours

total of cms  = 13

13/6 = 2.16

7 0
3 years ago
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