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lbvjy [14]
2 years ago
10

Write an inequality that represents the statement A number r is at most 11

Mathematics
1 answer:
Dmitrij [34]2 years ago
3 0

Answer:

a+r>11

Step-by-step explanation:

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Solve each equation.<br> Show all steps<br> 4) -8(-6-5k)=-232
Lubov Fominskaja [6]

the answer is k=4.6. hope this helps

6 0
3 years ago
I'll give 30 points and brianlist
Sedaia [141]

Answer:

80,10,66

Step-by-step explanation:

Hope this helps :)

6 0
3 years ago
A one day trip to new york, mary spent $6.25 for breakfast and 8.95 for lunch. She can spend no more than an average of $8.50. H
zysi [14]

Mary spent $10.30 on dinner to stay in budget.

Step-by-step explanation:

Amount spent on breakfast = $6.25

Amount spent on lunch = $8.95

Amount spent on dinner = x

Average = $8.50

Average = \frac{Sum\ of\ values}{No.\ of\ values}

8.50 = \frac{6.25+8.95+x}{3}\\8.50=\frac{15.20+x}{3}

Multiplying both sides by 3

8.50*3=\frac{15.20+x}{3}*3\\25.50=15.20+x\\25.50-15.20=x\\x=10.30

Mary spent $10.30 on dinner to stay in budget.

Keywords: Average, addition

Learn more about addition at:

  • brainly.com/question/4694425
  • brainly.com/question/4695279

#LearnwithBrainly

7 0
3 years ago
A. 16.0<br> b. 25.0<br> c. 9.0<br> d. 17.3
sveticcg [70]

Answer:

B 25.0 is the right answer

Step-by-step explanation:

8 0
3 years ago
Some body can help me with a geometric mean maze
Mars2501 [29]

Answer:

See explanation

Step-by-step explanation:

Theorem 1: The length of the altitude to the hypotenuse of a right triangle is the geometric mean of the lengths of the segments of the hypotenuse.

Theorem 2: The length of each leg of a right triangle is the geometric mean of the length of the hypotenuse and the length of the segment of the hypotenuse adjacent to that leg.

1. Start point: By the 1st theorem,

x^2=25\cdot (49-25)=25\cdot 24=5^2\cdot 2^2\cdot 6\Rightarrow x=5\cdot 2\cdot \sqrt{6}=10\sqrt{6}.

2. South-East point from the Start: By the 2nd theorem,

x^2=40\cdot (40+5)=4\cdot 5\cdot 2\cdot 9\cdot 5\Rightarrow x=2\cdot 5\cdot 3\cdot \sqrt{2}=30\sqrt{2}.

3. West point from the previous: By the 2nd theorem,

x^2=(32-20)\cdot 32=4\cdot 3\cdot 16\cdot 2\Rightarrow x=2\cdot 4\cdot \sqrt{6}=8\sqrt{6}.

4. West point from the previous: By the 1st theorem,

9^2=x\cdot 15\Rightarrow x=\dfrac{81}{15}=\dfrac{27}{5}=5.4.

5. West point from the previous: By the 2nd theorem,

10^2=8\cdot (8+x)\Rightarrow 8+x=12.5,\ x=4.5.

6. North point from the previous: By the 1st theorem,

x^2=48\cdot 6=6\cdot 4\cdot 2\cdot 6\Rightarrow x=6\cdot 2\cdot \sqrt{2}=12\sqrt{2}.

7. East point from the previous: By the 2nd theorem,

x^2=22.5\cdot 30=225\cdot 3\Rightarrow x=15\sqrt{3}.

8. North point from the previous: By the 1st theorem,

x^2=7.5\cdot 36=270\Rightarrow x=3\sqrt{30}.

8. West point from the previous: By the 2nd theorem,

x^2=12.5\cdot (12.5+13.5)=12.5\cdot 26=25\cdot 13\Rightarrow x=5\sqrt{13}.

9. North point from the previous: By the 1st theorem,

12^2=x\cdot 30\Rightarrow x=\dfrac{144}{30}=4.8.

101. East point from the previous: By the 1st theorem,

6^2=1.6\cdot (x-1.6)\Rightarrow x-1.6=22.5,\ x=24.1.

11. East point from the previous: By the 2nd theorem,

20^2=32\cdot (32-x)\Rightarrow 32-x=12.5,\ x=19.5.

12. South-east point from the previous: By the 2nd theorem,

18^2=x\cdot 21.6\Rightarrow x=15.

13. North point=The end.

6 0
3 years ago
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