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Gennadij [26K]
3 years ago
10

(3x-...)^(2)=9x^(2)-6x+1

Mathematics
1 answer:
kozerog [31]3 years ago
6 0

Answer:

You are missing a couple of numbers.

Step-by-step explanation:

Please provide more numbers or the right set of them,

-hamburger58

Thank you.

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Hunter's time for the 100-meter freestyle event was 57.9 seconds. His time for the second half of the race was 4.7 seconds slowe
Ratling [72]

Answer:

Step-by-step explanation:

Time for first race 57.9seconds

x=time for first half of race

x-4.7=second half of race

Equations is

x + (x-4.7)=57.9

2x-4.7=57.9

2x=57.9+4.7

2x=62.6

x=31.3

Check answer

31.3 + (31.3-4.7)=57.9

31.3 +26.6=57.9

57.9=57.9

6 0
4 years ago
Calculate the circumference of the circle. Use 3.14 for pi. I WILL MARK BRAINLIEST
Harrizon [31]

Answer:

The answer would be B

Step-by-step explanation:

C=2(pi)r

C=25.12

8 0
3 years ago
Read 2 more answers
Helppppppp i need help
Lunna [17]
Do you need the fraction? I don’t understand what you are asking
8 0
3 years ago
5 Read what Alicia says.
Maru [420]

Answer:

all the angles of the pentagon do not add up to 540

Step-by-step explanation:

total degree in a shape

180 × (n-2)          where n is the number of side

a pentagon has 5 sides so

180 × (5-2)

= 540

100+105+72+126+127= 530

all the angles of the pentagon do not add up to 540

4 0
1 year ago
What percentage of the semi circle is shaded? <br>​
mezya [45]

The percentage of the semicircle shaded section is approximately 23,606 %.

The percentage of the area of the semicircle is equal to the ratio of the semicircle area minus the half-cross area to the semicircle area. In other words, we have the following expression:

r = \left(\frac{A_{s}-A_{h}}{A_{s}} \right)\times 100\,\%

r = \left(1-\frac{A_{h}}{A_{s}} \right)\times 100\,\% (1)

Where:

  • A_{h} - Area of the half cross, in square centimeters.
  • A_{s} - Area of the semicircle, in square centimeters.
  • r - Percentage of the shaded section of the semicircle.

And the percentage of the shaded section is:

r = \left[1-\frac{4 \cdot (2\,cm)^{2}+4\cdot \left(\frac{1}{2} \right)\cdot (2\,cm)^{2}}{0.5\cdot \pi\cdot (16\,cm^{2}+4\,cm^{2})} \right]\times 100

r \approx 23.606\,\%

The percentage of the semicircle shaded section is approximately 23,606 %.

We kindly invite to check this question on percentages: brainly.com/question/15469506

3 0
3 years ago
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