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Phoenix [80]
3 years ago
12

Help me plzz plzzz plese2​

Mathematics
1 answer:
Dominik [7]3 years ago
7 0

Answer:

122m^2

Step-by-step explanation:

Awnser is in picture along with steps

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Can anybody help me with my homework please. and explain me how to got the answer. Thanks
natka813 [3]

In the problems A-F, You multiply each of the numbers by itself.  Like 9²=9×9=81

3³=3×3×3=27     6³=6×6×6=216 and so on for the other problems

G-I  is 36=6²  100=10² and so on the cube is the same except you multiply the number three times by itself.  I hope this helped a bit. Just ask if you still don't understand.

8 0
3 years ago
Find two consecutive positive integers such that the sum of their squares is 421 .
maria [59]
We assume the two numbers are x and (x+1)
so x^2+(x+1)^2=421
x^2+(x^2+2x+1)=421
2x^2+2x-420=0
According to the formula of quadratic 
x=14 or-15
cuz we know the two numbers are integers
so x=14
therefore the other number is 15
To make sure that's correct
14^2+15^2=421

Hope that helps you!!


4 0
3 years ago
What is 3/4 of 16/21? worked out please
timofeeve [1]
Of means multiplication

So

\frac{3}{4} \sf{of}  \frac{16}{21}  =  \frac{3}{4} \times  \frac{16}{21}

And then simplify the numerator ad denominator

= \frac{3 \times 16}{4 \times 21} =  \frac{\ 1 \times 16}{4 \times 7}

And simplify the numerator and denominator again

= \frac{\ 1 \times 16}{4 \times 7} = \frac{\ 1 \times 4}{1 \times 7}

And we are finally left with the answer which is \frac{4}{7}
5 0
4 years ago
Which mixed number is equivalent to the improper fraction 42/5
a_sh-v [17]

Answer:

The answer is C 8 2/5 im pretty sure.

5 0
3 years ago
Read 2 more answers
The diagram shows the sector of a circle with the centre O and radius 6cm.
cricket20 [7]

Answer:

Area of the shaded region = 1.92 cm²

Step-by-step explanation:

From the picture attached,

Area of the shaded region = Area of the sector OMN - Area of the triangle OMN

Area of sector OMN = \frac{\theta}{360}(\pi r^{2})

Here, θ = Central angle of the sector

r = radius of the sector

Area of sector OMN = \frac{50}{360}(\pi )(6)^2

                                  = 15.708 square cm

Area of ΔOMN = 2(ΔOPN)

Area of ΔOPN = \frac{1}{2}(OP)(PN)

Area of ΔOMN = OP × PN

In ΔOPN,

sin(25°) = \frac{PN}{ON}

PN = ONsin(25°)

     = 6sin(25°)

     = 2.536 cm²

cos(25°) = \frac{OP}{ON}

OP = ONcos(25°)

OP = 6cos(25°)

OP = 5.438 cm

Area of ΔOMN = 2.536 × 5.438

                         = 13.791 cm²

Area of the shaded region = 15.708 - 13.791 = 1.917 cm²

                                             ≈ 1.92 cm²

3 0
3 years ago
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