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2 years ago

11

What is the area of a circle with 20cm diameter

2 answers:

lesya [120]2 years ago

7 0

Well that's a 10cm radius and the formula for area of a circle is A= 3.14•R(squared). So to plug that in, that would be 3.14 • 10 squared or 3.14 • 100. 3.14•100=314

neonofarm [45]2 years ago

4 0

Answer:

area of the circle

area=7.54

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Please help ASAP, thank you and have a good night/day❤️

wolverine [178]

Answer:

1. 6 (3)^2 = 54 ft ^2

2. 6 (2)^2 = 24 ft ^2

3. 54+ 24 = 78 ft^2

5 0

3 years ago

Square of a binomial x^2+2x+1

garik1379 [7]

To put an equation into (x+c)^2, we need to see if the trinomial is a perfect square.
General form of a trinomial: ax^2+bx+c
If c is a perfect square, for example (1)^2=1, 2^2=4, that's a good indicator that it's a perfect square trinomial.
Here, it is, because 1 is a perfect square.
To ensure that it's a perfect square trinomial, let's look at b, which in this case is 2.
It has to be double what c is.
2 is the double of 1, therefore this is a perfect square trinomial.
Knowing this, we can easily put it into the form (x+c)^2.
And the answer is: (x+1)^2.
To do it the long way:
x^2+2x+1
Find 2 numbers that add to 2 and multiply to 1.
They are both 1.
x^2+x+x+1
x(x+1)+1(x+1)
Gather like terms
(x+1)(x+1)
or (x+1)^2.

3 0

3 years ago

PLEASE SHOW ALL THE STEPS THAT YOU USE TO SOLVE THIS PROBLEM

Mademuasel [1]

Answer:

{x = 1 , y=1, z=0

Step-by-step explanation:

Solve the following system:

{-2 x + 2 y + 3 z = 0 | (equation 1)

{-2 x - y + z = -3 | (equation 2)

{2 x + 3 y + 3 z = 5 | (equation 3)

Subtract equation 1 from equation 2:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x - 3 y - 2 z = -3 | (equation 2)

{2 x + 3 y + 3 z = 5 | (equation 3)

Multiply equation 2 by -1:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x+3 y + 2 z = 3 | (equation 2)

{2 x + 3 y + 3 z = 5 | (equation 3)

Add equation 1 to equation 3:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x+3 y + 2 z = 3 | (equation 2)

{0 x+5 y + 6 z = 5 | (equation 3)

Swap equation 2 with equation 3:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x+5 y + 6 z = 5 | (equation 2)

{0 x+3 y + 2 z = 3 | (equation 3)

Subtract 3/5 × (equation 2) from equation 3:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x+5 y + 6 z = 5 | (equation 2)

{0 x+0 y - (8 z)/5 = 0 | (equation 3)

Multiply equation 3 by 5/8:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x+5 y + 6 z = 5 | (equation 2)

{0 x+0 y - z = 0 | (equation 3)

Multiply equation 3 by -1:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x+5 y + 6 z = 5 | (equation 2)

{0 x+0 y+z = 0 | (equation 3)

Subtract 6 × (equation 3) from equation 2:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x+5 y+0 z = 5 | (equation 2)

{0 x+0 y+z = 0 | (equation 3)

Divide equation 2 by 5:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x+y+0 z = 1 | (equation 2)

{0 x+0 y+z = 0 | (equation 3)

Subtract 2 × (equation 2) from equation 1:

{-(2 x) + 0 y+3 z = -2 | (equation 1)

{0 x+y+0 z = 1 | (equation 2)

{0 x+0 y+z = 0 | (equation 3)

Subtract 3 × (equation 3) from equation 1:

{-(2 x)+0 y+0 z = -2 | (equation 1)

{0 x+y+0 z = 1 | (equation 2)

{0 x+0 y+z = 0 | (equation 3)

Divide equation 1 by -2:

{x+0 y+0 z = 1 | (equation 1)

{0 x+y+0 z = 1 | (equation 2)

{0 x+0 y+z = 0 | (equation 3)

Collect results:

Answer: {x = 1 , y=1, z=0

6 0

3 years ago

Triangles MOP and MNQ are similar. Find x.<br><br> 2<br><br> 3<br><br> 4<br><br> 5

Diano4ka-milaya [45]

Answer:

x = 3

Step-by-step explanation:

MQ : MN = 5:6

MP : MO = 5:6

MQ/MN = 5/6

Hence, MP/MO = 5/6

(5+x)/(6+(18/5)) = 5/6

5+x = (5/6)(48/5)

5+x = 8

x = 3

4 0

3 years ago

If a1 = 9 and and an-1 - 5 then find the value of a6.

SCORPION-xisa [38]

Answer:

a1 = 6 and an = an-1 - 5

a2 = a2-1 - 5 = a1 - 5 = 1

a3 = a3-1 - 5 = a2 - 5 = -4, etc

Step-by-step explanation:

8 0

3 years ago