This is because the radius of any circle is the distance from the middle of the circle to the edge of the circle. The diagram shows us 5cm as that distance.

#teamtrees #PAW (Plant And Water)

3 0

3 years ago

Find the area of the regular polygon 10, 4

Afina-wow [57]

With all the given context, I would assume it’s length times width, which would give us an answer of 40 :) hope this helped !

3 0

2 years ago

Find the missing side in the right triangle, Round to the nearest tenth,

N76 [4]

Hello!

$\large\boxed{c = 12.65}$

Use the Pythagorean Theorem to solve for the hypotenuse:

a² + b² = c²

Plug in the given values:

12² + 4² = c²

144 + 16 = c²

160 = c²

Take the square root of both sides:

c = √160

c ≈ 12.65.

3 0

2 years ago

The equation 2x^2 – 8x = 5 is rewritten in the form of 2(x – p)^2 + q = 0. what is the value of q?

QveST [7]

Answer:

q = -13

Step-by-step explanation:

The given equation is:

$2x^{2}-8x=5$

Taking 2 as common from left hand side, we get:

$2(x^{2}-4x)=5\\\\2[x^{2} - 2(x)(2)]=5$

The square of difference is written as:

$(a-b)^{2}=a^2 - 2ab + b^{2}$ Equation 1

If we compare the given equation from previous step to formula in Equation 1, we note that we have square of first term(x), twice the product of 1st term(x) and second term(2) and the square of second term(2) is missing. So in order to complete the square we need to add and subtract square of 2 to right hand side. i.e.

$2[x^{2}-2(x)(2)+(2)^2-(2)^{2}]=5\\\\ 2[x^{2}-2(x)(2)+(2)^2]-2(2)^{2}=5\\\\ 2(x-2)^{2}-2(4)=5\\\\ 2(x-2)^2-8=5\\\\ 2(x-2)^{2}-8-5=0\\\\ 2(x-2)^{2}-13=0$

Comparing the above equation with the given equation:

$2(x-p)^{2}+q=0$ , we can say:

p = 2 and q= -13

6 0

3 years ago