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2 years ago

How many times greater is the area of circle C than the area of circle A.

Mathematics

1 answer:

kicyunya [14]2 years ago

5 0

Answer:

I need a picture

Step-by-step explanation:

sorry but without a picture i don't know wht to do

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What are possible values of the number?

klio [65]

Answer:

I think is F

Step-by-step explanation:

3 0

2 years ago

Is this right??? if not helpp

Nonamiya [84]

Yupz u b right cuz 8+9 is ur answer Don’t be so hard on yourself take it easy

7 0

2 years ago

The triangular-shaped bases of a prism are 24 cm in length and have a height of 35 cm. Each side is 37 cm. The height of the pri

Savatey [412]

Answer:

The surface area of Triangular base Prism = 3682 cm²

Step-by-step explanation:

Given in question as :

For a Triangular base prism ,

The base of prism (b) = 24 cm

The height of prism (l) = 29 cm

Each side length (s) = 37 cm

The height of base triangle ( h ) = 35 cm

Hence , we know that The surface area of Triangular base prism is

= (b×h) + (2×l×s) + (l×b)

= (24×35) + (2×29×37) + (29×24)

= (840) + (2146) + (696)

= 3682 cm²

Hence The surface area of Triangular base Prism is 3682 cm² Answer

3 0

2 years ago

<img src="https://tex.z-dn.net/?f=%203%20000%20000%5E%7B2%7D%20%20" id="TexFormula1" title=" 3 000 000^{2} " alt=" 3 000 000^{2

Pachacha [2.7K]

${3000000}^{2} + {4000000}^{2} \\ \\ {10}^{6} ( {3}^{2} + {4}^{2} ) \\ \\ {10}^{6} \times 25 \\ \\ 25000000 \: \: \: \: \: Ans.$

8 0

2 years ago

What is a solution to the equation 3 / m + 3 - M / 3 - M equals m^2 + 9 / m^2-9?

Mnenie [13.5K]

Answer: Last option.

Step-by-step explanation:

Given the equation:

$\frac{3}{m+3}-\frac{m}{3-m}=\frac{m^2+9}{m^2-9}$

Follow these steps to solve it:

- Subtract the fractions on the left side of the equation:

$\frac{3(3-m)-m(m+3)}{(m+3)(3-m)}=\frac{m^2+9}{m^2-9}\\\\\frac{9-3m-m^2-3m}{(m+3)(3-m)}=\frac{m^2+9}{m^2-9}\\\\\frac{-m^2-6m+9}{(m+3)(3-m)}=\frac{m^2+9}{m^2-9}$

- Using the Difference of squares formula ( $a^2-b^2=(a+b)(a-b)$ ) we can simplify the denominator of the right side of the equation:

$\frac{-m^2-6m+9}{(m+3)(3-m)}=\frac{m^2+9}{(m+3)(m-3)}$

- Multiply both sides of the equation by $(m+3)(3-m)$ and simplify:

$\frac{(-m^2-6m+9)(m+3)(3-m)}{(m+3)(3-m)}=\frac{(m^2+9)(m+3)(3-m)}{(m+3)(m-3)}\\\\-m^2-6m+9=\frac{(m^2+9)(3-m)}{(m-3)}$

- Multiply both sides by $m-3$ :

$(-m^2-6m+9)(m-3)=\frac{(m^2+9)(3-m)(m-3)}{(m-3)}\\\\(-m^2-6m+9)(m-3)=(m^2+9)(3-m)$

- Apply Distributive property and simplify:

$(-m^2-6m+9)(m-3)=(m^2+9)(3-m)\\\\-m^3-6m^2+9m+3m^2+18m-27=3m^2+27-m^3-9m\\\\-m^3-3m^2+27m-27+m^3-3m^2+9m-27=0\\\\-6m^2+36m-54=0$

- Divide both sides of the equation by -6:

$\frac{-6m^2+36m-54}{-6}=\frac{0}{-6}\\\\m^2-6m+9=0$

- Factor the equation and solve for "m":

$(m-3)^2=0\\\\m=3$

In order to verify it, you must substitute $m=3$ into the equation and solve it:

$\frac{3}{3+3}-\frac{3}{3-3}=\frac{3^2+9}{3^2-9}\\\\\frac{3}{6}-\frac{3}{0}=\frac{18}{0}$

<em>NO SOLUTION</em>

7 0

2 years ago