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3 years ago

Which shows the expressions rewritten with the least common denominator? 5x+3/3x and 7x/2x^2

Mathematics

1 answer:

andrey2020 [161]3 years ago

4 0

Answer:

$\frac{10x^2+6x}{6x^2}$ and $\frac{21x}{6x^2}$ .

Step-by-step explanation:

The given expression is:

$\frac{5x+3}{3x}$ and $\frac{7x}{3x^2}$ .

The least common denominator is: $6x^2$ .

We collect LCM for the denominator to obtain;

$\frac{2x(5x+3)}{6x^2}$ and $\frac{3(7x)}{6x^2}$ .

We multiply out the parenthesis to obtain;

$\frac{10x^2+6x}{6x^2}$ and $\frac{21x}{6x^2}$ .

Therefore the correct answer is B or the second option.

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A circle has an area of 1075.21 m^2 <br> What is the radius to the nearest 1 decimal x.x

Sindrei [870]

Area of circle = πr²

1075.21 = πr²

1075.21/π = r²

r = √1075.21/π

r = 18.5

7 0

2 years ago

Rationalize the denominator.

Maslowich

If you're using the app, try seeing this answer through your browser: brainly.com/question/3166412

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$\mathsf{\dfrac{\sqrt{a}+2\sqrt{y}}{\sqrt{a}-2\sqrt{y}}}$

Multiply and divide by the conjugate of the denominator, which is $\mathsf{\sqrt{a}+2\sqrt{y}:}$

$\mathsf{=\dfrac{\big(\sqrt{a}+2\sqrt{y}\big)\cdot \big(\sqrt{a}+2\sqrt{y}\big)}{\big(\sqrt{a}-2\sqrt{y}\big)\cdot \big(\sqrt{a}+2\sqrt{y}\big)}}\\\\\\ \mathsf{=\dfrac{\big(\sqrt{a}+2\sqrt{y}\big)^2}{\big(\sqrt{a}-2\sqrt{y}\big)\cdot \big(\sqrt{a}+2\sqrt{y}\big)}}$

The numerator is a square of a sum, and the denominator is the product of a difference and a sum (product of two conjugates). Then, expand the common products:

$\mathsf{=\dfrac{\big(\sqrt{a}\big)^2+2\cdot \sqrt{a}\cdot 2\sqrt{y}+\big(2\sqrt{y}\big)^2}{\big(\sqrt{a}\big)^2-\big(2\sqrt{y}\big)^2}}\\\\\\ \mathsf{=\dfrac{a+4\cdot \sqrt{a}\cdot \sqrt{y}+2^2\cdot \big(\sqrt{y}\big)^2}{a-2^2\cdot \big(\sqrt{y}\big)^2}}$

$\mathsf{=\dfrac{a+4\sqrt{ay}+4y}{a-4y}}\quad\longleftarrow\quad\textsf{this is the answer.}$

I hope this helps. =)

8 0

2 years ago

In Diagram 1, identical transmitters are positioned at the vertices of an equilateral triangle. The

aksik [14]

Answer:

the answer is 21 i know right

4 0

2 years ago

The area of a circle, in terms of π, is 80π m2. Find the value of the radius. Give your answer as a simplified surd.

jolli1 [7]

Answer:

$r=4\sqrt{5}\:\: \sf m$

Step-by-step explanation:

$\textsf{Area of a circle}=\pi r^2 \quad \textsf{(where r is the radius)}$

Given:

Area = 80π m²

Substitute the given value into the formula and solve for r:

$\implies 80 \pi = \pi r^2$

$\implies \dfrac{80 \pi}{\pi} = \dfrac{\pi r^2}{\pi}$

$\implies 80=r^2$

$\implies \sqrt{r^2}=\sqrt{80}$

$\implies r=\sqrt{80}$

$\implies r=\sqrt{16 \cdot 5}$

$\implies r=\sqrt{16}\sqrt{5}$

$\implies r=\sqrt{4^2}\sqrt{5}$

$\implies r=4\sqrt{5}$

4 0

1 year ago

Read 2 more answers

3p^2-2p-5 factor completely

fiasKO [112]

$a x^{2} +bx+c$ Here is the basis equation. $3p ^{2} -2p-5$ Let us use the AC method. Multiple the A and C values, or 3 and -5. This is -15. So now ask yourself... what multiplies to -15 but adds to -2. that would be, -5 and 3. So now reconstruct your equation ... $(3p^2+3p)+(-5p-5)$ The larger number will go first (3>-5) Now you can factor each group. $3p(p+1)-5(p+1)$ Because what is now in the parentheses are qualm and they are being added, you can now make the coeffecients a group $(3p-5) (p+1)$ If you are solving for p, set equal to zero and then the answers are -1 and (5/3). If you are just factoring, then that is your answer.

7 0

2 years ago

Read 2 more answers