What is the distance from his house to school?

Compare the surface area-to-volume ratios of the moon and mars. express your answer using two significant figures.

antoniya [11.8K]

we know that

For a spherical planet of radius r, the volume V and the surface area SA is equal to

$V=\frac{4}{3} *\pi *r^{3} \\ \\ SA=4*\pi *r^{2}$

The
ratio of these two quantities may be written as

$SAV =\frac{(4*\pi*r^{2})}{(\frac{4}{3}*\pi*r^{3})} \\ \\ SAV =\frac{3}{r}$

we know

$rMoon=1,738Km\\ rMars=3,397 Km$

$\frac{SAV Moon}{SAV Mars} =\frac{3}{rMoon} *\frac{rMars}{3} \\ \\ \frac{SAV Moon}{SAV Mars} =\frac{rMars}{rMoon} \\ \\ \frac{SAV Moon}{SAV Mars} =\frac{3,397}{1,738} \\ \\ \frac{SAV Moon}{SAV Mars} =1.9545$

therefore

the answer is

$1.9$

5 0

3 years ago

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GIVING BRAINLY TO CORRECT answer

crimeas [40]

Answer:

False

Step-by-step explanation:

Range is defined as the greatest possible number minus the lowest possible number. The highest is 150 (in thousands) and the lowest is 25, meaning the range would be 125,000 and not 150,000.

Hope this helps!

8 0

2 years ago

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Type the missing number to complete the porportion

Nataliya [291]

34 x 2 = 17 x 4

answer = 4

6 0

3 years ago

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What is the quotient when 4x3 + 2x + 7 is divided by x + 3?

Arte-miy333 [17]

Answer:

The quotient of this division is $(4x^2 -12x + 38)$ . The remainder here would be $-26$ .

Step-by-step explanation:

The numerator $4x^3 + 2x + 7$ is a polynomial about $x$ with degree $3$ .

The divisor $x + 3$ is a polynomial, also about $x$ , but with degree $1$ .

By the division algorithm, the quotient should be of degree $3 - 1 = 2$ , while the remainder shall be of degree $1 - 1 = 0$ (i.e., the remainder would be a constant.) Let the quotient be $a\,x^2 + b\, x + c$ with coefficients $a$ , $b$ , and $c$ .

$4x^3 + 2x + 7 = \left(a\,x^2 + b\, x + c\right)(x + 3)$ .

Start by finding the first coefficient of the quotient.

The degree-three term on the left-hand side is $4 x^3$ . On the right-hand side, that would be $a\, x^3$ . Hence $a = 4$ .

Now, given that $a = 4$ , rewrite the right-hand side:

$\begin{aligned}&\left(4\,x^2 + b\, x + c\right)(x + 3) \cr =& \left(4x^2 + (b\, x + c)\right)(x + 3) \cr =& 4x^2(x + 3) + (bx + c)(x + 3) \cr =& 4x^3 + 12x^2 + (bx + c)(x + 3)\end{aligned}$ .

Hence:

$4x^3 + 2x + 7 = 4x^3 + 12x^2 + (b\,x + c)(x + 3)$

Subtract $\left(4x^3 + 12x^2\right$ from both sides of the equation:

$-12x^2 + 2x + 7 = (b\,x + c)(x + 3)$ .

The term with a degree of two on the left-hand side has coefficient $(-12)$ . Since the only term on the right hand side with degree two would have coefficient $b$ , $b = -12$ .

Again, rewrite the right-hand side:

$\begin{aligned}&\left(-12 x + c\right)(x + 3) \cr =& \left(-12 x+ c\right)(x + 3) \cr =& (-12x)(x + 3) + c(x + 3) \cr =& -12x^2 -36x + (bx + c)(x + 3)\end{aligned}$ .