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stepladder [879]

3 years ago

12

1. The most massive planet in the solar system, Jupiter, has a mass of about 1,898,000,000,000,000,000,000,000,000 kg. Estimate

Jupiter's mass using a one-digit whole number times a power of 10. Be sure to include units.
A. 1.9 X 10^18 kg
B. 2 x 10^18 kg
C. 1.9 x 10^27 kg
D. 2x 10^27 kg

1 answer:

UNO [17]3 years ago

6 0

Answer:

D. $2*10^{27} kg$

Step-by-step explanation:

Mass of Jupiter is given as 1,898,000,000,000,000,000,000,000,000 kg

Estimating this to the nearest whole number would be = 2,000,000,000,000,000,000,000,000,000 kg (rounding up the the "1,898")

To express this in scientific notation form, simply count how many zeros we have to the nearest whole number, 2.

We have 27 zeros there.

Therefore, we would multiply the whole number, 2, by 10 raise to the power of 27.

An estimate of the mass of Jupiter = $2*10^{27} kg$

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Is my answer correct? 10 points + brainleist!

Liono4ka [1.6K]

Answer:

your answer is incorrect. The correct answer is $h=-13$ and $k=13$ .

Step-by-step explanation:

If a quadratic function is $f(x)=ax^2+bx+c$ and a>0, then minimum value of the function at point $\left(-\dfrac{b}{2a},f(-\dfrac{b}{2a})\right)$ .

The given function is

$f(x)=x^2+bx+182$

Here, a=1, b=b and c=182. So.

$-\dfrac{b}{2a}=-\dfrac{b}{2(1)}=-\dfrac{b}{2}$

Put $x=-\dfrac{b}{2}$ in the given function to find the minimum value of the function.

$f(-\dfrac{b}{2})=(-\dfrac{b}{2})^2+b(-\dfrac{b}{2})+182$

We know that minimum value is 13. So,

$13=\dfrac{b^2}{4}-\dfrac{b^2}{2}+182$

$13-182=-\dfrac{b^2}{4}$

$-169=-\dfrac{b^2}{4}$

$169\times 4=b^2$

Taking square root on both sides.

$13\times 2=b$

$b=26$

The value of b is 26.

So, the given function is

$f(x)=x^2+26x+182$

Now, add and subtract square of half of coefficient of x.

$f(x)=x^2+26x+182+(13)^2-(13)^2$

$f(x)=(x^2+2(13)x+(13)^2)+182-169$

$f(x)=(x+13)^2+13$

On comparing with $f(x)=(x-h)^2+k$ , we get

$h=-13$

$k=13$

Therefore, your answer is incorrect.

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Answer:

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