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2 years ago

HELP I GIVE 15 POINTS

Mathematics

1 answer:

ExtremeBDS [4]2 years ago

8 0

Answer:

Alright well first off you start by converting everything to fractions, but decimals work better for me. So it would be 40.5 and 1.25, 40.5 being miles and the other gas. Divide 40.5 by 1.25, giving you 32.4

<u><em>His car get 32.4 miles per gallon</em></u>

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If there is a sphere with a volume of 1761.1 meters cubed, what is the radius?

Anarel [89]

$\bf \textit{volume of a sphere}\\\\
V=\cfrac{4\pi r^3}{3}\quad 
\begin{cases}
r=radius\\
-----\\
V=1761.1
\end{cases}\implies 1761.1=\cfrac{4\pi r^3}{3}
\\\\\\
3(1761.1)=4\pi r^3\implies \cfrac{3(1761.1)}{4\pi }=r^3\implies \sqrt[3]{\cfrac{3(1761.1)}{4\pi }}=r$

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3 years ago

Brian plays a video game in which he starts at 0 points. In round 1, he

fredd [130]

At this point, brian had lost 7 points.

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2 years ago

HELP ASAP!!!!!!!!!! Please

N76 [4]

Answer:

Step-by-step explanation:

Growth can be represented by the equation $A=A_01.02^{t}$ . We know that r=2% or 0.02 in this situation so we have 1.02. We know t=100 hours and the starting population A=200. We substitute these values and use a calculator to evaluate.

$A=A_01.02^{t}\\A=200(1.02)^{100}$

We now use our calculator to evaluate the exponent first and then multiply by 200.

$A=200(7.2446)=1448.93$

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2 years ago

(-4k⁴+14+3k²)+(3k⁴-14k²-8) write answer in standard form

Shkiper50 [21]

Dont click the link the bot sent you they can hack you

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2 years ago

In Exercises 40-43, for what value(s) of k, if any, will the systems have (a) no solution, (b) a unique solution, and (c) infini

svet-max [94.6K]

Answer:

If k = −1 then the system has no solutions.

If k = 2 then the system has infinitely many solutions.

The system cannot have unique solution.

Step-by-step explanation:

We have the following system of equations

$x - 2y +3z = 2\\x + y + z = k\\2x - y + 4z = k^2$

The augmented matrix is

$\left[\begin{array}{cccc}1&-2&3&2\\1&1&1&k\\2&-1&4&k^2\end{array}\right]$

The reduction of this matrix to row-echelon form is outlined below.

$R_2\rightarrow R_2-R_1$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\2&-1&4&k^2\end{array}\right]$

$R_3\rightarrow R_3-2R_1$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&3&-2&k^2-4\end{array}\right]$

$R_3\rightarrow R_3-R_2$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&0&0&k^2-k-2\end{array}\right]$

The last row determines, if there are solutions or not. To be consistent, we must have k such that

$k^2-k-2=0$

$\left(k+1\right)\left(k-2\right)=0\\k=-1,\:k=2$

Case k = −1:

$\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-1-2\\0&0&0&(-1)^2-(-1)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-3\\0&0&0&-2\end{array}\right]$

If k = −1 then the last equation becomes 0 = −2 which is impossible.Therefore, the system has no solutions.

Case k = 2:

$\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&2-2\\0&0&0&(2)^2-(2)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&0\\0&0&0&0\end{array}\right]$

This gives the infinite many solution.

5 0

3 years ago