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3 years ago

The graph shows how the values of two used cars changed during a 10 year period?

Mathematics

1 answer:

KonstantinChe [14]3 years ago

7 0

Answer:

$9,000

Step-by-step explanation:

If you look at the graph the lines intercept just below the 10 which is $10,000 so that means the answer would be $9,000

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A motorcycle zoomed along the freeway, traveling 10 meters in

Nana76 [90]

Answer:

30 meters per second

Step-by-step explanation:

$speed =\frac{distance}{time}\\speed =\frac{10}{\frac{1}{3} }\\speed =10*3\\speed = 30mps$

4 0

3 years ago

NEED HELP PLEASE SOMEONE!!!!

weqwewe [10]

Let me work this out really quick and I'll get back to you

4 0

3 years ago

Helppppppp pleasssseeeeee

Svet_ta [14]

$14mi-5\frac{7}{10}mi=[(14-5)-\frac{7}{10}]mi=(9-\frac{7}{10})mi=(8+1-\frac{7}{10})mi\\\\=(8+\frac{10}{10}-\frac{7}{10})mi=(8+\frac{10-7}{10})mi=(8+\frac{3}{10})mi=\boxed{8\frac{3}{10}mi}\\\\other\ method\\\\14mi-5\frac{7}{10}mi=14mi-5.7mi=(14-5-0.7)mi\\\\=(9-0.7)mi=\boxed{8.3mi}$

6 0

2 years ago

Is (-3,-9) a solution of the equation y = 3x ?<br> O YES<br> O NO<br> O NOT A SOLUTION

melomori [17]

Answer:

Yes, (-3, -9) is a solution.

Step-by-step explanation:

y = 3x (-3,-9)

-9 = 3(-3)

-9 = -9

4 0

2 years ago

Find a polynomial P(x) with real coefficients having a degree 4, leading coefficient 4, and zeros 3-i and 4i.

Alisiya [41]

now, this polynomial has roots of 3-i and 4i, namely 3 - i and 0 + 4i.

let's bear in mind that a complex root never comes all by her lonesome, her sibling is always with her, the conjugate, so if 3 - i is there, 3 + i is also coming along, likewise if 0 + 4i is there, her sibling 0 - 4i is also there.

$\bf \begin{cases} x=3-i\implies &x-3+i=0\\ x=3+i\implies &x-3-i=0\\ x=4i\implies &x-4i=0\\ x=-4i\implies &x+4i=0 \end{cases}\\\\[-0.35em] ~\dotfill\\\\ (x-3+i)(x-3-i)(x-4i)(x+4i)=\stackrel{y}{0} \\[2em] \underset{\textit{difference of squares}}{[(x-3)+i][(x-3)-i]}\underset{\textit{difference of squares}}{[x-4i][x+4i]}=0$

$\bf [(x-3)^2-i^2][x^2-(4i)^2]=y\implies [(x-3)^2-(-1)][x^2-(4^2i^2)]=0 \\[2em] [(x-3)^2-(-1)][x^2-(16(-1))]=0\implies [(x-3)^2+1][x^2+16]=0 \\[2em] [(x^2-6x+9)+1][x^2+16]=y\implies (x^2-6x+10)(x^2+16)=0 \\\\\\ x^4-6x^3+10x^2+16x^2-96x+160=0 \\\\\\ x^4-6x^3+26x^2-96x+160=0 \\\\\\ \stackrel{\textit{multiplying both sides by 4}}{4(x^4-6x^3+26x^2-96x+160)=4(0)} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill 4x^4-24x^3+104x^2-384x+640=y~\hfill$

3 0

3 years ago