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3 years ago

HURRY!! THIS ASSIGNMENT IS DUE IN 5MIN!!! THANK YOU!!!!!!

1 answer:

5 0

Since Paula left home an hour ago driving at 40mph, it means she has already gone 40 miles by the time Dan left. Dan is driving at 50mph, which is 10 greater than 40. This means that every hour, Dan will drive 10 more miles than Paula, lessening their difference by 10. 40 divided by 10 is 4, so iit will take Dan 4 hours to catch up to Paula:)

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X squared+ y squared = 2 y = 2x squared – 3 Which of the following describes the system?

cluponka [151]

Answer:

$x=-1,1,-\sqrt{\frac{7}{4} },\sqrt{\frac{7}{4}}$ and $y=-1,\frac{1}{2}$

The ordered pair solutions are $(-\sqrt{\frac{7}{4}},0.5)$ , $(\sqrt{\frac{7}{4}},0.5)$ , $(-1,-1)$ , and $(1,-1)$ .

Step-by-step explanation:

I'm assuming the system is $\left \{ {x^2+y^2=2} \atop {y=2x^2-3}} \right.$ :

$x^2+y^2=2$

$x^2+(2x^2-3)^2=2$

$x^2+(4x^4-12x^2+9)=2$

$x^2+4x^4-12x^2+9=2$

$4x^4-11x^2+9=2$

$4x^4-11x^2+7=0$

$x^4-11x^2+28=0$

$(x^2-7)(x^2-4)=0$

$(4x^2-7)(x^2-1)=0$

$4x^2-7=0$

$4x^2=7$

$x^2=\frac{7}{4}$

$x=\pm\sqrt{\frac{7}{4}}$

$x^2-1=0$

$x^2=1$

$x=\pm1$

$y=2x^2-3$

$y=2(\pm\sqrt{\frac{7}{4}})^2-3$

$y=2({\frac{7}{4}})-3$

$y=\frac{7}{2}-3$

$y=\frac{1}{2}$

$y=2x^2-3$

$y=2(\pm1)^2-3$

$y=2(1)-3$

$y=2-3$

$y=-1$

Therefore, $x=-1,1,-\sqrt{\frac{7}{4} },\sqrt{\frac{7}{4}}$ and $y=-1,\frac{1}{2}$

The ordered pair solutions are $(-\sqrt{\frac{7}{4}},0.5)$ , $(\sqrt{\frac{7}{4}},0.5)$ , $(-1,-1)$ , and $(1,-1)$ .

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0.1376 is the answer.

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Answer:

hope it helps you see the attachment for further information

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When Joseph first starts working at a grocery store, his hourly rate is $10. For each year he works at the grocery store, his ho

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Answer: R(t) = 10 + 0.50t

Step-by-step explanation:

From the question, we've been informed that Joseph's hourly rate is $10 and that for every year he works at the grocery store, his hourly rate increases by $0.50.

Therefore, the function will be

R(t) = 10 + 0.50t

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