PLEASE HELP !! 36 POINTS

Mathematics

2 answers:

nika2105 [10]3 years ago

7 0

1)d 2)a 3)d 4)b 5)a 6)d

Alexandra [31]3 years ago

3 0

Answer:

Hope this helps you. look it once.

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How do you solve -4+p +2

klasskru [66]

You just combine like terms.....

- 4 + p + 2

- 4 + 2 + p

The answer is p - 2

3 0

4 years ago

School is making digital backups of old reels of film in its library archives the table shown approximate run Times of the films

Gemiola [76]

One technique that you can apply when solving such a problem is trial and error. We try to use each equation to prove that a given value of x on the table given will correspond to the value of y on the table.

a) Let's try to put x = 3 for the first equation and we must get an answer equal to 2.25.

$y=7.72(3)-29.02=-5.86_{}$

Since the value of y is not equal to 2.25 and the deviation is too large. this equation is not a good model,

b) We put x = 3 on the second equation and solve for y

$y=-7.52(3)^2+0.19(3)+3.26=-63.85$

Since the value of y is not equal to 2.25 and the deviation is too large. this equation is not a good model,

c) We put x = 3 on the third equation and solve for y,

$y=0.4(3)^2+0.79(3)-4.93=1.04$

Again, the value that we get is not equal to 2.25, hence, this equation is not a good model. But since its value is close to 2.25, we try to other values of x. If x = 5, we get

$y=0.4(5)^2+0.79(5)-4.93=9.02$

which has a slight deviation on the given value of y on the table for x = 5. let's try for x = 7. We have

$y=0.4(7)^2+0.79(7)-4.93=20.2$

and the answer has a small deviation compared to the actual value given. The other values of x can again be put on the equation and check their corresponding value of y, and the resulting values are as follows

$\begin{gathered} y=0.4(8)^2+0.79(8)-4.93=26.99 \\ y=0.4(12)^2+0.79(12)-4.93=62.15 \\ y=0.4(14)^2+0.79(14)-4.93=84.53 \end{gathered}$

And as you can see, the deviation of values from the table to calculated becomes smaller. Hence, this is the best model.

d) We put x = 3 on the third equation and solve for y,

$y=4.19(1.02)^3=4.45_{}_{}$

Again, the value that we get is not equal to 2.25, hence, this equation is not a good model. But since its value is close to 2.25, we try to other values of x. If x = 5, we get