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2 years ago

What is the value of 7z^2 + 4 x 3w when w= 8 and z= -3

Mathematics

2 answers:

zepelin [54]2 years ago

6 0

64+96=160. I just plugged in the numbers!

Lesechka [4]2 years ago

5 0

So the answer would be 32x^3+63 right?

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Three packages of pens and one box of paper cost $26. Two packages of pens and 7 boxes of paper cost $49. How much does one box

algol13

Answer:

5 dollars

Step-by-step explanation:

Let's call the price of a package of pens n, and the cost of a box of paper r.

3n+r=26

2n+7r=49

Multiply the first equation by 7:

21n+7r=182

Subtract the second equation from the first:

19n=133

Divide both sides by 19 to isolate n:

n=7

Plug this back in to find the value of a box of paper:

3(7)+r=26

21+r=26

r=5

Hope this helps!

5 0

2 years ago

Complete the two column proof<br> Given angle to an angle five are supplementary <br> Prove: I || m

Oksanka [162]

Answer:

Step-by-step explanation:

8 0

2 years ago

Can someone help????????????

kirill115 [55]

Answer:

I believe it is the top right scatterplot, I can't tell which letter, I think B though.

Step-by-step explanation:

If you were to put a line of best fit for each graph, the top right scatterplot would be the only one that has a straight line of best fit.

Hope this helps :)

3 0

2 years ago

Read 2 more answers

What am I doing sumbody please help

djyliett [7]

Answer:

um, i think its d

Step-by-step explanation:

8 0

2 years ago

At 8:00 am, here's what we know about two airplanes: Airplane #1 has an elevation of 80870 ft and is decreasing at the rate of 4

wel

Let's begin by listing out the information given to us:

8 am

airplane #1: x = 80870 ft, v = -450 ft/ min

airplane #2: x = 5000 ft, v = 900ft/min

We must note that the airplanes are moving at a constant speed. The equation for the airplanes is given by:

$\begin{gathered} E=x_1+vt----1 \\ E=x_2+vt----2 \\ where\colon E=elevation,ft;x=InitialElevation,ft; \\ v=velocity,ft\text{/}min;t=time,min \\ x_1=80,870ft,v=-450ft\text{/}min \\ E=80870-450t----1 \\ x_2=5,000ft,v=900ft\text{/}min \\ E=5000+900t----2 \end{gathered}$

We equate equations 1 & 2 to get the time both airlanes will be at the same elevation. We have:

$\begin{gathered} 5000+900t=80870-450t \\ \text{Add 450t to both sides, we have:} \\ 900t+450t+5000=80870-450t+450t \\ 1350t+5000=80870 \\ \text{Subtract 5000 from both sides, we have:} \\ 1350t+5000-5000=80870-5000 \\ 1350t=75870 \\ \text{Divide both sides by 1350, we have:} \\ \frac{1350t}{1350}=\frac{75870}{1350} \\ t=56.2min \\ \\ \text{After }56.2\text{ minutes, both airplanes will be at the same elevation} \end{gathered}$

The elevation at that time (when the elevations of the two airplanes are the same) is given by substituting the value of time into equations 1 & 2. We have:

$\begin{gathered} E_1=80870-450t \\ E_1=80870-450(56.2) \\ E_1=80870-25290 \\ E_1=55580ft \\ \\ E_2=5000+900t \\ E_2=5000+900(56.2) \\ E_2=5000+50580 \\ E_2=55580ft \\ \\ \therefore E_1\equiv E_2=55580ft \end{gathered}$

6 0

4 months ago