Ask question

Ask question

All categories

alexandr1967 [171]

3 years ago

7

The graph shows the relationship between the cost of making cardboard boxes and the number of boxes made , what rate does the po

int (35,7) represent ? Y=$ and X=boxes
7$/35boxes
7$/1$
35$/7boxes
35$/1box

I'm confused between 1 and 3 for the answers please help !

1 answer:

Digiron [165]3 years ago

5 0

The Answer is A because 7 is in the $ side and 35 in the boxes.

$7 / 35 boxes.

You might be interested in

PLEASE HURRY WITH THIS

Snezhnost [94]

Answer:

Step-by-step explanation:

100.5 in ^2

5 0

2 years ago

What is the measure of MN?<br><br> A. 45<br> B. 29<br> C. 74<br> D. 37

Diano4ka-milaya [45]

Answer:

45°

Step-by-step explanation:

There are sets of formulas you can refer to for different situations like this. In your case here, you need to apply the formula and solve for the unknown. My work is in the attachment.

7 0

3 years ago

Read 2 more answers

Brainiest If Correct + 50pts ---The cylinder with base radius 5 and height 12 is inscribed in a right rectangular prism. Find th

Romashka [77]

Answer:

SA = 680 units^2

V =1200 unit^3

Step-by-step explanation:

If the radius is 5 then the diameter is 2r = 2*5 = 10

The height would remain the same

Surface area of a prism is given by

SA = 2 (lw+lh+hw)

l =d = 10 w = d = 10 h = 12

SA = 2 ( 10*10 + 10 * 12+ 12* 10)

SA = 2( 100+120+120)

SA = 2(340)

SA = 680 units^2

V = l*w*h = 10*10*12 =1200 unit^3

4 0

3 years ago

I have been asking and asking for help and not one hleped me all they did was troll and say dumb stuff that has nothing to do wi

Andreas93 [3]

Answer:

Try 120

Step-by-step explanation:

There are five dashes in between 100 and 200. Count up by 20s

6 0

2 years ago

A 140 gallon tank initially contains 5 lbs salt dissolved in 40 gallons of water. Brine containing 1 lb salt per gallon begins t

laiz [17]

Answer:

96.46

Step-by-step explanation:

Volume of the Gallon =140 gallons

At time t=0, V(0)=40 gallons

$\frac{dV}{dt}=Rate \:In - Rate \:Out \\\frac{dV}{dt}=3\frac{gal}{min} - 1\frac{gal}{min} =2\frac{gal}{min}$

V(t)=2t+k, Where K is a constant.

Since V(0)=40

V(0)=2(0)+k=40

k=40

V(t)=40+2t

The Change in Salt Concentration

$\frac{dC}{dt}=Rate \:In - Rate \:Out \\\frac{dC}{dt}=1\frac{lb}{gallon}*3\frac{gallon}{minute}-1\frac{lb}{gallon}*\frac{C}{40+2t} \frac{gallon}{minute}\\\frac{dC}{dt}=3-\frac{C}{40+2t}$

Rearranging

$\frac{dC}{dt}+\frac{C}{40+2t}=3$

Using Integrating Factor: $e^{\int\frac{1}{40+2t} dt} =e^{ln|40+2t|}=40+2t$

$\frac{dC(40+2t)}{dt}=3(40+2t)$

$\int d[C(40+2t)]=\int(120+6t)dt\\C(40+2t)=120t+3t^2+k_0\\C(t)=\frac{120t+3t^2}{40+2t}+\frac{k_0}{40+2t}$

At t=0, C(0)=0.125, k_0=5

$C(t)=\frac{120t+3t^2}{40+2t}+\frac{5}{40+2t}\\C(t)=\frac{120t+3t^2+5}{40+2t}$

At V(t)=140

V(t)=40+2t=140

t=50

$C(50)=96.46$

Just the tank overflows, C(t)=96.46

8 0

3 years ago