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3 years ago

6

The width of a rectangular painting is 5 inches shorter than it's length. Creat a table of values to include possible widths,len

gths,and areas of the painting.

1 answer:

Misha Larkins [42]3 years ago

6 0

A = L * W
W = L - 5

these are in inches...
length width area
10 5 50 in^2
9 4 36 "
8 3 24 "
7 2 14 "
6 1 6 "

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In the triangle below, what is the measure of ZC?

Diano4ka-milaya [45]

Answer:

C

Step-by-step explanation:

The triangle is an equilateral triangle, which is equal on all three sides in both side length and angle. since one of them of 60 degrees, then all of them would also be 60 degrees. Since the total amount of degrees in a triangle is 180, 60*3=180 so the answer is C.

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2 years ago

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3 years ago

Only number one please help

Alex73 [517]

Answer:

C) 75 + 50x = 25 + 70x

Step-by-step explanation:

ABC Tech:

$50 an hour. So,for x hours= 50 * x = 50x

Service charge = 75 + 50x

Tech Squad:

$70 an hour. So,for x hours= 70 * x = 70x

Service charge = 25 + 70x

C) 75 + 50x = 25 + 70x

50x - 70x = 25 - 75

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6 0

3 years ago

A 75-gallon tank is filled with brine (water nearly saturated with salt; used as a preservative) holding 11 pounds of salt in so

Debora [2.8K]

Let $A(t)$ = amount of salt (in pounds) in the tank at time $t$ (in minutes). Then $A(0) = 11$ .

Salt flows in at a rate

$\left(0.6\dfrac{\rm lb}{\rm gal}\right) \left(3\dfrac{\rm gal}{\rm min}\right) = \dfrac95 \dfrac{\rm lb}{\rm min}$

and flows out at a rate

$\left(\dfrac{A(t)\,\rm lb}{75\,\rm gal + \left(3\frac{\rm gal}{\rm min} - 3.25\frac{\rm gal}{\rm min}\right)t}\right) \left(3.25\dfrac{\rm gal}{\rm min}\right) = \dfrac{13A(t)}{300-t} \dfrac{\rm lb}{\rm min}$

where 4 quarts = 1 gallon so 13 quarts = 3.25 gallon.

Then the net rate of salt flow is given by the differential equation

$\dfrac{dA}{dt} = \dfrac95 - \dfrac{13A}{300-t}$

which I'll solve with the integrating factor method.

$\dfrac{dA}{dt} + \dfrac{13}{300-t} A = \dfrac95$

$-\dfrac1{(300-t)^{13}} \dfrac{dA}{dt} - \dfrac{13}{(300-t)^{14}} A = -\dfrac9{5(300-t)^{13}}$

$\dfrac d{dt} \left(-\dfrac1{(300-t)^{13}} A\right) = -\dfrac9{5(300-t)^{13}}$

Integrate both sides. By the fundamental theorem of calculus,

$\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac1{(300-t)^{13}} A\bigg|_{t=0} - \frac95 \int_0^t \frac{du}{(300-u)^{13}}$

$\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac{11}{300^{13}} - \frac95 \times \dfrac1{12} \left(\frac1{(300-t)^{12}} - \frac1{300^{12}}\right)$

$\displaystyle -\dfrac1{(300-t)^{13}} A = \dfrac{34}{300^{13}} - \frac3{20}\frac1{(300-t)^{12}}$

$\displaystyle A = \frac3{20} (300-t) - \dfrac{34}{300^{13}}(300-t)^{13}$

$\displaystyle A = 45 \left(1 - \frac t{300}\right) - 34 \left(1 - \frac t{300}\right)^{13}$

After 1 hour = 60 minutes, the tank will contain

$A(60) = 45 \left(1 - \dfrac {60}{300}\right) - 34 \left(1 - \dfrac {60}{300}\right)^{13} = 45\left(\dfrac45\right) - 34 \left(\dfrac45\right)^{13} \approx 34.131$

pounds of salt.

7 0

2 years ago

Edward was running errands on his bike first to be like from his house to the post office to pick up a package the post office i

Karo-lina-s [1.5K]

Answer:

Edward drove altogether of 16.1 miles.

Step-by-step explanation:

Given:

Edward was Running around on his bike.

Distance from home to post office = $4 \frac{4}{5} \ miles$

$4 \frac{4}{5} \ miles$ can be Rewritten as $4.8 \ miles$

Distance from home to post office = $4.8 \ miles$

Now Given:

Distance from Post office to market = 3.25 miles.

Now we will calculate Distance from Home to market.

Distance from Home to market can be calculated by adding Distance from home to post office and Distance from Post office to market.

Framing in equation form we get;

Distance from Home to market = $4.8+3.25 = 8.05\ miles$

Also Given:

his mother picked him up at the market and drove him home.

We need to find the total miles Edward drive altogether.

total miles Edward drive altogether will be equal to Sum of Distance from Home to market and Distance from market to home.

framing in equation form we get;

Total Distance Edward drove = $8.05+8.05 = 16.1\ miles$

Hence Edward drove altogether of 16.1 miles.

5 0

3 years ago