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3 years ago

Which is NOT the formula for an area of a rectangle?

Mathematics

1 answer:

Mademuasel [1]3 years ago

7 0

The answer that is NOT the formula for an area of a rectangle is A= 1/2X h X (Base 1 + Base 2)

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Find the value of r. Note: Write your answer as a decimal.

Solnce55 [7]

Answer:

$\huge\boxed{\sf r = 10.5}$

Step-by-step explanation:

Since it is a right-angled triangle, it has base, perpendicular and hypotenuse.

Base = r

Perpendicular = 14

Hypotenuse = r + 7

Using Pythagoras theorem:

$(Hypotenuse)^2 = (Base)^2+(Perp)^2$

Put the values

(r + 7)² = (r)² + (14)²

Using formula a² + 2ab + b² = (a + b)²

r² + 14r + 49 = r² + 196

Subtract r² to both sides

14r + 49 = 196

Subtract 49 to both sides

14r = 196 - 49

14r = 147

Divide 14 to both sides

r = 147 / 14

r = 10.5

$\rule[225]{225}{2}$

8 0

2 years ago

you roll two dice. let event A be "the first die shows a 1, a 2 , a 3 or a 6" and Event B be " The second die shows an even numb

maxonik [38]

A- 4/6, B- 3/6
If both events happen, the probability is A*B= 4/6 * 3/6 = 12/36= 1/3
P(AnB)= 1/3

4 0

3 years ago

F(x) = 12 + x X f(x) -3 -1 0 4 6

katen-ka-za [31]

9514 1404 393

Answer:

(-3, 9), (-1, 11), (0, 12), (4, 16), (6, 18)

Step-by-step explanation:

The function definition tells you that adding 12 to the x-value will give you the value of f(x).

-3 +12 = 9, for example

The (x, f(x)) values for the table are shown above.

7 0

3 years ago

A rectangle tabletop has an area of 28 1/8 square feet and a length of 6 1/4 feet. What is the width of the tabletop?

tatuchka [14]

I believe it’s 4 1/2 . Sorry if I’m wrong

4 0

3 years ago

Read 2 more answers

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