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3 years ago

What is the measure of the length of AB

Mathematics

2 answers:

Gnom [1K]3 years ago

5 0

Answer:

Step-by-step explanation:

AC = AB + BC

16 = x+1 + x+7

Combine like terms

16 =2x+8

Subtract 8 from each side

16-8 = 2x+8-8

8 = 2x

Divide by 2

8/2 = 2x/2

4 =x

We want to find AB

AB = x+1 = 4+1 = 5

marusya05 [52]3 years ago

4 0

The measuring is equal to 5 do to the whole length being 16 when adding the two equations together and equaling it to 16 you can solve the equation to equal 5.

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Find the volume of a right circular cone that has a height of 17.7 cm and a base with a diameter of 15.8 cm. Round your answer t

Lelu [443]

Answer:

$1156.8cm^3$

Step-by-step explanation:

The volume of a cone is given as:

$V =\frac{1}{3} \pi r^2h$

where r = radius

h = height of cone

The height of the cone is 17.7 cm and its base radius is 7.9 cm (diameter is 15.8 cm).

Its volume is:

$V = \frac{1}{3} * \pi = 7.9^2 * 17.7\\ \\V = 1156.8cm^3$

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3 years ago

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Why does the tail approaches the x-axis but never touches it

Brut [27]

Because most of the area under any normal curve falls within a limited range of the number line

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3 years ago

50.4 is what percent of 280

shusha [124]

Us the is over of method
50.4•100=5,040
5,040/280=x
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3 years ago

A 500500-gallon tank initially contains 200200 gallons of brine containing 100100 pounds of dissolved salt. brine containing 11

mixer [17]

Let $A(t)$ denote the amount of salt in the tank at time $t$ . We're given that the tank initially holds $A(0)=100$ lbs of salt.

The rate at which salt flows in and out of the tank is given by the relation

$\dfrac{\mathrm dA}{\mathrm dt}=\underbrace{\dfrac{11\text{ lb}}{1\text{ gal}}\times\dfrac{44\text{ gal}}{1\text{ min}}}_{\text{rate in}}-\underbrace{\dfrac{A(t)}{200+(44-11)t}\times\dfrac{11\text{ gal}}{1\text{ min}}}_{\text{rate out}}$
$\implies A'(t)+\dfrac{11}{200+33t}A(t)=484$

Find the integrating factor:

$\mu(t)=\exp\left(\displaystyle\int\frac{11}{200+33t}\,\mathrm dt\right)=(200+33t)^{1/3}$

Distribute $\mu(t)$ along both sides of the ODE:

$(200+33t)^{1/3}A'(t)+11(200+33t)^{-2/3}A(t)=484(200+33t)^{-1/3}$
$\bigg((200+33t)^{1/3}A(t)\bigg)'=484(200+33t)^{-1/3}$
$A(t)=484\displaystyle\int(200+33t)^{-1/3}\,\mathrm dt$
$A(t)=22(200+33t)^{2/3}+C$

Since $A(0)=100$ , we get

$100=22(200)^{2/3}+C\implies C\approx-652.39$

so that the particular solution for $A(t)$ is

$A(t)=22(200+33t)^{2/3}-652.39$

The tank becomes full when the volume of solution in the tank at time $t$ is the same as the total volume of the tank:

$200+(44-11)t=500\implies 33t=300\implies t\approx9.09$

at which point the amount of salt in the solution would be

$A(9.09)\approx733.47\text{ lb}$

4 0

3 years ago

2. The troupe’s account balance first reached $500 on Day 4. The last day the account balance was $500 was Day 8.

timofeeve [1]

Answer:

2. c is the answer hope it helps ^_^

3 0

2 years ago