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3 years ago

I’m confused about this question

Mathematics

1 answer:

arsen [322]3 years ago

7 0

Angle B is 102° bc 180-78=102 and if sides DB and BC are congruent, angles D and C must also be, so you can use the original 78 and divide by 2 to get 39° for angle C

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Determine the value of the variable in the simplified expressions. What is the value of x in this simplified expression? x = Wha

arsen [322]

Answer:x=7 y=5

Step-by-step explanation:

I just took it, sorry I'm not good at math and just guessed until I got it right for this answer.

8 0

3 years ago

Simplify each expression, #1: 4(f+2g) - 4g +f. #2: 5g(9m-2) + 3g

Serjik [45]

Answer:

1) 5f

2) 45gm-7g

Step-by-step explanation:

hope this helped!

7 0

2 years ago

an inverted conical water tank with a height of 20 ft and a radius of 8 ft is drained through a hole in the vertex (bottom) at a

viktelen [127]

Answer:

the rate of change of the water depth when the water depth is 10 ft is; $\mathbf{\dfrac{dh}{dt} = \dfrac{-25}{100 \pi} \ \ ft/s}$

Step-by-step explanation:

Given that:

the inverted conical water tank with a height of 20 ft and a radius of 8 ft is drained through a hole in the vertex (bottom) at a rate of 4 ft^3/sec.

We are meant to find the rate of change of the water depth when the water depth is 10 ft.

The diagrammatic expression below clearly interprets the question.

From the image below, assuming h = the depth of the tank at a time t and r = radius of the cone shaped at a time t

Then the similar triangles ΔOCD and ΔOAB is as follows:

$\dfrac{h}{r}= \dfrac{20}{8}$ ( similar triangle property)

$\dfrac{h}{r}= \dfrac{5}{2}$

$\dfrac{h}{r}= 2.5$

h = 2.5r

$r = \dfrac{h}{2.5}$

The volume of the water in the tank is represented by the equation:

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

$V = \dfrac{1}{3} \pi (\dfrac{h^2}{6.25}) h$

$V = \dfrac{1}{18.75} \pi \ h^3$

The rate of change of the water depth is :

$\dfrac{dv}{dt}= \dfrac{\pi r^2}{6.25}\ \dfrac{dh}{dt}$

Since the water is drained through a hole in the vertex (bottom) at a rate of 4 ft^3/sec

Then,

$\dfrac{dv}{dt}= - 4 \ ft^3/sec$

Therefore,

$-4 = \dfrac{\pi r^2}{6.25}\ \dfrac{dh}{dt}$

the rate of change of the water at depth h = 10 ft is:

$-4 = \dfrac{ 100 \ \pi }{6.25}\ \dfrac{dh}{dt}$

$100 \pi \dfrac{dh}{dt} = -4 \times 6.25$

$100 \pi \dfrac{dh}{dt} = -25$

$\dfrac{dh}{dt} = \dfrac{-25}{100 \pi}$

Thus, the rate of change of the water depth when the water depth is 10 ft is; $\mathtt{\dfrac{dh}{dt} = \dfrac{-25}{100 \pi} \ \ ft/s}$

4 0

3 years ago

30. Tina went to batting cages with $20. She rented a bat for $3. 25 and

aliya0001 [1]

Answer:

Step-by-step explanation:

Tina started with $20, but had to spend $3.25 on the bat.

$20-$3.25 = $16.75, which is all she has left for rounds.

Each round costs $3.00

$16.75 divided by $3.00 comes out to 5.58 rounds, but since she can't play 1/2 a round, that means she could only play 5 rounds

5 rounds X $3.00 per round comes to $15.00

8 0

2 years ago

The frequency of musical note B4 us about 493.88 Hz.

Ivahew [28]

Answer: A

Step-by-step explanation:493.88/5

6 0

2 years ago

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