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OverLord2011 [107]
4 years ago
12

PLEASE ANSWER ASAP 20 POINTS

Mathematics
1 answer:
Schach [20]4 years ago
6 0
X=7. The two segments on the bottom as well as on the diagonal are the same length. This shows that the entire triangle and the inner triangle on the right are similar. So if we call the length of the bottom 2y, then 84/2y=6x/y. Solving this we get 84=12x, and so x=7
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What is the value of x?
mel-nik [20]

Answer:

x = 24

Step-by-step explanation:

1.) put the equations together and equal them to 180

2x + 4 + 6x - 16 = 180

2.) combine like terms

8x - 12 = 180

3.) divide both sides by 8

8x = 192

4.) simplify

x = 24

7 0
3 years ago
If a1 = 9 and and an-1 - 5 then find the value of a6.
SCORPION-xisa [38]

Answer:

a1 = 6 and an = an-1 - 5

a2 = a2-1 - 5 = a1 - 5 = 1

a3 = a3-1 - 5 = a2 - 5 = -4, etc

Step-by-step explanation:

8 0
3 years ago
Find the value of x. Round your answer to the nearest tenth
Sati [7]

x=13.8

<u><em>Step-by-step explanation:</em></u>

  • Since the triangle is right use the tangent ratio to solve for x

tan41^o=\frac{opposite}{adjacent}=\frac{12}{x}

  • Multiply both sides by x

x × tan41^o=12 ( divide both sides by tan41^o )

x =  ≈  ( nearest tenth )

6 0
2 years ago
I need the answer ASAP
3241004551 [841]

Answer:

Hey there!

The answer is 2 1/4, which is 2.25

Let me know if you have any other questions!

5 0
3 years ago
Read 2 more answers
A punch glass is in the shape of a hemisphere with a radius of 5 cm. If the punch is being poured into the glass so that the cha
Galina-37 [17]

Answer:

28.27 cm/s

Step-by-step explanation:

Though Process:

  • The punch glass (call it bowl to have a shape in mind) is in the shape of a hemisphere
  • the radius r=5cm
  • Punch is being poured into the bowl
  • The height at which the punch is increasing in the bowl is \frac{dh}{dt} = 1.5
  • the exposed area is a circle, (since the bowl is a hemisphere)
  • the radius of this circle can be written as 'a'
  • what is being asked is the rate of change of the exposed area when the height h = 2 cm
  • the rate of change of exposed area can be written as \frac{dA}{dt}.
  • since the exposed area is changing with respect to the height of punch. We can use the chain rule: \frac{dA}{dt} = \frac{dA}{dh} . \frac{dh}{dt}
  • and since A = \pi a^2 the chain rule above can simplified to \frac{da}{dt} = \frac{da}{dh} . \frac{dh}{dt} -- we can call this Eq(1)

Solution:

the area of the exposed circle is

A =\pi a^2

the rate of change of this area can be, (using chain rule)

\frac{dA}{dt} = 2 \pi a \frac{da}{dt} we can call this Eq(2)

what we are really concerned about is how a changes as the punch is being poured into the bowl i.e \frac{da}{dh}

So we need another formula: Using the property of hemispheres and pythagoras theorem, we can use:

r = \frac{a^2 + h^2}{2h}

and rearrage the formula so that a is the subject:

a^2 = 2rh - h^2

now we can derivate a with respect to h to get \frac{da}{dh}

2a \frac{da}{dh} = 2r - 2h

simplify

\frac{da}{dh} = \frac{r-h}{a}

we can put this in Eq(1) in place of \frac{da}{dh}

\frac{da}{dt} = \frac{r-h}{a} . \frac{dh}{dt}

and since we know \frac{dh}{dt} = 1.5

\frac{da}{dt} = \frac{(r-h)(1.5)}{a}

and now we use substitute this \frac{da}{dt}. in Eq(2)

\frac{dA}{dt} = 2 \pi a \frac{(r-h)(1.5)}{a}

simplify,

\frac{dA}{dt} = 3 \pi (r-h)

This is the rate of change of area, this is being asked in the quesiton!

Finally, we can put our known values:

r = 5cm

h = 2cm from the question

\frac{dA}{dt} = 3 \pi (5-2)

\frac{dA}{dt} = 9 \pi cm/s// or//\frac{dA}{dt} = 28.27 cm/s

5 0
3 years ago
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