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Arlecino [84]
3 years ago
11

Given the figure below, find the values of x and z.

Mathematics
1 answer:
N76 [4]3 years ago
5 0

Answer:

x=9

z=73

Step-by-step explanation:

Angle 107 and 8x+35 are vertical angles, so they are equal

107 = 8x+35

Subtract 35 from each side

107-35 = 8x+35

72 = 8x

Divide by 8

72/8 = 8x/8

9 =x

We also know that 107 and z are supplementary angles since they form a straight line.  This means they add to 180 degrees

107+z = 180

Subtract 107 from each side

107-107+z=180-107

z =73

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If f (x)=3^5-x + 6, what is the value of f (5), to the nearest tenth?
Amiraneli [1.4K]

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240

Step-by-step explanation:

3^5 - 5 + 6

243 - 5 = 238+6= 244

240?

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What is the slope of y= x + 7
Dimas [21]

Answer:

1

I hope this helps!

8 0
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I need help with this so could you give me the answers with an explanation if you can.​
Katarina [22]

Answer:

2.44745763 \times  {10}^{6}

Step-by-step explanation:

R =  \frac{ {x}^{2} }{y}  \\  \\  =  \frac{ {(3.8 \times  {10}^{5}) }^{2} }{5.9 \times  {10}^{4} }  \\  \\  =  \frac{14.44 \times  {10}^{10} }{5.9 \times  {10}^{4} }  \\  \\  = 2.44745763 \times  {10}^{10 - 4}  \\ = 2.44745763 \times  {10}^{6}

4 0
3 years ago
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
2 years ago
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