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umka2103 [35]
3 years ago
12

Geometry math question please help

Mathematics
1 answer:
stiks02 [169]3 years ago
5 0

You need this key result: in any polygon, the sum of the interior angles is

(n-2)\times 180

where n is the number of sides of the polygon.

Your polygon has seven sides, and the measures of all angles are given. So, we can write

2x+4x+4x+4x+4x+2x+10x = (7-2)\times 180

Again, we've simply written that the sum of all interior angles is the number of sides, minus two, times 180.

We can simplify both sides: we sum like terms on the left hand side, and we easily compute the right hand side:

30x = 5\times 180 = 900

and we can solve this equation for x by dividing both sides by 30:

x = \cfrac{900}{30} = 30

Finally, the required angles is 2x = 2\times 30 = 60

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What is the average rate of change of f over the interval [3,4]?
masya89 [10]

Answer:

4

Step-by-step explanation:

The average rate of change of f(x) in the closed interval [ a, b ] is

\frac{f(b)-f(a)}{b-a}

Here [ a, b ] = [ 3, 4 ] , then

f(b) = f(4) = - 1 ← from table

f(a) = f(3) = - 5 ← from table , then

average rate of change = \frac{-1-(-5)}{4-3} = \frac{-1+5}{1} = 4

3 0
3 years ago
Find the area of the shape shown below
ra1l [238]

Answer:

18 MARK AS BRAINLIEST OR ELSE!!!!

Step-by-step explanation:

area = 1/2h(sum of bases)

area = 1/2x3(3+9)

area = 1/2 of 36

area = 18

3 0
3 years ago
Use the arithmetic sequence of numbers 1, 3, 5, 7, 9,.to find the following:
Rasek [7]
a)\\\\
a_{n+1}-a_n=d\\
for\ n=1\\
a_2-a_1=d\\
d=3-1=2\ - difference\ between\ two\ consecutive\ terms\\\\
b)\\
a_n=a_1+(n-1)d\\
n=101\\
a_{101}=1+(101-1)*2\\
a_{101}=1+100*2=200+1=201\\\\
c)\\
Sn=\frac{a_1+a_n}{2}*n\\
n=20\\
a_n=a_{20}=1+(20-1)*2=1+38=39\\
Sn=\frac{1+39}{2}*20=400\ \ Sum\ of\ the\ first\ 20 \ numbers.
6 0
3 years ago
1. (5pts) Find the derivatives of the function using the definition of derivative.
andreyandreev [35.5K]

2.8.1

f(x) = \dfrac4{\sqrt{3-x}}

By definition of the derivative,

f'(x) = \displaystyle \lim_{h\to0} \frac{f(x+h)-f(x)}{h}

We have

f(x+h) = \dfrac4{\sqrt{3-(x+h)}}

and

f(x+h)-f(x) = \dfrac4{\sqrt{3-(x+h)}} - \dfrac4{\sqrt{3-x}}

Combine these fractions into one with a common denominator:

f(x+h)-f(x) = \dfrac{4\sqrt{3-x} - 4\sqrt{3-(x+h)}}{\sqrt{3-x}\sqrt{3-(x+h)}}

Rationalize the numerator by multiplying uniformly by the conjugate of the numerator, and simplify the result:

f(x+h) - f(x) = \dfrac{\left(4\sqrt{3-x} - 4\sqrt{3-(x+h)}\right)\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{\left(4\sqrt{3-x}\right)^2 - \left(4\sqrt{3-(x+h)}\right)^2}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{16(3-x) - 16(3-(x+h))}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{16h}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)}

Now divide this by <em>h</em> and take the limit as <em>h</em> approaches 0 :

\dfrac{f(x+h)-f(x)}h = \dfrac{16}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ \displaystyle \lim_{h\to0}\frac{f(x+h)-f(x)}h = \dfrac{16}{\sqrt{3-x}\sqrt{3-x}\left(4\sqrt{3-x} + 4\sqrt{3-x}\right)} \\\\ \implies f'(x) = \dfrac{16}{4\left(\sqrt{3-x}\right)^3} = \boxed{\dfrac4{(3-x)^{3/2}}}

3.1.1.

f(x) = 4x^5 - \dfrac1{4x^2} + \sqrt[3]{x} - \pi^2 + 10e^3

Differentiate one term at a time:

• power rule

\left(4x^5\right)' = 4\left(x^5\right)' = 4\cdot5x^4 = 20x^4

\left(\dfrac1{4x^2}\right)' = \dfrac14\left(x^{-2}\right)' = \dfrac14\cdot-2x^{-3} = -\dfrac1{2x^3}

\left(\sqrt[3]{x}\right)' = \left(x^{1/3}\right)' = \dfrac13 x^{-2/3} = \dfrac1{3x^{2/3}}

The last two terms are constant, so their derivatives are both zero.

So you end up with

f'(x) = \boxed{20x^4 + \dfrac1{2x^3} + \dfrac1{3x^{2/3}}}

8 0
2 years ago
15. The height and yolume of a cylinder are 4cm and 616cm respectively. Calculate the diameter of the base. (take t = 27 쪽 A. 7c
irakobra [83]

Step-by-step explanation:

Given that,

  • Height of cylinder = 4 cm
  • Volume of cylinder = 616 cm³

To find,

  • Diameter of the base = ?

Firstly we'll find the base radius of the cylinder.

\longmapsto\rm{V_{(Cylinder)} = \pi r^2h}\\

<u>Accordi</u><u>ng</u><u> </u><u>to</u><u> </u><u>the</u><u> </u><u>qu</u><u>estion</u><u>,</u>

\longmapsto\rm{616= \dfrac{22}{7} \times r^2 \times 4}\\

\longmapsto\rm{616 \times 7 = 22 \times r^2 \times 4}\\

\longmapsto\rm{4312 = 88 \times r^2 }\\

\longmapsto\rm{\cancel{\dfrac{4312}{88}} =  r^2 }\\

\longmapsto\rm{49 =  r^2 }\\

\longmapsto\rm{\sqrt{49} =  r }\\

\longmapsto\rm{7 \; cm =  r }\\

Now,

\longmapsto\rm{Diameter = 2r }\\

\longmapsto\rm{Diameter = 2(7 \; cm) }\\

\longmapsto\bf{Diameter = 14 \; cm}\\

The required answer is 14 cm.

4 0
3 years ago
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