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Evgesh-ka [11]
3 years ago
14

Using the polygon angle sum theorem, find x in the graphic above

Mathematics
1 answer:
Aliun [14]3 years ago
6 0

Answer:

D) 120

Step-by-step explanation:

720- 135-115-110-140-100=120

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OlgaM077 [116]

Answer:

For it to be parallel the answer is B (3)

7 0
3 years ago
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If you could show as much work as possible that would be amazing!!
Nikitich [7]

Answers:

<u>Reduce:</u>

Here we gave to simplify the expressions:

9) x^{2}+7x+\frac{12}{x^{2}}+11x+28

Grouping similar terms:

(x^{2}+\frac{12}{x^{2}})+(18x+28)

Applying common factor x^{2} in the first parenthesis and common factor 2 in the second parenthesis:

x^{2}(1+\frac{12}{x^{4}})+2(9x+14) This is the answer

11) y^{3}+\frac{27}{y^{2}}+2y-3

Rearranging the terms:

(y^{3}+2y)+(\frac{27}{y^{2}}-3)

Applying common factor y in the first parenthesis and common factor 3 in the second parenthesis:

y(y^{2}+2)+3(\frac{9}{y^{2}}-1) This is the answer

<u>Multiply:</u>

19) (\frac{12a^{9}u^{7}}{15 c})(\frac{3c^{4}}{21a^{13 u^{8}}})

Multiplying both fractions:

\frac{36 a^{9}u^{7}c^{4}}{315 c a^{13}u^{8}}

Dividing numerator and denominator by 3 and simplifying:

\frac{12 c^{3}}{105 c a^{4}u} This is the answer

21) (x-\frac{3}{x}-7)(x^{2}-9x+\frac{35}{x^{2}}-18)

(\frac{x^{2}-3-7x}{x})(\frac{x^{4}-9x^{3}+35-18x^{2}}{x^{2}})

Operating with cross product:

x^{2} (x^{2}-3-7x) x(x^{4}-9x^{3}+35-18x^{2})

x^{9} -9x^{8} -18x^{7}+35x^{5}-7x^{8}  +63x^{7}+126x^{6}-245x^{4}-3x^{7}+27x^{6}+54x^{5}-105x^{3}

Grouping similar terms and factoring:

x^{9}-2(8x^{8}+21x^{7} )+153x^{6}+89x^{5}-5(49x^{4}+21x^{3}) This is the answer

<u>Divide:</u>

29) \frac{\frac{k^{6}}{x^{2}}}{\frac{2k^{4}}{3x^{6}}}

\frac{3k^{6}x^{6}}{2x^{2}k^{4}}

Simplifying:

\frac{3}{2} k^{2}x^{4} This is the answer

33) \frac{\frac{x+5}{x+1}}{\frac{x^{2}+11x+30}{x^{2}+3x+2}}

\frac{(x+5)(x^{2}+3x+2)}{(x+1)(x^{2}+11x+30)}

Factoring numerator and denominator:

\frac{(x+5)(x+2)(x+1)}{(x+1)(x+6)(x+5)}

Simplifying:

\frac{x+2}{x+6} This is the answer

37) \frac{\frac{x-10}{x+13}}{\frac{x^{3}-1000}{x^{2}+15x+21}}

\frac{(x-10)(x^{2}+15x+21)}{(x+13)(x^{3}-1000)}

Applying the distributive property in numerator and denominator:

\frac{x^{3}+15x^{2}+21x-10x^{2}-150x-210}{(x+13)(x^{4}-1000x+13x^{3}-13000)}

Grouping similar terms and factoring by common factor:

-\frac{5x(x^{2}-129)(x^{2}-42)}{1000x^{3} (x+13)(x-13)}

Dividing by 5 in numerator and denominator and simplifying:

-\frac{(x^{2}-129)(x^{2}-42)}{200x^{2}(x+13)(x-13)} This is the answer

3 0
3 years ago
Find the following:
Butoxors [25]

Answer:

Step-by-step explanation:

Limit refers to the value that the function approaches as the input approaches some value.

We say \displaystyle \lim_{x\rightarrow a}f(x)=L, if f(x) approaches L as x approaches 'a'.

(a)

\displaystyle \lim_{x\rightarrow 5}\left ( \frac{f(x)-8}{x-5} \right )=4\\\frac{\displaystyle \lim_{x\rightarrow 5}f(x)-\displaystyle \lim_{x\rightarrow 5}8}{\displaystyle \lim_{x\rightarrow 5}x-\displaystyle \lim_{x\rightarrow 5}5}=4\\

\frac{\displaystyle \lim_{x\rightarrow 5}f(x)-8}{\displaystyle \lim_{x\rightarrow 5}x-5}=4\\\displaystyle \lim_{x\rightarrow 5}f(x)-8=4\left ( \displaystyle \lim_{x\rightarrow 5}x-5 \right )\\\displaystyle \lim_{x\rightarrow 5}f(x)-8=4\displaystyle \lim_{x\rightarrow 5}x-4(5)\\\displaystyle \lim_{x\rightarrow 5}f(x)-8=4(5)-4(5)\\

\displaystyle \lim_{x\rightarrow 5}f(x)-8=20-20=0\\\displaystyle \lim_{x\rightarrow 5}f(x)=8

(b)

\displaystyle \lim_{x\rightarrow 5}\left ( \frac{f(x)-8}{x-5} \right )=7\\\frac{\displaystyle \lim_{x\rightarrow 5}f(x)-\displaystyle \lim_{x\rightarrow 5}8}{\displaystyle \lim_{x\rightarrow 5}x-\displaystyle \lim_{x\rightarrow 5}5}=7\\\frac{\displaystyle \lim_{x\rightarrow 5}f(x)-8}{\displaystyle \lim_{x\rightarrow 5}x-5}=7\\

\displaystyle \lim_{x\rightarrow 5}f(x)-8=7\left ( \displaystyle \lim_{x\rightarrow 5}x-5 \right )\\\displaystyle \lim_{x\rightarrow 5}f(x)-8=7\displaystyle \lim_{x\rightarrow 5}x-7(5)\\\displaystyle \lim_{x\rightarrow 5}f(x)-8=7(5)-7(5)\\\displaystyle \lim_{x\rightarrow 5}f(x)-8=35-35=0\\\displaystyle \lim_{x\rightarrow 5}f(x)=8

3 0
3 years ago
I need help with this​
Schach [20]
8- 1/3 quarts
4- 2/3 quarts
7 0
3 years ago
The cost of a daily newspaper varies from city to city. however, the variation among prices remains steady with a standard devia
Tanya [424]

First, we establish our hypothesis:

<span>Null hypothesis H0: μ = $1.00  </span>

Alternative hypothesis Ha: μ ≠ $1.00

<span>Let’s say X  = the sample average cost of a daily newspaper = 0.96</span>

u = population mean cost = 1.00

S = sample standard deviation = 0.18

Calculating for z value:

z = (X – u) / S

z = (0.96 – 1) / 0.18

z = – 0.222

From the standard distribution table at this z value, p-value = 0.4129

Since alpha = 0.01, the decision therefore is:

<span>Do not reject the null hypothesis because the p-value is greater than 0.01. There is enough evidence to support the claim that the mean cost of newspapers is $1. </span>

6 0
3 years ago
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