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dusya [7]
3 years ago
8

Find the value of the expression below for x=10 x ^2 - 2 (x + 5)

Mathematics
2 answers:
zheka24 [161]3 years ago
8 0
Hopefully this helps
Thepotemich [5.8K]3 years ago
7 0

Answer:

20

Step-by-step explanation:

your welcome!!

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Write RS in component form, with R(7, 8) and S(0,5),
german

Answer:

<-7,-3>

Step-by-step explanation:

To write RS in component form we need to know how far to move over horizontally and then how many to move vertically.

Since horizontal movement is the x values, we subtract the x values of R and S first.

0 - 7 = -7

Since vertical movement is in the y values we subtract those next.

5 - 8 = -3

So written in component form we have <-7,-3>

4 0
3 years ago
Read 2 more answers
<img src="https://tex.z-dn.net/?f=%5Ccolor%7Bred%7D%7B%5Ctt%20%7BIn%20%5C%3A%20%20%CE%94%20%5C%3A%20%20ABC%2C%20%E2%88%A0B%3D90%
makkiz [27]

Use Pythagoras theorem

Hypotenuse²=Base²+ Height ²

H²=(8)²+(6)²

H²=64+36

H²= 910

H²=10√91 cm Solution

5 0
2 years ago
Read 2 more answers
Evaluate: Four To The Third Power write the numerical answer only
Alja [10]
The answer is 4^3 and I believe that this is right
7 0
2 years ago
Cory found a job listed in a classified that pays a yearly salary of $54,013. What is the biweekly
Anon25 [30]

Answer:

2077.42

Step-by-step explanation:

There are 52 weeks in a year, so there are 26 "biweekly" weeks in a year.

Cory receieves $54013 per year, so he gets 54013 / 26 dollars biweekly.

We can compute that number, getting 2077.42 dollars.

5 0
3 years ago
How to solve logarithmic equations as such
Serga [27]

\bf \textit{exponential form of a logarithm} \\\\ \log_a b=y \implies a^y= b\qquad\qquad a^y= b\implies \log_a b=y \\\\\\ \begin{array}{llll} \textit{Logarithm of exponentials} \\\\ \log_a\left( x^b \right)\implies b\cdot \log_a(x) \end{array} ~\hspace{7em} \begin{array}{llll} \textit{Logarithm Cancellation Rules} \\\\ log_a a^x = x\qquad \qquad \stackrel{\textit{we'll use this one}}{a^{log_a x}=x} \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}

\bf \log_2(x-1)=\log_8(x^3-2x^2-2x+5) \\\\\\ \log_2(x-1)=\log_{2^3}(x^3-2x^2-2x+5) \\\\\\ \log_{2^3}(x^3-2x^2-2x+5)=\log_2(x-1) \\\\\\ \stackrel{\textit{writing this in exponential notation}}{(2^3)^{\log_2(x-1)}=x^3-2x^2-2x+5}\implies (2)^{3\log_2(x-1)}=x^3-2x^2-2x+5

\bf (2)^{\log_2[(x-1)^3]}=x^3-2x^2-2x+5\implies \stackrel{\textit{using the cancellation rule}}{(x-1)^3=x^3-2x^2-2x+5} \\\\\\ \stackrel{\textit{expanding the left-side}}{x^3-3x^2+3x-1}=x^3-2x^2-2x+5\implies 0=x^2-5x+6 \\\\\\ 0=(x-3)(x-2)\implies x= \begin{cases} 3\\ 2 \end{cases}

5 0
3 years ago
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