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3 years ago

Find the value of x so that the function has the givin value n(x)=2x+7, n(x)=17

Mathematics

1 answer:

Sunny_sXe [5.5K]3 years ago

6 0

Answer: x=5

Step-by-step explanation:

Substitute 17 for n(x)

17=2x+7

Subtract 7 from both sides

10=2x

Divide each side by 2 to get the x by itself

x=5

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Solve the following and explain your steps. Leave your answer in base-exponent form. (3^-2*4^-5*5^0)^-3*(4^-4/3^3)*3^3 please st

Naily [24]

Answer:

$\boxed{2^{\frac{802}{27}} \cdot 3^9}$

Step-by-step explanation:

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$(3^{-2} \cdot 4^{-5} \cdot 5^0)^{-3} \cdot (4^{-\frac{4}{3^3} })\cdot 3^3$

Note that

$\boxed{a^{-b} = \dfrac{1}{a^b}, a\neq 0 }$

The denominator can't be 0 because it would be undefined.

So, we can solve the expression inside both parentheses.

$\left(\dfrac{1}{3^2} \cdot \dfrac{1}{4^5} \cdot 5^0 \right)^{-3} \cdot \left(\dfrac{1}{4^{\frac{4}{3^3} } }\right)\cdot 3^3$

Also,

$\boxed{a^{0} = 1, a\neq 0 }$

$\left(\dfrac{1}{9} \cdot \dfrac{1}{1024} \cdot 1 \right)^{-3} \cdot \left(\dfrac{1}{4^{\frac{4}{27} } }\right)\cdot 27$

Note

$\boxed{\dfrac{1}{a} \cdot \dfrac{1}{b}= \frac{1}{ab} , a, b \neq 0}$

$\left(\dfrac{1}{9216} \right)^{-3} \cdot \left(\dfrac{1}{4^{\frac{4}{27} } }\right)\cdot 27$

$\left(\dfrac{1}{9216} \right)^{-3} \cdot \left(\dfrac{27}{4^{\frac{4}{27} } }\right)$

$\left( \dfrac{1}{\left(\dfrac{1}{9216}\right)^3} \right)\cdot \left(\dfrac{27}{4^{\frac{4}{9} } }\right)$

$\left( \dfrac{1}{\left(\dfrac{1}{9216}\right)^3} \right)\cdot \left(\dfrac{27}{4^{\frac{4}{27} } }\right)$

Note

$\boxed{\dfrac{1}{\dfrac{1}{a} } = a}$

$9216^3\cdot \left(\dfrac{27}{4^{\frac{4}{9} } }\right)$

$\left(\dfrac{ 9216^3\cdot 27}{4^{\frac{4}{27} } }\right)$

Once

$9216=2^{10}\cdot 3^2 \implies 9216^3=2^{30}\cdot 3^6$

$\boxed{(a \cdot b)^n=a^n \cdot b^n}$

And

$4^{\frac{4}{27}} = 2^{\frac{8}{27}$

We have

$\left(\dfrac{ 2^{30} \cdot 3^6\cdot 27}{2^{\frac{8}{27} } }\right)$

Also, once

$\boxed{\dfrac{c^a}{c^b}=c^{a-b}}$

$2^{30-\frac{8}{27}} \cdot 3^6\cdot 27$

$30-\dfrac{8}{27} = \dfrac{30 \cdot 27}{27}-\dfrac{8}{27} =\dfrac{802}{27}$

$2^{30-\frac{8}{27}} \cdot 3^6\cdot 27 = 2^{\frac{802}{27}} \cdot 3^6 \cdot 3^3$

$2^{\frac{802}{27}} \cdot 3^9$

4 0

3 years ago

Please I really need help with this

ipn [44]

You would do 4-2= 2 then 8/2 =4 .. final answer being 4

4 0

3 years ago

Read 2 more answers

Find the GCF. 14abc and 28a2b2c3

Zarrin [17]

The greatest common factor is the largest factor that both terms share.

First, start with the coefficients, 14 and 28. The largest factor they both share is 14. 14 x 1 = 14, 14 x 2 = 28.

Next we move to the variables, abc and a²b²c³. The largest factor the both share is abc, all to the power of 1.

So, your GCF and final answer is 14abc

5 0

3 years ago

Is the equation y=-2x+5 in standard form? Explain

anyanavicka [17]

The equation, y = -2x + 5, is not in standard form. The standard form of a linear line is ax + by = c. This equation is in slope - intercept form, which is y = mx + b.

7 0

3 years ago

Can someone solve and explain this question plz

oee [108]

firstly let's convert the mixed fraction to improper fraction, then hmmm let's see we have two denominators, 5 and 3, and their LCD will simply be 15, so we'll multiply both sides by that LCD to do away with the denominators, let's proceed,

$\bf \stackrel{mixed}{2\frac{1}{3}}\implies \cfrac{2\cdot 3+1}{3}\implies \stackrel{improper}{\cfrac{7}{3}} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{z}{5}-4=\cfrac{7}{3}\implies \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{15}}{15\left( \cfrac{z}{5}-4 \right)=15\left( \cfrac{7}{3} \right)}\implies 3z-60=35 \\\\\\ 3z=95\implies z=\cfrac{95}{3}\implies z = 31\frac{2}{3}$

4 0

3 years ago