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3 years ago

Find the 50th term of 2, 5, 8, 11 its +3 every time.

Mathematics

1 answer:

vova2212 [387]3 years ago

8 0

Answer:

the answer is 149

2,5,8,11,14,17,20,23,26,29,32,35,38,41,44,47,50,53,56,59,62,65,68,71,74,77,80,83,86,89,92,95,98,101,104,107,110,113,116,119,122,125,128,131,134,137,140,143, 146,149

Step-by-step explanation:

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Solve 2 log x = log 64. (1 point)

Vladimir79 [104]

Answer:

Step-by-step explanation:

2log(x)=log(64)

By logarithms properties

$xlog_{a}(b)=log_{a}(b^{x} ) \\$

log(x²)=log(64)

When you have log or ln in both sites, you delete them and...

x²=64

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7 0

3 years ago

Izara built a fence for her horses. The fence was 210 feet long. She put a fence post in the ground at the start of the fence an

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3 years ago

F(x)=4x+1 and g(x)=x^2-5find(f-g)(x)

Vinvika [58]

Answer:

Step-by-step explanation:

Given f(x)=4x+1 and g(x)=x²-5

(f-g)(x) is derived by taking the difference of both functions

(f-g)(x) = f(x)-g(x)

(f-g)(x) = 4x+1 - (x²-5)

(f-g)(x) = 4x+1-x²+5

(f-g)(x) = -x²+4x +6

This gives the requires expression

4 0

3 years ago

I need help with multiplying and dividing powers

Tamiku [17]

You need to know these two formulas:

$\huge a^b \times a^c = a^{b+c}\\\\\huge \frac{a^b}{a^c}=a^{b-c}$

For example,

$5^2 \times 5^3 = 5^{2+3} = 5^5$

$\frac{2^6}{2^4} = 2^{6-4} = 2^2$

6 0

3 years ago

What value of b will cause the system to have an infinite number of solutions?

irga5000 [103]

b must be equal to -6 for infinitely many solutions for system of equations $y = 6x + b$ and $-3 x+\frac{1}{2} y=-3$

<u>Solution: </u>

Need to calculate value of b so that given system of equations have an infinite number of solutions

$\begin{array}{l}{y=6 x+b} \\\\ {-3 x+\frac{1}{2} y=-3}\end{array}$

Let us bring the equations in same form for sake of simplicity in comparison

$\begin{array}{l}{y=6 x+b} \\\\ {\Rightarrow-6 x+y-b=0 \Rightarrow (1)} \\\\ {\Rightarrow-3 x+\frac{1}{2} y=-3} \\\\ {\Rightarrow -6 x+y=-6} \\\\ {\Rightarrow -6 x+y+6=0 \Rightarrow(2)}\end{array}$

Now we have two equations

$\begin{array}{l}{-6 x+y-b=0\Rightarrow(1)} \\\\ {-6 x+y+6=0\Rightarrow(2)}\end{array}$

Let us first see what is requirement for system of equations have an infinite number of solutions

If $a_{1} x+b_{1} y+c_{1}=0$ and $a_{2} x+b_{2} y+c_{2}=0$ are two equation

$\Rightarrow \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$ then the given system of equation has no infinitely many solutions.

In our case,

$\begin{array}{l}{a_{1}=-6, \mathrm{b}_{1}=1 \text { and } c_{1}=-\mathrm{b}} \\\\ {a_{2}=-6, \mathrm{b}_{2}=1 \text { and } c_{2}=6} \\\\ {\frac{a_{1}}{a_{2}}=\frac{-6}{-6}=1} \\\\ {\frac{b_{1}}{b_{2}}=\frac{1}{1}=1} \\\\ {\frac{c_{1}}{c_{2}}=\frac{-b}{6}}\end{array}$

As for infinitely many solutions $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$

$\begin{array}{l}{\Rightarrow 1=1=\frac{-b}{6}} \\\\ {\Rightarrow6=-b} \\\\ {\Rightarrow b=-6}\end{array}$

Hence b must be equal to -6 for infinitely many solutions for system of equations $y = 6x + b$ and $-3 x+\frac{1}{2} y=-3$

8 0

3 years ago