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2 years ago

Can sb please help me

Mathematics

2 answers:

snow_lady [41]2 years ago

6 0

Hey there!

Given rule is 2x + 7

For x = 6 we get,

2(6)+7 = 19.

Mekhanik [1.2K]2 years ago

4 0

Answer:

the answer is 19

Step-by-step explanation:

hence the output is 19 hope it help u

putting the value of x in rule

2x+7

2*6+7

=19

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3 years ago

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Brrunno [24]

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3 years ago

Describe the position of the function below relative to the graph of the indicated basic function.

matrenka [14]

From the given the option moved left 1 unit; reflected across the x axis is the correct.

what is function?

A function from a set X to a set Y in mathematics assigns each element of X exactly one element of Y. The set X is known as the function's domain, and the set Y is known as the function's codomain.

We have to describe the position of the function $f(x) = -2^x^+^1$ relative to the basic function $2^x$ .

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After reflection in x axis it becomes, $-2^x^+^1$ .

Hence, from the given the option moved left 1 unit; reflected across the x axis is the correct.

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7 0

1 year 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

Factor the expression completely -20x -40

Alenkasestr [34]

X=2

its x=2 because you have to divide -20x by -40 so the two negatives cancel out each other and then you get 2.

6 0

3 years ago

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