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3 years ago

Solve the system of equations. 2x+8y=6 -5x-20y+-15

Mathematics

1 answer:

kondor19780726 [428]3 years ago

7 0

Answer:

2x+8y=6

Step 1: Add -8y to both sides.

2x+8y+−8y=6+−8y

2x=−8y+6

Step 2: Divide both sides by 2

$\frac{2x}{2}$ = $\frac{-8y+6}{2}$

x=−4y+3

_________________________________________________________

−5x−20y−15

There are no like terms.

Answer:

=−5x−20y−15

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3 years ago

Which of these statements is true for f(x)=(1/10)^x

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Step-by-step explanation:

Considering the function

$f\left(x\right)=\:\left(\frac{1}{10}\right)^x$

Analyzing option A)

Considering the function

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Putting $x = 1$ in the function

$f\left(1\right)=\:\left(\frac{1}{10}\right)^1$

$f\left(1\right)=\:\left\frac{1}{10}\right$

So, it is TRUE that when $x = 1$ then the out put will be $f\left(1\right)=\:\left\frac{1}{10}\right$

Therefore, the statement that '' The graph contains $\left(1,\:\frac{1}{10}\right)$ '' is TRUE.

Analyzing option B)

Considering the function

$f\left(x\right)=\:\left(\frac{1}{10}\right)^x$

The range of the function is the set of values of the dependent variable for which a function is defined.

$\mathrm{The\:range\:of\:an\:exponential\:function\:of\:the\:form}\:c\cdot \:n^{ax+b}+k\:\mathrm{is}\:\:f\left(x\right)>k$

$k=0$

$f\left(x\right)>0$

Thus,

$\mathrm{Range\:of\:}\left(\frac{1}{10}\right)^x:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)>0\:\\ \:\mathrm{Interval\:Notation:}&\:\left(0,\:\infty \:\right)\end{bmatrix}$