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3 years ago

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Solve the equation. Then check your solution. – two-fifths + a = One-fifth a. Three-fifths c. StartFraction 3 Over 10 EndFractio

n b. Negative one-fifth d. Negative three-fifths

1 answer:

VLD [36.1K]3 years ago

5 0

Answer:

a = Negative one-fifth

Step-by-step explanation:

The given equation is :

$\dfrac{2}{5}+a=\dfrac{1}{5}$

We need to find the value of a.

Subtract 2/5 from both sides.

$\dfrac{2}{5}+a-\dfrac{2}{5}=\dfrac{1}{5}-\dfrac{2}{5}\\\\a=\frac{1}{5}-\frac{2}{5}\\\\a=\dfrac{-1}{5}$

So, the correct answer is Negative one-fifth. Hence, the correct option is (b).

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The boundary lines for the system of inequalities is given in the graph.

diamong [38]

Answer:

C)

region C

Step-by-step explanation:

We have to use what is called the zero-interval test [test point] in order to figure out which portion of the graph these inequalities share:

$\displaystyle y ≤ -x + 2$

0 ≤ 2 ☑ [We shade the portion of the graph that CONTAIN THE ORIGIN, which is the bottom portion.]

$\displaystyle y ≥ 2x - 3$

0 ≥ −3 ☑ [We shade the portion of the graph that CONTAINS THE ORIGIN, which is the left side.]

So, now that we got that all cleared up, we can tell that both graphs share a region where the ORIGIN IS VISIBLE. Therefore region C matches the above inequalities.

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3 years ago

Read 2 more answers

Merlin wants to make more money selling fake beards this month than he did last month. All black beards in his shop cost the sam

sukhopar [10]

We know that all black beards cost the same money

And all red beards cost $25

And B represents number of black beards

And R represents number of red beards

We have been given the inequality 20B+25R > 350 to show the target for this month.

The above inequality shows that Merlin needs to make more than 350, hence he made 350 last month

Further, it shows that B has been multiplied by 20, hence the black beard's cost was 20

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3 years ago

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The Square sand pit in park lot has an area of 16 ft. There is a rope along one side of the pit. How long is the rope?

slavikrds [6]

Answer:

ydhhh

Step-by-step explanation:

1)tiff

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4 0

3 years ago

Please answer this question as soon as possible

Taya2010 [7]

Answer: The graph is shifted vertically, but not horizontally

Step-by-step explanation:

Since the added constant is outside the x term, the constant will only shift the graph vertically.

6 0

3 years ago

A. Use composition to prove whether or not the functions are inverses of each other. B. Express the domain of the compositions u

Kryger [21]

Given: $f(x) = \frac{1}{x-2}$

$g(x) = \frac{2x+1}{x}$

A.)Consider

$f(g(x))= f(\frac{2x+1}{x} )$

$f(\frac{2x+1}{x} )=\frac{1}{(\frac{2x+1}{x})-2}$

$f(\frac{2x+1}{x} )=\frac{1}{\frac{2x+1-2x}{x}}$

$f(\frac{2x+1}{x} )=\frac{x}{1}$

$f(\frac{2x+1}{x} )=1$

Also,

$g(f(x))= g(\frac{1}{x-2} )$

$g(\frac{1}{x-2} )= \frac{2(\frac{1}{x-2}) +1 }{\frac{1}{x-2}}$

$g(\frac{1}{x-2} )= \frac{\frac{2+x-2}{x-2} }{\frac{1}{x-2}}$

$g(\frac{1}{x-2} )= \frac{x }{1}$

$g(\frac{1}{x-2} )= x$

Since, $f(g(x))=g(f(x))=x$

Therefore, both functions are inverses of each other.

B.

For the Composition function $f(g(x)) = f(\frac{2x+1}{x} )=x$

Since, the function $f(g(x))$ is not defined for $x=0$ .

Therefore, the domain is $(-\infty,0)\cup(0,\infty)$

For the Composition function $g(f(x)) =g(\frac{1}{x-2} )=x$

Since, the function $g(f(x))$ is not defined for $x=2$ .

Therefore, the domain is $(-\infty,2)\cup(2,\infty)$

8 0

3 years ago