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2 years ago

Which point is a solution to the linear inequality y < x + 2?

2 answers:

7 0

First you can try substituting (0,0) to see if that works because it is the easiest.

0<0+2
0<2

That is true so (0,0) is solution of y

Charra [1.4K]2 years ago

7 0

Answer:

Step-by-step explanation:

When we are looking for a solution of inequality , there are various possibilities. We can have no solution to it or a finite solution of many/infinite solutions . Lets try to figure out one point that is asked to find for this inequality .

The inequality we are dealing with is y < x+2

Since , y= x+2 is an equation of line , it means the given inequality will represent one side of the line on a cartesian axis .

The very basic and easiest point we use to find the solution is (0,0) also known as the origin test . so lets try to figure out if the point (0,0) is a solution or not .

To find if (0,0) is a solution or not we need to plug it in the given inequality and check if it holds true . so when we plug (0,0) we get

0 <0 +2

0< 2

Which is true so we can say that (0,0) is a point which is solution of the given inequality .

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4. Write 0.5% as a fraction.

11111nata11111 [884]

Answers
4. A
5. D
6. D

7 0

2 years ago

Read 2 more answers

What is the value of x?<br>x=____

Setler79 [48]

The answer is x = 6

To solve this you would use the pythagorean theorem which is a^2 + b^2 = c^2

In this problem a = x b = 8 and c = 10

You would then square these numbers
x^2 + 64 = 100

Next, you would subtract 64 from both sides.

x^2 = 36

All that’s left is the find the square root, which is x = 6.

So, therefore, you’re answer is x = 6

6 0

2 years ago

For the piecewise function, find the values h( - 6), h(0), h(5), and h(9).- 5x – 13, for x < -3h(x) = { 5, for - 35x<5for

sleet_krkn [62]

Given the piecewise function h(x):

$h(x)=\begin{cases}-5x-13,x$

-When x is less than "-3", h(x)=-5x-13

-When x is between -3 and 5, h(x)=5

-When x is greater than or equal to 5, h(x)=x+1

1) For h(-6), this notation indicates that you have to determine the value of h(x) when x=-6

-6 is less than -3, which means that for this value of x, the function has the following shape

$h(x)=-5x-13$

Replace the expression with x=-6 and calculate the corresponding value of x:

$\begin{gathered} h(-6)=-5(-6)-13 \\ h(-6)=30-13 \\ h(-6)=17 \end{gathered}$

2) For h(0), you have to determine the value of h(x) when x=0. Zero is between -3 and 5, for this value of x, the function h(x) has the following shape:

$h(x)=5$

This equation represents a horizontal line, which means that for every value within the interval of definition -3≤x<5, the function always has the same value h(x)=5

We can conclude that:

$h(0)=5$

3) For h(5), you have to determine the value of h(x) for x=5, for values of x greater than or equal to 5, h(x) has the following shape:

$h(x)=x+1$

Replace the expression with x=5 and calculate the corresponding value of h(x):

$\begin{gathered} h(5)=5+1 \\ h(5)=6 \end{gathered}$

4) For h(9), you have to determine the value of h(x) when x=9, 9 is greater than 5, for this value of x, the function has the following shape:

$h(x)=x+1$

Replace the expression with x=9, and calculate the corresponding value of h(x):

$\begin{gathered} h(9)=9+1 \\ h(9)=10 \end{gathered}$

So, to sum up:

$undefined$

5 0

10 months ago

Pls help Question On the photo

Mrac [35]

Answer:

C. In step 2, Cindy should have multiplied both sides of the equation by 2.

Step-by-step explanation:

I'll go through the equation correctly, and then point out what would have happened.

3 = x/2 - 5

Add 5 to both sides. -5 and 5 cancel out.

8 = x/2

What Cindy should have done here is multiply by 2, not divide by 2. It is important to remember that you do the INVERSE operation to cancel out a number.

What would have happened here is that we have an extra step (though either way we still end up with x being 16.)

Correct Step 2:

Multiply both sides by 2. x/2 and 2x cancel out.

x = 16

Incorrect Step 2:

8/2 = (x/2)/2

4 = x/4

Multiply both sides by 4 (this would be the extra step!).

16 = x

8 0

2 years ago

ewrite the rational exponent as a radical by extending the properties of integer exponents. 2 to the 3 over 4 power, all over 2

yawa3891 [41]

$\bf ~\hspace{7em}\textit{rational exponents} \\\\ a^{\frac{ n}{ m}} \implies \sqrt[ m]{a^ n} ~\hspace{10em} a^{-\frac{ n}{ m}} \implies \cfrac{1}{a^{\frac{ n}{ m}}} \implies \cfrac{1}{\sqrt[ m]{a^ n}} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \cfrac{2^{\frac{3}{4}}}{2^{\frac{1}{2}}}\implies 2^{\frac{3}{4}}\cdot 2^{-\frac{1}{2}}\implies 2^{\frac{3}{4}-\frac{1}{2}}\implies 2^{\frac{3-2}{4}}\implies 2^{\frac{1}{4}}\implies \sqrt[4]{2^1}\implies \sqrt[4]{2}$

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2 years ago