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2 years ago

The solutions to the inequality y ≤ −x + 1 are shaded on the graph. Which point is a solution?

Mathematics

2 answers:

algol [13]2 years ago

7 0

Just a matter of subbing answer choices into ur inequality to see which is true.
answer is : (3,-2)

Delvig [45]2 years ago

5 0

we know that

If a point is a solution of the inequality

then

The point must satisfy the inequality

we have

$y\leq -x+1$

The solution of the inequality is the shaded area below the solid red line

see the attached figure to better understand the problem

Step 1

Point $A(2,3)$

Substitute the value of x and the value of y in the inequality

$x=2\ y=3$

$3\leq -2+1$

$3\leq -1$ -------> is not true

therefore

The point A is not a solution for the inequality

See the attached figure------> the point A is not on the shaded area

Step 2

Point $B(3,-2)$

Substitute the value of x and the value of y in the inequality

$x=3\ y=-2$

$-2\leq -3+1$

$-2\leq -2$ -------> is true

therefore

The point B is a solution for the inequality

See the attached figure------> the point B is on the shaded area

Step 3

Point $C(2,1)$

Substitute the value of x and the value of y in the inequality

$x=2\ y=1$

$1\leq -2+1$

$1\leq -1$ -------> is not true

therefore

The point C is not a solution for the inequality

See the attached figure------> the point C is not on the shaded area

Step 4

Point $D(-1,3)$

Substitute the value of x and the value of y in the inequality

$x=-1\ y=3$

$3\leq 1+1$

$3\leq 2$ -------> is not true

therefore

The point D is not a solution for the inequality

See the attached figure------> the point D is not on the shaded area

<u>the answer is</u>

$B(3,-2)$

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$\lim_{n\to\infty} (1+ \frac{2}{n} +\frac{1}{n^2})$

And when we apply the limit we got that:

$\lim_{n\to\infty} (1+ \frac{2}{n} +\frac{1}{n^2}) =1$

Step-by-step explanation:

Assuming this complete problem: "The following formula for the sum of the cubes of the first n integers is proved in Appendix E. Use it to evaluate the limit . 1^3+2^3+3^3+...+n^3=[n(n+1)/2]^2"

We have the following formula in order to find the sum of cubes:

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We can express this formula like this:

$\lim_{n\to\infty} \sum_{n=1}^{\infty}i^3 =\lim_{n\to\infty} [\frac{n(n+1)}{2}]^2$

And using this property we need to proof that: 1^3+2^3+3^3+...+n^3=[n(n+1)/2]^2

$\lim_{n\to\infty} [\frac{n(n+1)}{2}]^2$

If we operate and we take out the 1/4 as a factor we got this:

$\lim_{n\to\infty} \frac{n^2(n+1)^2}{n^4}$

We can cancel $n^2$ and we got

$\lim_{n\to\infty} \frac{(n+1)^2}{n^2}$

We can reorder the terms like this:

$\lim_{n\to\infty} (\frac{n+1}{n})^2$

We can do some algebra and we got:

$\lim_{n\to\infty} (1+\frac{1}{n})^2$

We can solve the square and we got:

$\lim_{n\to\infty} (1+ \frac{2}{n} +\frac{1}{n^2})$

And when we apply the limit we got that:

$\lim_{n\to\infty} (1+ \frac{2}{n} +\frac{1}{n^2}) =1$

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2 years ago

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Answer:

Option F is correct, i.e. "rotating 90° counterclockwise around point C."

Step-by-step explanation:

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We can not use Translations, Reflections, and Dilations because they will change size and coordinates.

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faust18 [17]

Answer:

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The player has already batted 134 times and will still bat 46 times. So in the end of the year, he is going to have 134 + 46 = 180 at bats.

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