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3 years ago

Good morning!! can someone help?!:)

Mathematics

2 answers:

Fudgin [204]3 years ago

6 0

Answer:

Option A. {-4, 2, 4}

Step-by-step explanation:

There is <em>no other options</em> because<em> none of them </em>are bigger than -5 except Option A.

Firdavs [7]3 years ago

6 0

A gggggggggggggggggggggggg

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△TVX ~ △WUX

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2 years ago

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4 years ago

4. If y - 2x = 3 and y - x = 1, what are the values of x and y?

ioda

y - x = 1.

or, y= 1+x

y - 2x = 3.

=>1+x -2x= 3 (substituting the value of y)

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3 years ago

What is the measure of angle C?

Lapatulllka [165]

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4 years ago

LOTS OF POINTS!!!!!!!! HELP PLEASE RSM IS DUE TOMORROW!!!!!!!

elena55 [62]

Answer: Anything larger than the fraction 21/127

21/127 = 0.16535 approximately

=========================================================

Explanation:

Let's solve each inequality for x

We'll start with the first inequality

$3(a-5x) < 1+x\\\\3a-15x < 1+x\\\\3a-1 < x+15x\\\\3a-1 < 16x\\\\16x > 3a-1\\\\x > \frac{3a-1}{16}$

The variable x is larger than (3a-1)/16. If we knew what the value of 'a' was, then we could nail down the range of values for x more concretely.

Do the same for the second inequality

$2 - \frac{x}{2} > 3 + 5(x-a)\\\\2 - \frac{1}{2}x > 3 + 5(x-a)\\\\2 - 0.5x > 3 + 5(x-a)\\\\2 - 0.5x > 3 + 5x-5a\\\\2 + 5a-3 > 5x+0.5x\\\\5a-1 > 5.5x\\\\5.5x < 5a-1\\\\x < \frac{5a-1}{5.5}\\\\$

The variable x is also less than (5a-1)/(5.5)

--------------------------------------

At this point, we have found that

$x > \frac{3a-1}{16}$ and $x < \frac{5a-1}{5.5}$

This means x is between those endpoints, excluding each endpoint

So we can form this compound inequality

$\frac{3a-1}{16} < x < \frac{5a-1}{5.5}$

This tells us the range of where x spans from.

---------------------------------------

If the endpoints are equal, aka the same value, then there's no way to have x have any solutions

So let's equate the endpoints and see what happens

$\frac{3a-1}{16} = \frac{5a-1}{5.5}\\\\5.5(3a-1) = 16(5a-1)\\\\16.5a - 5.5 = 80a - 16\\\\-5.5 + 16 = 80a - 16.5a\\\\10.5 = 63.5a\\\\63.5a = 10.5\\\\a = \frac{10.5}{63.5}\\\\a = \frac{105}{635}\\\\a = \frac{21*5}{127*5}\\\\a = \frac{21}{127}\\\\a \approx 0.16535\\\\$

If 'a' is equal to that value, then the two endpoints of that compound inequality are completely identical; therefore, in that situation, we wouldn't have any solutions for x.

---------------------------------------

The question is: what happens if 'a' is some number smaller than 21/127?

Let's say a = 0

The left endpoint would be (3a-1)/16 = (3*0-1)/16 = -0.0625
The right endpoint would be (5a-1)/(5.5) = (5*0-1)/(5.5) = -0.1818 approximately

So we see that -0.0625 < x < -0.1818; however, upon closer inspection, you should find that such an interval makes no sense. Why not? Because -0.0625 is not smaller than -0.1818. It should be the other way around. No number exists that is between -0.0625 and -0.1818 (I recommend using a number line to help see why this is the case).

Therefore, anything smaller than a = 21/127 will result in having no solutions for x in the original system of inequalities.

--------------------------------------

Now let's try something larger than a = 21/127

Let's say we picked a = 1

The left endpoint would be (3a-1)/16 = (3*1-1)/16 = 0.125
The right endpoint would be (5a-1)/(5.5) = (5*1-1)/(5.5) = 0.7273 approximately

We end up with 0.125 < x < 0.7273 which is now valid because 0.125 is indeed smaller than 0.7273

Therefore, if a > 21/127, then we'll have those original system of inequalities lead to more than one solution for x.

7 0

3 years ago