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3 years ago

Which of the following values of x is NOT a solution to the following inequality?

Mathematics

1 answer:

vova2212 [387]3 years ago

8 0

Answer:

A. 1

Step-by-step explanation: You have to substitute the numbers in for x

A. 1:

-5x -1 < -9

-5(1) - 1 < -9

-5 - 1 < -9

-6 < -9

B. 2:

-5x - 1 < -9

-5(2) - 1 < -9

-10 - 1 < -9

-11 < -9

C. 3:

-5x - 1 < -9

-5(3) - 1 < - 9

-15 - 1 < -9

-16 < -9

D. 4:

-5x - 1 < -9

-5(4) - 1 < -9

-20 - 1 < -9

-21 < -9

Remember, a negative plus another negative is a negative

Hope this helps you!!! :)

You might be interested in

Share £200 between Alice and Brian in the ratio 3 : 5<br>

Pepsi [2]

Answer:

75:125=3:5

Step-by-step explanation:

1. Add the total number of parts

3+5=8

2. Find out how much one part is

200/8=25

3. Multiply each ratio by 25

3*25=75

5*25=125

4. Check they add up to 200

75+125=200

8 0

2 years ago

Read 2 more answers

Can someone please help me with this?

jok3333 [9.3K]

The first answer is 2
The Second is 8
The third is, if the w changes then the p would change as well resulting to multiplication

Hope this helps!!!

6 0

2 years ago

You are a lifeguard and spot a drowning child 60 meters along the shore and 40 meters from the shore to the child. You run along

sukhopar [10]

Answer:

The lifeguard should run across the shore a distance of 48.074 m before jumpng into the water in order to minimize the time to reach the child.

Step-by-step explanation:

This is a problem of optimization.

We have to minimize the time it takes for the lifeguard to reach the child.

The time can be calculated by dividing the distance by the speed for each section.

The distance in the shore and in the water depends on when the lifeguard gets in the water. We use the variable x to model this, as seen in the picture attached.

Then, the distance in the shore is d_b=x and the distance swimming can be calculated using the Pithagorean theorem:

$d_s^2=(60-x)^2+40^2=60^2-120x+x^2+40^2=x^2-120x+5200\\\\d_s=\sqrt{x^2-120x+5200}$

Then, the time (speed divided by distance) is:

$t=d_b/v_b+d_s/v_s\\\\t=x/4+\sqrt{x^2-120x+5200}/1.1$

To optimize this function we have to derive and equal to zero:

$\dfrac{dt}{dx}=\dfrac{1}{4}+\dfrac{1}{1.1}(\dfrac{1}{2})\dfrac{2x-120}{\sqrt{x^2-120x+5200}} \\\\\\\dfrac{dt}{dx}=\dfrac{1}{4} +\dfrac{1}{1.1} \dfrac{x-60}{\sqrt{x^2-120x+5200}} =0\\\\\\ \dfrac{x-60}{\sqrt{x^2-120x+5200}} =\dfrac{1.1}{4}=\dfrac{2}{7}\\\\\\ x-60=\dfrac{2}{7}\sqrt{x^2-120x+5200}\\\\\\(x-60)^2=\dfrac{2^2}{7^2}(x^2-120x+5200)\\\\\\(x-60)^2=\dfrac{4}{49}[(x-60)^2+40^2]\\\\\\(1-4/49)(x-60)^2=4*40^2/49=6400/49\\\\(45/49)(x-60)^2=6400/49\\\\45(x-60)^2=6400\\\\$

$x$

As $d_b=x$ , the lifeguard should run across the shore a distance of 48.074 m before jumpng into the water in order to minimize the time to reach the child.