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1 year ago

Quick algebra 1 question for 15 points!

Mathematics

1 answer:

Mamont248 [21]1 year ago

6 0

The desired linear functions are given as follows:

a) x = a.

b) y = b.

<h3>What is a linear function?</h3>

A linear function is modeled by:

y = mx + b

In which:

m is the slope, which is the rate of change, that is, by how much y changes when x changes by 1.

b is the y-intercept, which is the value of y when x = 0, and can also be interpreted as the initial value of the function.

A vertical line has undetermined slope, hence it is defined by x = a, containing point (a,b).

A horizontal line has slope of 0, hence it is defined by y = b, containing point (a,b).

More can be learned about linear functions at brainly.com/question/24808124

#SPJ1

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Draw a line representing the "rise" and a line representing the "run" of the line. State the slope of the line in simplest form.

Nat2105 [25]

Answer:

Step-by-step explanation:

First choose two points on the line,

Let the points are A and B.

From point A to B we are moving downwards by (5 - 1 = 4 units)

Therefore, Rise = -4

Horizontal distance between A and B = 2 units

Therefore, Run = 2

Since, slope of a line = $\frac{\text{Rise}}{\text{Run}}$

Slope = $\frac{-4}{2}$

= -2

7 0

2 years ago

Harry bought a sweater that only cost $5 when it normally costs $35. What percent was the discount? Provide an explanation. A. 1

erastova [34]

Answer:

D. 85.7%

Step-by-step explanation:

its figuring out the percentage decrease

so you would go

35-5= 30

30/35 x 100 = 85.7

8 0

2 years ago

Read 2 more answers

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.