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marta [7]
3 years ago
13

100 POINTS PLEASE PLEASE HELP I NEED THE ANSWER ASAPP ;(

Mathematics
2 answers:
Anestetic [448]3 years ago
8 0

Answer:

y

Step-by-step explanation:

First, let's find the equation of the line.

From the graph, we can pick out two points to use: (0,1) (the y-intercept) and (4,0) (the x-intercept).

Find the slope. Let (0,1) be x₁ and y₁ and let (4,0) be x₂ and y₂

Find the slope:

m=\frac{y_2-y_1}{x_2-x_1}

Substitute:

m=\frac{0-1}{4-0}\\m=-1/4

Use the slope-intercept form:

y=mx+b

Where m is the slope and b is the y-intercept.

We already know the slope is -1/4 and the y-intercept is 1. So, substitute:

y=-\frac{1}{4}x+1

Now, replace the equal sign with an inequality. Note that the line is dotted, so we won't use "or equal to." Also note that the blue, shaded area is <em>below</em> the line. Thus, we will use less than. Therefore:

y

The sentences:

First, we picked out two points from the graph. Next, we used those two points to find the slope. We used the slope and one of the points to write our slope-intercept equation. Afterwards, we can determine from the graph that since the shaded area is below the dotted line, the equation must have a less than.

Naddik [55]3 years ago
7 0

Answer:

y < -1/4 x +1

Step-by-step explanation:

Find two points on the dotted line

( 0,1) and  (4,0)

We can use the slope formula

m = ( y2-y1)/(x2-x1)

    = (0-1)/(4-0)

    = -1/4

The slope is -1/4

The y intercept is where it crosses the y axis  or 1

The equation of the line, using slope intercept form

y = mx+b

y = -1/4x +1

The line is dotted which means there is no equals, just greater than or less than.  It is shaded below which means less than

y < -1/4 x +1

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A particular isotope has a​ half-life of 74 days. If you start with 1 kilogram of this​ isotope, how much will remain after 150
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\bf \stackrel{300~days}{\textit{Amount for Exponential Decay using Half-Life}} \\\\ A=P\left( \frac{1}{2} \right)^{\frac{t}{h}}\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{initial amount}\dotfill &1\\ t=\textit{elapsed time}\dotfill &300\\ h=\textit{half-life}\dotfill &74 \end{cases} \\\\\\ A=1\left( \frac{1}{2} \right)^{\frac{300}{74}}\implies A=1\left( \frac{1}{2} \right)^{\frac{150}{37}}\implies \boxed{A\approx 0.060202}

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Mattie uses the discriminant to determine the number of zeros the quadratic equation 0 = 3x2 – 7x + 4 has. Which best describes
Eduardwww [97]
For ax^2+bx+c=0 the discriminant is b^2-4ac

there are 3 basic cases of what happens for different discriminants
1. if the discriminant is less than 0, then there are no real zeroes
2. if the discriminant is 0, then it has 1 zero
3. if the discriminant is greater than 0, it has 2 zeroes


so given
0=3x^2-7x+4
a=3,b=-7,c=4
thus the discriminant is (-7)^2-4(3)(4)=49-48=1
the discriminant is 1. 1 is positive, thus the equation has 2 zeroes because the discriminant is greater than 0

the answer is the equation has two zeroes because the discriminant is greater than 0
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