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Alenkasestr [34]
2 years ago
13

Find the slope of the line graphed below.

Mathematics
2 answers:
AleksandrR [38]2 years ago
6 0

Answer:

2 is the slope i think

Step-by-step explanation:

Ganezh [65]2 years ago
5 0

Answer:

slope = \frac{3}{2}

Step-by-step explanation:

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 2x- 3y=9   5x+3y=12   How would i solve this? I dont understand this and i really need help with it!Also I have to graph the an
dedylja [7]
\begin{cases} 2x- 3y=9 \\5x+3y=12 \end{cases}\\ + \ \ ------- \\ 7x=21 \ \ /: 7\\ \\x=\frac{21}{3}\\ \\ x=3 \\ \\2x- 3y=9

2*3-3y=9\\ \\6-3y =9\\ \\-3y=9-6\\ \\-3y=3 \ \ / :(-3)\\ \\x=-1\\ \\ \begin{cases} x=3\\y= -1 \end{cases}

graph\\ \\\begin{cases} 2x- 3y=9 \\5x+3y=12 \end{cases}\\ \\ \begin{cases} - 3y=-2x+9\ \ /:(-3) \\ 3y=-5x+12 \ \ /:3 \end{cases}\\ \\ \begin{cases} y= \frac{2}{3}x-3 \\ y=-\frac{5}{3}x+4 \end{cases}


7 0
3 years ago
David found and factored out the GCF of the polynomial 80b4 – 32b2c3 + 48b4c. His work is below.
Vikentia [17]

Answer:

Given Polynomial:

80b^4-32b^2c^3+48b^4c

Factors of Coefficient of terms

80 = 5 × 16

32 = 2 × 16

48 = 3 × 16

Common factor of the coefficient of all term is 16.

Each term contain variable. So the Minimum power of b is common from all terms.

Common from all variable part comes b².

So, Common factor of the polynomial = 16b²

⇒ 16b² ( 5b² ) - 16b² ( 2c³ ) + 16b² ( 3b²c )

⇒ 16b² ( 5b² - 2c³ + 3b²c )

Therefore, Statements that are true about David's word are:

The GCF of the coefficients is correct.

The variable c is not common to all terms, so a power of c should not have been factored out.

In step 6, David applied the distributive property

8 0
3 years ago
Read 2 more answers
HELP!! Algebra help!! Will give stars thank u so much <333
Anna35 [415]

Answers:

  • Part a)  \bf{\sqrt{x^2+(x^2-3)^2}
  • Part b)  3
  • Part c)   2.24
  • Part d)  1.58

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

Work Shown:

Part (a)

The origin is the point (0,0) which we'll make the first point, so let (x1,y1) = (0,0)

The other point is of the form (x,y) where y = x^2-3. So the point can be stated as (x2,y2) = (x,y). We'll replace y with x^2-3

We apply the distance formula to say...

d = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\\\\d = \sqrt{(0-x)^2+(0-y)^2}\\\\d = \sqrt{(0-x)^2+(-y)^2}\\\\d = \sqrt{x^2 + y^2}\\\\d = \sqrt{x^2 + (x^2-3)^2}\\\\

We could expand things out and combine like terms, but that's just extra unneeded work in my opinion.

Saying d = \sqrt{x^2 + (x^2-3)^2} is the same as writing d = sqrt(x^2-(x^2-3)^2)

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

Part (b)

Plug in x = 0 and you should find the following

d(x) = \sqrt{x^2 + (x^2-3)^2}\\\\d(0) = \sqrt{0^2 + (0^2-3)^2}\\\\d(0) = \sqrt{(-3)^2}\\\\d(0) = \sqrt{9}\\\\d(0) = 3\\\\

This says that the point (x,y) = (0,3) is 3 units away from the origin (0,0).

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

Part (c)

Repeat for x = 1

d(x) = \sqrt{x^2 + (x^2-3)^2}\\\\d(1) = \sqrt{1^2 + (1^2-3)^2}\\\\d(1) = \sqrt{1 + (1-3)^2}\\\\d(1) = \sqrt{1 + (-2)^2}\\\\d(1) = \sqrt{1 + 4}\\\\d(1) = \sqrt{5}\\\\d(1) \approx 2.23606797749979\\\\d(1) \approx 2.24\\\\

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

Part (d)

Graph the d(x) function found back in part (a)

Use the minimum function on your graphing calculator to find the lowest point such that x > 0.

See the diagram below. I used GeoGebra to make the graph. Desmos probably has a similar feature (but I'm not entirely sure). If you have a TI83 or TI84, then your calculator has the minimum function feature.

The red point of this diagram is what we're after. That point is approximately (1.58, 1.66)

This means the smallest d can get is d = 1.66 and it happens when x = 1.58 approximately.

6 0
2 years ago
Find the slope of the function f ( x ) = 5x/2 + 3, by the definition of limit. Express answer as a fraction using the "/" key as
musickatia [10]

\huge{\boxed{\frac{5}{2}}}

Slope-intercept form is f(x)=mx+b, where m is the slope and b is the y-intercept.

\frac{5x}{2}=\frac{5}{2}x, so change the equation to represent this. f(x)=\frac{5}{2}x+3

Great! Now the function is in slope-intercept form, so we can just see that m, or the slope, is equal to \boxed{\frac{5}{2}}

3 0
3 years ago
Which line plot displays a data set with an outlier?<br><br> Plz answer quickly!
iren2701 [21]
The answer is the first data set because remember, an outlier is a number in a data set that is extremely far away from all of the other numbers.

You can easily tell which one is the outlier, the first data set's outlier is 13, because it is was placed so far away from the other data points.

We know that the first data set is correct because all of the other data sets have numbers that are clustered together.

~Hope I helped!~
7 0
3 years ago
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