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3 years ago

Equation that represents the line that passes through the point (-2,-1) and has a slope of 5

1 answer:

5 0

Straight line:
y = mx + c
m is the slope of the graph and c is the y-intercept
In this case, m = 5 as stated in the question, so...
y = 5x + c
By substituting the given co-ordinates (-2,-1) into this equation, we can find c
-1 = 5(-2) + c
-1 = -10 + c
-1 + 10 = -10 + c + 10 (Add 10 to both sides)
9 = c
c = 9
Put c = 9 into the equation:
y = 5x + 9

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Classify each polynomial according to its degree and type. Look at the screenshot below!

Ahat [919]

Answer:

Monomial Binominal Trimoninal

Degree 1: -9x x + 6

Degree 2: -4 $x^{2}$ $4x^{2} -x$ $x -3x^{2} + 1$

Degree 3: 4 $x^{3}$ 5 - 2 $x^{3}$ $3x^{2} + 3x^{3} -10$

Step-by-step explanation:

I do not know how to explain how I got to this answer, but here is the answer.

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1 year ago

What is the median of this data? 43, 28, 39, 19, 24, 47, 31, 20, 37

kotegsom [21]

32 Just add all the numbers up and divide by how many numbers there are (9).

7 0

3 years ago

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The graph from the US Department of Agriculture shows the growth of farmers markets in the United States. What was the first yea

Helen [10]

The answer is C.) 2002
The graph shows that the previous years have been under that, the number of markets is shown on the y axis

3 0

3 years ago

Area of a triangle with points at (-9,5), (6,10), and (2,-10)

Ann [662]

First we are going to draw the triangle using the given coordinates.
Next, we are going to use the distance formula to find the sides of our triangle.
Distance formula: $d= \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

Distance from point A to point B:
$d_{AB}= \sqrt{[6-(-9)]^2+(10-5)^2}$
$d_{AB}= \sqrt{(6+9)^2+(10-5)^2}$
$d_{AB}= \sqrt{(15)^2+(5)^2}$
$d_{AB}= \sqrt{225+25}$
$d_{AB}= \sqrt{250}$
$d_{AB}=15.81$

Distance from point A to point C:
$d_{AC}= \sqrt{[2-(-9)]^2+(-10-5)^2}$
$d_{AC}= \sqrt{(2+9)^2+(-10-5)^2}$
$d_{AC}= \sqrt{11^2+(-15)^2}$
$d_{AC}= \sqrt{121+225}$
$d_{AC}= \sqrt{346}$
$d_{AC}= 18.60$

Distance from point B from point C
$d_{BC}= \sqrt{(2-6)^2+(-10-10)^2}$
$d_{BC}= \sqrt{(-4)^2+(-20)^2}$
$d_{BC}= \sqrt{16+400}$
$d_{BC}= \sqrt{416}$
$d_{BC}=20.40$

Now, we are going to find the semi-perimeter of our triangle using the semi-perimeter formula:
$s= \frac{AB+AC+BC}{2}$
$s= \frac{15.81+18.60+20.40}{2}$
$s= \frac{54.81}{2}$
$s=27.41$

Finally, to find the area of our triangle, we are going to use Heron's formula:
$A= \sqrt{s(s-AB)(s-AC)(s-BC)}$
$A=\sqrt{27.41(27.41-15.81)(27.41-18.60)(27.41-20.40)}$
$A= \sqrt{27.41(11.6)(8.81)(7.01)}$
$A=140.13$

We can conclude that the perimeter of our triangle is 140.13 square units.

3 0

2 years ago

What two numbers plus 35 equal 180

Andrews [41]

X^2+35=180?
Subtract 35 on each side

x^2=145
Take the square root

x= -sqrt(145)
+sqrt 145...

I hope this helps!!

4 0

2 years ago

Read 2 more answers