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2 years ago

Someone complete this. Do I shave above the line or below the line. It’s it soils or dotted.

Mathematics

1 answer:

den301095 [7]2 years ago

8 0

Answer:

So this one is dotted and this here is 2 is on the line right point 3 and -1 is below the line after 2

Step-by-step explanation:

-Remember:

+solid is >= or <=

+Dotted is > or <

-Slope is -1/2 and y-intercept: 3.

<h3>Finally Select a point on one side of the line test it, and then shade it </h3>

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What is the product of (-2x)(-6x^4)?

yaroslaw [1]

Answer:

12x^5

Step-by-step explanation:

(-2x)(-6x^4)

Multiply the numbers

-2 * -6 = 12

Multiply the variables

x* x^4 = x^5

12x^5

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2 years ago

Answer right with proof please.

Diano4ka-milaya [45]

Answer:

the answer is a

Step-by-step explanation:

you times the 1/6 by 2 to get 2/12 then 3/4 by 3 to get 9/12 so all your fractions have the same denominator and the do the rest of the math with the new fractions and the same whole numbers that went with them. hope this helps

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2 years ago

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To which of the following sets of numbers does ⅞ belong?

artcher [175]

Answer:

A. Rational number

Step-by-step explanation:

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2 years ago

The temperature in the Arctic Circle at 9:00am was 22.8° below zero. By 6:00pm on the same day, the temperature had decreased by

rosijanka [135]

4.65 is the new temperature.

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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