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2 years ago

Help me with this please

Mathematics

1 answer:

finlep [7]2 years ago

3 0

50+20=70
70-100=30

12+34=46
100-46=54

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I just want to make sure I am right.<br> Its Mathematics

docker41 [41]

It seems about right

8 0

3 years ago

Rectangle JKLM is shown.

ad-work [718]

Answer:

Step-by-step explanation:

We can use the Pythagorean theorem to solve this.

(13)^2 = (8)^2+x^2

169 = 64 + x^2

x^2 = 105

x is approximately 10.2, so B

6 0

3 years ago

A rectangle has an area of 36 square units. As the length and the width change, what do you know about their product? Is this le

m_a_m_a [10]

Area = 36 = length*width

Then, while the area is constant the product length*width is constant.

Length and area are inversely related.

That means that to keep the same product (area), length vary in inverse proportion to width.

If you increase length, widht has to decrease to satisfy this relation:

width = 36/length

Of course, you can also say: length = 36 / width.

5 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

Anyone know this? Help?

alisha [4.7K]

maybe a? because it’s reflecting off of the x axis

8 0

3 years ago