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2 years ago

Find the area I will mark as brainless thank u and rate 5 starts

Mathematics

1 answer:

Veronika [31]2 years ago

7 0

Answer:

The area is 10

Step-by-step explanation:

The blue area can be divided into a square and a number of triangles, but it is just as easy to subtract the areas of the triangles around the outside:

The blue area is ...

16 -1 -1 -1 -2 -1 = 10 . . . . . . (CCW from upper left)

Also you can count the number of dots between each curve

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Find the solution to the system of equations

Neko [114]

Answer:

x=1

y=-4

Step-by-step explanation:

Since -x-3=y subsitue that in at the top equation

$- x - 3 = - 7x + 3$

$- x = - 7x + 6$

$6x = 6$

$x = 1$

Plug x=1 back into the equation to solve for y.

$y = - 7(1) + 3$

$y = - 4$

3 0

2 years ago

What is 1/9 * 1/10= *please answer fast!

mamaluj [8]

Answer: 1/90

multiply 1/10 and 1/9 across, and you get 1/90

$\frac{1}{9}$ × $\frac{1}{10}$ = $\frac{1}{90}$

hope it helps!!!

6 0

2 years ago

Read 2 more answers

What is the length side of a triangle that has vertices at (-5, -1), (-5, 5), and (3, -1)?

Rasek [7]

The side lengths of triangle are 6 units, 8 units and 10 units.

SOLUTION:

Given that, we have to find what is the length side of a triangle that has vertices at (-5, -1), (-5, 5), and (3, -1)

We know that, distance between two points $P\left(x_{1}, y_{1}\right) \text { and } Q\left(x_{2}, y_{2}\right)$ is given by

$P Q=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

Now,

$\begin{array}{l}{\text { Distance between }(-5,-1) \text { and }(-5,5)=\sqrt{(-5-(-5))^{2}+(5-(-1))^{2}}} \\\\ {\qquad \begin{array}{l}{=\sqrt{(-5-(-5))^{2}+(5-(-1))^{2}}} \\\\ {=\sqrt{(0)^{2}+(5+1)^{2}}=\sqrt{(6)^{2}}=6} \\\\ {=\sqrt{(-5)^{2}+(5+1)^{2}+(5-(-1))^{2}}} \\\\ {=\sqrt{(-8)^{2}+(5+1)^{2}}=\sqrt{64+36}=\sqrt{100}=10} \\\\ {=\sqrt{(3-1)^{2}+(-1-(-1))^{2}}} \\\\ {=\sqrt{(5+3)^{2}+(0)^{2}}=\sqrt{(8)^{2}}=8}\end{array}}\end{array}$