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2 years ago

Find the area. plzz help

Mathematics

1 answer:

12345 [234]2 years ago

6 0

The square has a side length of 9.

The area of the full square would be 9^2 = 9 x 9 = 81 square cm.

The area of the triangle is 1/2 x 4 x 3 = 6 square cm.

The area of the shape = 81 -6 = 75 square cm.

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What is the value of x so that the line segment with endpoints W(x, −2) and X(5, −4) is parallel to the line segment with endpoi

coldgirl [10]

Answer:

x equals six start fraction one over two end fraction

Step-by-step explanation:

Segments which are parallel have the same slope. Find the slope of of YZ. Then using that value, find the slope WX and solve for the value of x.

Slope of YZ is:

$m = \frac{y_2-y_1}{x_2-x_1}=\frac{2-6}{2-5}=\frac{-4}{-3}=\frac{4}{3}$

Since they are parallel, then WX has a slope of 4/3 too.

Slope of WX is:

$m = \frac{y_2-y_1}{x_2-x_1}\\\\\frac{4}{3}=\frac{-2--4}{x-5}\\\\\frac{4}{3}=\frac{2}{x-5}\\\\\frac{4}{3} (x-5) = 2\\\\4x -20=6\\\\4x = 26\\\\x = 6\frac{1}{2}$

5 0

2 years ago

The segments shown below could form a triangle.<br> help please :)

leonid [27]

That’s false because two lines are to short so it’s false

8 0

2 years ago

Read 2 more answers

Mr. Martinez walks 5 miles in 2 hours at a steady pace. How far can he walk in 1 hour and 15 minutes

otez555 [7]

Junaid Mirza
Apr 5, 2018
2.8125 mile

Explanation:
Speed
=
Distance
Time
Speed
=
5 mile
2 hr
=
2.25 miles/hr
1 hr 15 min
=
1 hr
+
1
4

hr
=
1.25 hr
Distance = Speed × Time
=
2.25

miles
hr
×
1.25
hr
=
2.8125 mile

5 0

3 years ago

Use the table of integrals, or a computer or calculator with symbolic integration capabilities, to find the indefinite integral.

andriy [413]

Answer:

$\frac{2}{3}(\frac{1}{5}ln|3x-5|-\frac{1}{5}ln|x|)+C$

Step-by-step explanation:

We have been given a indefinite integral $\int \frac{2}{3x\left(3x-5\right)}dx$ . We are asked to find the indefinite integral.

We will use partial fraction formula to solve our given problem.

$\frac{2}{3x\left(3x-5\right)}=\frac{3}{5(3x-5)}-\frac{1}{5x}$

$\int \frac{2}{3x\left(3x-5\right)}dx=\frac{2}{3}\int \frac{1}{x\left(3x-5\right)}dx$

$\frac{2}{3}\int \frac{1}{x\left(3x-5\right)}dx=\frac{2}{3}\int \frac{3}{5(3x-5)}-\frac{1}{5x}dx$

Using difference rule of integrals, we will get:

$\frac{2}{3}(\int \frac{3}{5(3x-5)}dx-\int \frac{1}{5x}dx)$

Now, we need to use u-substitution as:

Let $u=3x-5$ .

$\frac{du}{dx}=3$

$dx=\frac{1}{3}du$

$\int \frac{3}{5(3x-5)}dx= \frac{3}{5}\int \frac{1}{(u)}*\frac{1}{3}du=\frac{3}{5}*\frac{1}{3}\int \frac{1}{(u)}du=\frac{1}{5}ln|u|=\frac{1}{5}ln|3x-5|$

$\int \frac{1}{5x}dx=\frac{1}{5}\int \frac{1}{x}dx=\frac{1}{5}ln|x|$

Substitute back these values:

$\frac{2}{3}(\int \frac{3}{5(3x-5)}dx-\int \frac{1}{5x}dx)=\frac{2}{3}(\frac{1}{5}ln|3x-5|-\frac{1}{5}ln|x|)$

Let us add a constant C.

$\frac{2}{3}(\frac{1}{5}ln|3x-5|-\frac{1}{5}ln|x|)+C$

Therefore, our required integral would be $\frac{2}{3}(\frac{1}{5}ln|3x-5|-\frac{1}{5}ln|x|)+C$ .

5 0

3 years ago

Someonnnne helllppppp !!!

WARRIOR [948]

First, remove the parentheses:

7x+7=1

Then, subtract 7 from both sides of the problem.

7x-6

Then, divide 7 from 7x, and divide 7 to -6.

x= -6/7 or -0.85

6 0

3 years ago

Find the area. plzz help​

Find the area. plzz help