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neonofarm [45]
3 years ago
7

What is the area of the figure?​

Mathematics
2 answers:
Alex73 [517]3 years ago
8 0

Answer:5.4 squared inched

Step-by-step explanation:

the shape is a trapezium

Area of trapezium=0.5x(sum of parallel sides) x height

Area of trapezium=0.5x(3.4+2)x2

Area of trapezium=0.5x5.4x2

Area of trapezium=5.4

Oksi-84 [34.3K]3 years ago
6 0

Answer:

5.4 in.

Step-by-step explanation:

Figuring out the area of shapes like these are quite simple, you first have to break apart this shape to make solving this easier. If you draw a line and break off the triangle from the square you will get 2 different shapes. A square with all the sides being 2 inches, and a triangle that is 2 inches tall and 1.4 inches across (you subtract 3.4 by 2). Next you just use the equation (2 * 2) + ((1.4 * 2)/2). Multiply 2 by 2 (which is 4) and you get the area of the square (you multiply the base by the width). And for the triangle you multiple 1.4 by 2 (you get 2.8)... But because it's a triangle you have to divide that number by 2 since the triangle is half of a square. So 2.8 / 2 is going to be 1.4. After that you now have the equation 4 + 1.4 and the answer is going to be 5.4.

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Help with all please
seropon [69]

a) First, draw a graph. The x axis should be numbered: 0, 1, 2, 3, 4, 5, 6, 7, 8. Each number is the last digit of the year. For each year, there is one average score that has to be plotted. Number the y axis of the graph from 0-600, by increments of 50, or something around there. Now, with each year (each x value) place the dot as high as the average score. Repeat with all years. DO NOT draw a line to connect them. A scatter plot is a bunch of points that are not connected.

b) Use the form (y2-y1)/(x2-x1) where (x1,y1) (x2,y2) are any two points from 2001-2006. It does not matter which points, pick any, and assign each of the numbers x1,y1 and x2,y2. Plug it in to the equation, and simplify. This is your slope, also called m. Now plug m into the equation y-y1=m(x-x1) where y1 and x1 are the x and y coordinates of any point from 2001 to 2006. x and y dont have values themselves, they stay there. Now distribute the m into the parenthesis, add y1 to both sides (to cancel it out on the left), and you should be left with the same equation, but now in y=mx+b form. B is where the line crosses the y axis, so put a point on the vertical axis where b is. M is your slope, so every time you go to the right one number, go up M numbers. Draw another point. Repeat this until you can connect these new dots into a line. This line should be on the same graph as your other points, but might not touch all of the scatter plot points. That's still okay, just leave it as is.

c) Find the point where x=6 (meaning the year is 2006) on your line. Find the average test score for that year by seeing where the point lines up along the y axis (draw a straight line from the point to the left to see, if you have to). Take that number and add 1m (the m is the number you used in step B) to it. This predicts what the average score might be after 1 (that's why the 1 is there) year. This is a predicted value, and might not be perfectly correct!! Write/record how close this new, predicted, score is to the actual number, 515, which can be found on the scatter plot.

d) repeat step C exactly as before (still use the year 2006) , but add 4m instead of 1m, to get the predicted score 4 years after 2006 (2010), instead of 1 year after (2007).

I hope I have been of some help! Best of luck!!! :)

3 0
3 years ago
What are the endpoint coordinates for the midsegment of △PQR that is parallel to PQ¯¯¯¯¯?
andriy [413]

Answer:

M(x₄ ,y₄) = (-3.5 , 0.5)  and

N (x₅ ,y₅) = ( -1 , -0.5 )

Step-by-step explanation:

Let the endpoint coordinates for the mid segment of △PQR that is parallel to PQ be

M(x₄ ,y₄) and N(x₅ ,y₅) such that MN || PQ

point P( x₁ , y₁) ≡ ( -3 ,3 )

point Q( x₂ , y₂) ≡ (2 , 1 )

point R( x₂ , y₂) ≡ (-4 , -2)  

To Find:

M(x₄ ,y₄) = ?  and

N (x₅ ,y₅) = ?

Solution:

We have Mid Point Formula as

Mid\ point(x,y)=(\frac{x_{1}+x_{2} }{2}, \frac{y_{1}+y_{2} }{2})

As M is the mid point of PR and N is the mid point of RQ so we will have

Mid\ pointM(x_{4} ,y_{4})=(\frac{x_{1}+x_{3} }{2}, \frac{y_{1}+y_{3} }{2})

Mid\ pointN(x_{5} ,y_{5})=(\frac{x_{2}+x_{3} }{2}, \frac{y_{2}+y_{3} }{2})

Substituting the given value in above equation we get

Mid\ pointM(x_{4} ,y_{4})=(\frac{-3+-4 }{2}, \frac{3+-2} }{2})

∴ Mid\ pointM(x_{4} ,y_{4})=(\frac{-7} }{2}, \frac{1}{2})

∴ Mid\ pointM(x_{4} ,y_{4})=(-3.5, 0.5)

Similarly,

Mid\ pointN(x_{5} ,y_{5})=(\frac{2+-4 }{2}, \frac{1+-2 }{2})

∴ Mid\ pointN(x_{5} ,y_{5})=(\frac{-2 }{2}, \frac{-1}{2})

∴ Mid\ pointN(x_{5} ,y_{5})=(-1, -0.5)

∴ M(x₄ ,y₄) = (-3.5 , 0.5)  and

  N (x₅ ,y₅) = ( -1 , -0.5 )

3 0
3 years ago
Solve for y ; -2y - 2x =4
Deffense [45]
I think this is what you want to do

Your welcome

4 0
3 years ago
Read 2 more answers
Given that the expression x³_ ax² + bx+c leaves the same remainder when divided by x + 1 or x-2, find 'a' in terms of 'b'.​
goblinko [34]

Answer:

Hi,

a=b+3

Step-by-step explanation:

\begin{array}{c||c|c|c|c}&x^3&x^2&x&1\\---&---&---&---&---\\&1&-a&b&c\\x=-1&&-1&a+1&-a-b-1\\---&---&---&---&---\\&1&-a-1&a+b+1&c-a-b-1\\\end{array}

\begin{array}{c||c|c|c|c}&x^3&x^2&x&1\\---&---&---&---&---\\&1&-a&b&c\\x=2&&2&4-2a&2b+8-4a\\---&---&---&---&---\\&1&2-a&b+4-2a&c+2b+8-4a\\\end{array}

Thus:

c+2b+8-4a=c-a-b-1

3b-3a=-9

<u>a=b+3</u>

5 0
2 years ago
A ball is thrown from an initial height of 1 meter with an initial upward velocity of 15m/s. The ball's height h (in meters) aft
lakkis [162]

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\bf t=\cfrac{-(-15)\pm\sqrt{(-15)^2-4(5)(5)}}{2(5)}\implies t=\cfrac{15\pm\sqrt{225-100}}{10} \\\\\\ t=\cfrac{15\pm\sqrt{125}}{10}\implies t=\cfrac{15\pm\sqrt{5^2 \cdot 5}}{10}\implies t=\cfrac{\stackrel{3}{~~\begin{matrix} 15 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}\pm ~~\begin{matrix} 5 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~\sqrt{5}}{\underset{2}{~~\begin{matrix} 10 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}}

\bf t=\cfrac{3\pm \sqrt{5}}{2}\implies t= \begin{cases} \frac{3+ \sqrt{5}}{2} \approx 2.618\\\\ \frac{3- \sqrt{5}}{2}\approx 0.382 \end{cases}

8 0
3 years ago
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