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3 years ago

Area of trapezoid? i keep getting it wrong lol the shape is confusing to me please help

Mathematics

1 answer:

Rina8888 [55]3 years ago

6 0

Answer:

Area= $154m^{2}$

Step-by-step explanation:

A= $\frac{a+b}{2} h=\frac{12+16}{2} (11)=154$

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Find the sum of the first 31 terms of the following series, to the nearest integer. 5,14,23,..

UkoKoshka [18]

Answer:

Step-by-step explanation:

5+14 is 19

19+23 is 4242

5 0

4 years ago

What is 0.54 repeating

Phoenix [80]

If you mean as a fraction:0.54 repeating as a fraction is 54/99.

7 0

3 years ago

Bianca is making scarves to sell. She has 33 peices of blue fabric, 37 peices of green fabric, and 41 peices of red fabric. Supp

Masja [62]

Answer:

I did my work, and I got 37 scarves. I feel really bad bc I am not 100% sure, and I hate when I am not 100% sure.

Step-by-step explanation:

For if you need to show your work:

33 pieces of blue fabric --> it can make 11 scarves exactly

37 pieces of green fabric --> it can make 12 scarves, with 1 piece of fabric left over

41 pieces of red fabric --> it can make 13 scarves, with 2 pieces of fabric left over

Left over fabric --> there are 3 pieces of fabric left over; 1 green and 2 red. This results in 1 more scarf

So lets add up our scarves: 11 scarves + 12 scarves + 13 scarves + 1 more scarf

That equals 37 scarves!

I hope this helps. Please mark brainliest, I would really appreciate it if you did

8 0

4 years ago

Show that ( 2xy4 + 1/ (x + y2) ) dx + ( 4x2 y3 + 2y/ (x + y2) ) dy = 0 is exact, and find the solution. Find c if y(1) = 2.

fredd [130]

$\dfrac{\partial\left(2xy^4+\frac1{x+y^2}\right)}{\partial y}=8xy^3-\dfrac{2y}{(x+y^2)^2}$

$\dfrac{\partial\left(4x^2y^3+\frac{2y}{x+y^2}\right)}{\partial x}=8xy^3-\dfrac{2y}{(x+y^2)^2}$

so the ODE is indeed exact and there is a solution of the form $F(x,y)=C$ . We have

$\dfrac{\partial F}{\partial x}=2xy^4+\dfrac1{x+y^2}\implies F(x,y)=x^2y^4+\ln(x+y^2)+f(y)$

$\dfrac{\partial F}{\partial y}=4x^2y^3+\dfrac{2y}{x+y^2}=4x^2y^3+\dfrac{2y}{x+y^2}+f'(y)$

$f'(y)=0\implies f(y)=C$

$\implies F(x,y)=x^2y^3+\ln(x+y^2)=C$

With $y(1)=2$ , we have

$8+\ln9=C$

$\boxed{x^2y^3+\ln(x+y^2)=8+\ln9}$

8 0

3 years ago

What is the midpoint of the segment shown below?

aleksandr82 [10.1K]

Answer:

$Mid\ points = (\frac{5}{2} , \frac{7}{2})$ .

Step-by-step explanation:

Given : End points of line are (3, 5) and ( 2 , 2) .

To find : What is the midpoint of the line .

Solution : We have given

End points(3, 5) and ( 2 , 2) .

$Mid\ points = (\frac{x_{1}+x_{2}}{2} , \frac{y_{1}+y_{2}}{2})$ .

Here ,

$x_{1} = 3.\\x_{2} = 2.\\y_{1} = 5.\\y_{2} = 2$ .

The plug the values in given formula.

$Mid\ points = (\frac{3+ 2}{2} , \frac{ 5+2}{2})$

$Mid\ points = (\frac{5}{2} , \frac{7}{2})$ .

Therefore, D). $Mid\ points = (\frac{5}{2} , \frac{7}{2})$ .

4 0

4 years ago