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2 years ago

Please hurry up .......

Mathematics

2 answers:

Xelga [282]2 years ago

7 0

The answer is most likely D.

andre [41]2 years ago

4 0

Answer:

I think the answer is D.

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Plzzz help me with this I’ll give brainliest

scoray [572]

Answer18:

The quadrilateral ABCD is not a parallelogram

Answer19:

The quadrilateral ABCD is a parallelogram

Step-by-step explanation:

For question 18:

Given that vertices of a quadrilateral are A(-4,-1), B(-4,6), C(2,6) and D(2,-4)

The slope of a line is given m= $\frac{Y2-Y1}{X2-X1}$

Now,

The slope of a line AB:

m= $\frac{Y2-Y1}{X2-X1}$

m= $\frac{6-(-1)}{(-4)-(-4)}$

m= $\frac{7}{0}$

The slope is 90 degree

The slope of a line BC:

m= $\frac{Y2-Y1}{X2-X1}$

m= $\frac{6-6}{(-4)-(-1)}$

m= $\frac{0}{(-3)}$

The slope is zero degree

The slope of a line CD:

m= $\frac{Y2-Y1}{X2-X1}$

m= $\frac{(-4)-6}{2-2}$

m= $\frac{-10}{0}$

The slope is 90 degree

The slope of a line DA:

m= $\frac{Y2-Y1}{X2-X1}$

m= $\frac{(-1)-(-4)}{(-4)-(2)}$

m= $\frac{3}{-6}$

m= $\frac{-1}{2}$

The slope of the only line AB and CD are the same.

Thus, The quadrilateral ABCD is not a parallelogram

For question 19:

Given that vertices of a quadrilateral are A(-2,3), B(3,2), C(2,-1) and D(-3,0)

The slope of a line is given m= $\frac{Y2-Y1}{X2-X1}$

Now,

The slope of a line AB:

m= $\frac{Y2-Y1}{X2-X1}$

m= $\frac{2-3}{3-(-2)}$

m= $\frac{-1}{5}$

The slope of a line BC:

m= $\frac{Y2-Y1}{X2-X1}$

m= $\frac{(-1)-2}{2-3}$

m= $\frac{-3}{-1}$

m=3

The slope of a line CD:

m= $\frac{Y2-Y1}{X2-X1}$

m= $\frac{0-(-1)}{(-3)-2}$

m= $\frac{-1}{5}$

The slope of a line DA:

m= $\frac{Y2-Y1}{X2-X1}$

m= $\frac{3-0}{(-2)-(-3)}$

m=3

The slope of the line AB and CD are the same

The slope of the line BC and DA are the same

Thus, The quadrilateral ABCD is a parallelogram

3 0

3 years ago

Set up but do not solve for the appropriate particular solution yp for the differential equation y′′+4y=5xcos(2x) using the Meth

taurus [48]

Answer:

So, solution of the differential equation is

$y(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos 2x+c_1e^{-2it}+c_2e^{2it}\\$

Step-by-step explanation:

We have the given differential equation: y′′+4y=5xcos(2x)

We use the Method of Undetermined Coefficients.

We first solve the homogeneous differential equation y′′+4y=0.

$y''+4y=0\\\\r^2+4=0\\\\r=\pm2i\\\\$

It is a homogeneous solution:

$y_h(t)=c_1e^{-2i t}+c_2e^{2i t}$

Now, we finding a particular solution.

$y_p(t)=A5x\cos 2x\\\\y'_p(t)=A5\cos 2x-A10x\sin 2x\\\\y''_p(t)=-A20\sin 2x-A20x\cos 2x\\\\\\\implies y''+4y=5x\cos 2x\\\\-A20\sin 2x-A20x\cos 2x+4\cdot A5x\cos 2x=5x\cos 2x\\\\-A20\sin 2x=5x\cos 2x\\\\A=-\frac{x}{4} \cot 2x\\$

we get

$y_p(t)=A5\cos 2x\\\\y_p(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos 2x\\\\\\y(t)=y_p(t)+y_h(t)\\\\y(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos 2x+c_1e^{-2it}+c_2e^{2it}\\$

So, solution of the differential equation is

$y(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos 2x+c_1e^{-2it}+c_2e^{2it}\\$

7 0

3 years ago

Which graph shows the equation V = 4 + 2t, where V is the total volume of water in a bucket and t is the elapsed time in minutes

kodGreya [7K]

1. Subtract 4 from both sides
v - 4 = 2t
2. Divide both sides by 2
v - 4/2 = t
3. Switch sides
t = v - 4/2

8 0

3 years ago

Read 2 more answers

The sum of two numbers is 21 . One number is 12 more than the other one. Find the numbers.

vaieri [72.5K]

The 2 numbers would be 9 and 12

8 0

2 years ago

Read 2 more answers

Help meeeeedkxkdjsjs

Alex

18 - (8 - 3 • (2t + 5)) = 0

6t + 25 = 0

3.1 Solve : 6t+25 = 0

Subtract 25 from both sides of the equation :
6t = -25
Divide both sides of the equation by 6:
t = -25/6 = -4.167

Final answer is

t = -25/6 = -4.167

5 0

3 years ago