7.Show that quadrilateral FGHJ is a trapezoid, but is not a parallelogram.

PLEASE HELP I GOT IT WRONG EARLIER AND THE WRONG ANSWER WAS H PLS HELP!!

dybincka [34]

Answer:

H. 39.6

Step-by-step explanation:

i promise this is the right answer!

there is a mean, mode, range etc. calculator that can help you!

4 0

3 years ago

Graph the line that passes through the points (6, -4) and (7, -4) and determine the equation of the line.

Fantom [35]

Answer:

the equation in slope intercept is y = -4.

ot your points and it is a straight line across..no slope.

8 0

3 years ago

NEED IT ASAP What is the value of x in the diagram below? A.18/ ROOT 3 B.18 ROOT 2/ROOT 3 C.18 ROOT 3/ ROOT 2 D. 18 ROOT 2

Ostrovityanka [42]

Answer:

x = B. 18 ROOT 2/ROOT 3

Step-by-step explanation:

There are two Triangles in the diagram above.

Step 1

We would solve for Triangle A First

Using Trigonometric function Cosine

cos θ = adjacent /hypotenuse

θ = 30°

Adjacent = 9 units

Hypotenuse = unknown

cos 30= 9/ Hypothenuse

Cross Multiply

cos 30 × Hypotenuse = 9

Hypotenuse = 9 / cos 30

cos 30 in surd form = √3/2

Hypotenuse = 9/√3/2

Hypotenuse = 9 × 2/√3

= 18/√3 units

Step 2

We would solve for the upper triangle = Triangle B

We are looking for x

θ = 45°

For Triangle B, the Hypotenuse we solved for in Triangle A is equivalent to the adjacent in Triangle B

Therefore ,

Hypotenuse for Triangle A = Adjacent side for Triangle B

θ = 45°

Adjacent = 18/√3 units

Hypotenuse = x = unknown

We would solve for this using Trigonometric function cosine

cos 45 = 18/√3 units / Hypothenuse

Cross Multiply

cos 45 × Hypotenuse = 18/√3 units

Hypotenuse = 18/√3 units / cos 45

cos 45 in surd form = 1/√2

Subtituting, we have

Hypotenuse (x) = 18/√3 units / 1/√2

Hypotenuse (x) = 18/√3 units ×√2/1

= 18× √2/ √3 units

= 18 √2/√3 units

Therefore x = Option B. 18 ROOT 2/ROOT 3

3 0

3 years ago

Wilhema bought 6 bars of soap for $12. The next day, Sophia bought 10 bars of the same kind of soap for $20. What is the cost of

kaheart [24]

Answer:

$2

Step-by-step explanation:

To get this number you can divide the cost by the amount of bars of soap.

12/6=2

20/10=2

Hope this helps!

8 0

3 years ago

3y''-6y'+6y=e*x sexcx

Simora [160]

From the homogeneous part of the ODE, we can get two fundamental solutions. The characteristic equation is

$3r^2-6r+6=0\iff r^2-2r+2=0$

which has roots at $r=1\pm i$ . This admits the two fundamental solutions

$y_1=e^x\cos x$
$y_2=e^x\sin x$

The particular solution is easiest to obtain via variation of parameters. We're looking for a solution of the form

$y_p=u_1y_1+u_2y_2$

where

$u_1=-\displaystyle\frac13\int\frac{y_2e^x\sec x}{W(y_1,y_2)}\,\mathrm dx$
$u_2=\displaystyle\frac13\int\frac{y_1e^x\sec x}{W(y_1,y_2)}\,\mathrm dx$

and $W(y_1,y_2)$ is the Wronskian of the fundamental solutions. We have

$W(e^x\cos x,e^x\sin x)=\begin{vmatrix}e^x\cos x&e^x\sin x\\e^x(\cos x-\sin x)&e^x(\cos x+\sin x)\end{vmatrix}=e^{2x}$

and so

$u_1=-\displaystyle\frac13\int\frac{e^{2x}\sin x\sec x}{e^{2x}}\,\mathrm dx=-\int\tan x\,\mathrm dx$
$u_1=\dfrac13\ln|\cos x|$

$u_2=\displaystyle\frac13\int\frac{e^{2x}\cos x\sec x}{e^{2x}}\,\mathrm dx=\int\mathrm dx$
$u_2=\dfrac13x$

Therefore the particular solution is

$y_p=\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x$

so that the general solution to the ODE is

$y=C_1e^x\cos x+C_2e^x\sin x+\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x$

7 0

3 years ago