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2 years ago

Solve similar triangles solve for x can someone answer please help

Mathematics

1 answer:

zlopas [31]2 years ago

8 0

Answer:

Step-by-step explanation:

ΔABC & ΔADE are similar triangles

$\frac{AD}{AB}=\frac{DE}{BC}\\\\\frac{12}{4}=\frac{12}{x}\\$

Cross multiply,

12*x = 12*4

$x=12*\frac{4}{12}\\$

x = 1 * 4

x = 4

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HELP PLEASEEEE. The question is attached. I would appreciate if someone could help me.

Blizzard [7]

a counterclockwise rotation about the origin of 90°

The coordinates of P(3, 3), Q(5, 3), R(5, 7)

The coordinates of P'(- 3, 3 ), Q'(- 3, 5), R'(- 7, 5)

Note that the y-coordinate of the image is the negative of the original, while the x-coordinate of the original becomes the y-coordinate of the image

The rotation which does this is a counterclockwise rotation about the origin of 90°

a point (x, y ) → (- y, x )

8 0

3 years ago

A norman window is constructed by adjoining a semicircle to the top of an ordinary rectangular. Find the dimensions of a norman

Yanka [14]

Answer:

$W\approx 8.72$ and $L\approx 15.57$ .

Step-by-step explanation:

Please find the attachment.

We have been given that a norman window is constructed by adjoining a semicircle to the top of an ordinary rectangular. The total perimeter is 38 feet.

The perimeter of the window will be equal to three sides of rectangle plus half the perimeter of circle. We can represent our given information in an equation as:

$2L+W+\frac{1}{2}(2\pi r)=38$

We can see that diameter of semicircle is W. We know that diameter is twice the radius, so we will get:

$2L+W+\frac{1}{2}(2r\pi)=38$

$2L+W+\frac{\pi}{2}W=38$

Let us find area of window equation as:

$\text{Area}=W\cdot L+\frac{1}{2}(\pi r^2)$

$\text{Area}=W\cdot L+\frac{1}{2}(\pi (\frac{W}{2})^2)$

$\text{Area}=W\cdot L+\frac{\pi}{2}(\frac{W}{2})^2)$

$\text{Area}=W\cdot L+\frac{\pi}{2}(\frac{W^2}{4})$

$\text{Area}=W\cdot L+\frac{\pi}{8}W^2$

Now, we will solve for L is terms W from perimeter equation as:

$L=38-(W+\frac{\pi }{2}W)$

Substitute this value in area equation:

$A=W\cdot (38-W-\frac{\pi }{2}W)+\frac{\pi}{8}W^2$

Since we need the area of window to maximize, so we need to optimize area equation.

$A=W\cdot (38-W-\frac{\pi }{2}W)+\frac{\pi}{8}W^2$

$A=38W-W^2-\frac{\pi }{2}W^2+\frac{\pi}{8}W^2$

Let us find derivative of area equation as:

$A'=38-2W-\frac{2\pi }{2}W+\frac{2\pi}{8}W$

$A'=38-2W-\pi W+\frac{\pi}{4}W$

$A'=38-2W-\frac{4\pi W}{4}+\frac{\pi}{4}W$

$A'=38-2W-\frac{3\pi W}{4}$

To find maxima, we will equate first derivative equal to 0 as:

$38-2W-\frac{3\pi W}{4}=0$

$-2W-\frac{3\pi W}{4}=-38$

$\frac{-8W-3\pi W}{4}=-38$

$\frac{-8W-3\pi W}{4}*4=-38*4$

$-8W-3\pi W=-152$

$8W+3\pi W=152$

$W(8+3\pi)=152$

$W=\frac{152}{8+3\pi}$

$W=8.723210$

$W\approx 8.72$

Upon substituting $W=8.723210$ in equation $L=38-(W+\frac{\pi }{2}W)$ , we will get:

$L=38-(8.723210+\frac{\pi }{2}8.723210)$

$L=38-(8.723210+\frac{8.723210\pi }{2})$

$L=38-(8.723210+\frac{27.40477245}{2})$

$L=38-(8.723210+13.70238622)$

$L=38-(22.42559622)$

$L=15.57440378$

$L\approx 15.57$

Therefore, the dimensions of the window that will maximize the area would be $W\approx 8.72$ and $L\approx 15.57$ .

8 0

3 years ago

.!.!..!.!..!.!..!.!.!..!.!!..!9

vladimir1956 [14]

Answer:

47 degrees

Step-by-step explanation:

total degrees in a triangle is 180

since this is a right triangle there is a 90 degree in there and it also gives the 43 degree

subtract 90 and 43 from 180 and you get 47

3 0

2 years ago

I need help with 3-4 quick

Agata [3.3K]

3-4 = -1

Hope this helps!

:)

6 0

3 years ago

Solve the system of linear equations using the

MArishka [77]

Answer:

no solution

Step-by-step explanation:

y = -2x + 3

6x + 3y = -3

the substitution method means you plug one equation into the next, because the first equation gives us a solution for y we can go ahead and plug that into y of the second equation

6x + 3(-2x + 3) = -3

6x - 6x + 9 = -3

9 = -3

which is false meaning that there are no solutions and the lines don't touch at any point

8 0

3 years ago