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3 years ago

Why isn’t it reading the problem right

Mathematics

2 answers:

Andrej [43]3 years ago

8 0

what do we do??? im not understanding the question..lol

Vanyuwa [196]3 years ago

4 0

Just make sure you type is correctly

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The midpoint of is (14, 26). One endpoint of is A(32, 40). What are the coordinates of B?

cupoosta [38]

For x,
(32+x)/2=14
32+x=14*2
32+x=28
x=28-32
x=-4

For y,
(40+y)/2=26
40+y=26*2
40+y=52
y=52-40
y=12

Therefore coordinates of B(-4,12)
Hope this helps!

7 0

3 years ago

What is the final step in solving the inequality –2(5 – 4x) < 6x – 4?

mrs_skeptik [129]

Answer:

x <3

Step-by-step explanation:

–2(5 – 4x) < 6x – 4

Distribute

-10 +8x < 6x-4

Subtract 6x from each side

-10+8x-6x < 6x-6x-4

-10 +2x < -4

Add 10 to each side

-10+2x+10 < -4+10

2x< 6

Divide by 2

2x/2 < 6/2

x <3

4 0

3 years ago

HURRRYYYY PLEASE ANSWER SUPER QUICK The table shows the relationship between x and y.

Alex17521 [72]

Answer:

2 times

Step-by-step explanation:

According to the table, each value of y (dependent variable) is twice the corresponding value of x (independent variable).

This can be shown as

y = 2x

Correct choice is 2 times

5 0

3 years ago

HELP ON THE GRAPH ONE

velikii [3]

The answer is B because you have a positive slope (m in y = mx + b) so that eliminates C and D. And you have a positive y-intercept (b in y = mx + b). This eliminates option A as well, which leaves you with option B.

7 0

3 years ago

A Norman window is a rectangle with a semicircle on top. Suppose that the perimeter of a particular Norman window is to be 56 ft

harkovskaia [24]

Answer:

Dimensions of the window in order to allow maximum light is $x=\dfrac{56}{4+\pi}$ and $y=\dfrac{56}{4+\pi}$

Step-by-step explanation:

Consider following Norman window, assuming ABCD as rectangle and arc AD as semicircle with center at E and radius r. (Refer attachment)

Given that perimeter of window is 56 ft. Therefore perimeter of window is given as,

$Perimerter = AB + BC + CD + arc\:AD$

Calculate arc AD as follows,

Let, x denote radius of semi circle. That is, r=x

Since AD is the diameter of semi circle

So AD = 2 r = 2 x.

Now perimeter of semicircle is equal to circumference of semicircle, so calculate circumference of semicircle as follows

Circumference of circle is $C=2\pi r$ . So half of it will be

$C=\dfrac{2\pi r}{2}$

$C=\dfrac{2\pi x}{2}$

$C=\pi x$

So, $arc\:AD = \pi x$

Calculate AB , BC and AD as follows,

Consider rectangle ABCD,

Since, AD is the diameter of semi circle which is also one of the side of the rectangle.

So AD = 2 r = 2 x.

Since AD is parallel to BC. Therefore, AD=BC=2x. Also length of rectangle be y

$\therefore BC=2x,AB=CD=y$

Substituting the value,

Perimeter = AB + BC + CD + arc AD

$56=y+2x+y+\pi x$

$56=2y+2x+\pi x$

To calculate value of y,subtracting 2x and $\pi x$ on both sides,

$56-2x-\pi x=2y$

Dividing by 2,

$28-x-\dfrac{\pi x}{2}=y$

Now calculate area of window.

Area = Area of rectangle + Area of semicircle

From diagram,

$Area = width\times length + \dfrac{1}{2}\times\pi r^{2}$

$Area = 2x\times y + \dfrac{1}{2}\times\pi x^{2}$

$Area = 2xy + \dfrac{\pi x^{2}}{2}$

Substituting value of y in above equation,

$Area = 2x\left (28-x-\dfrac{\pi x}{2} \right ) + \dfrac{\pi x^{2}}{2}$

Simplifying,

$Area = 56x-2x^{2}-\pi x^{2}+ \dfrac{\pi x^{2}}{2}$

$Area = 56x-2x^{2}- \dfrac{\pi x^{2}}{2}$

In order to find the maximum of area function, differentiate the equation with respect to x and find the critical points.

Applying difference rule of derivative,

$\dfrac{dA}{dx}=\dfrac{d}{dx}\left(56x\right)-\dfrac{d}{dx}\left ( 2x^{2} \right )-\dfrac{d}{dx}\left ( \dfrac{\pi x^{2}}{2} \right )$

Applying constant multiple rule of derivative,

$\dfrac{dA}{dx}=56\dfrac{d}{dx}\left(x\right)-2\dfrac{d}{dx}\left ( x^{2} \right )-\dfrac{\pi}{2}\dfrac{d}{dx}\left (x^{2}\right )$

Applying power rule of derivative,

$\dfrac{dA}{dx}=56\left(1x^{1-1}\right)-2\left(2x^{2-1}\right)- \dfrac{\pi}{2}\dfrac{d}{dx}\left (2x^{2-1}\right )$

$\dfrac{dA}{dx}=56\left(1\right)-2\left(2x\right)- \dfrac{\pi}{2}\dfrac{d}{dx}\left (2x\right )$

$\dfrac{dA}{dx}=56-4x-\pi x$

Now find the critical number by solving as follows,

$\dfrac{dA}{dx}=0$

$56-\left(4+\pi\right)x =0$

$56=\left(4+\pi \right)x$

$\dfrac{56}{4+\pi} =x$

Since there is only one critical point, directly substitute the value of x into equation of A. If value of A is greater than 0, then the area is maximum at critical point.

$Area = 56\left (\dfrac{56}{4+\pi} \right )-2\left (\dfrac{56}{4+\pi} \right )^{2}- \dfrac{\pi \left (\dfrac{56}{4+\pi} \right )^{2}}{2}$