All categories

2 years ago

Which of the following is a perfect square trinomial?

Mathematics

1 answer:

Bezzdna [24]2 years ago

3 0

Answer:

$D.~4x^2-20xy+25y^2$

Step-by-step explanation:

$4x^2-20xy+25y^2\\\\=(2x)^2-2\cdot 2x \cdot 5y+(5y)^2\\\\=(2x-5y)^2$

You might be interested in

Once a commercial plane reaches the desired altitudes, the pilot often travels at a crushing speed. On average, the cruising spe

sweet [91]

I know it may be too late, but maybe it will eventually help you or someone else.

If a plane travels with a speed of 570 miles per hour for 7 hours, then he will travel 570 * 7 = 3990 miles in that time.

6 0

3 years ago

Read 2 more answers

H(x)=-2x-5, find h(-2)

Vedmedyk [2.9K]

Answer:

h(-2) = -1

Step-by-step explanation:

h(x)=-2x-5

Let x= -2

h(-2)=-2*-2-5

= 4 -5

h(-2) = -1

5 0

3 years ago

The midpoint of AB is (3,2). If point is is located at(8,4) what are the coordinates of point B

Dmitry [639]

Step-by-step explanation:

8-3 = 5,

3-5 = -2 = the x coordinate of B

-2-4 =-6,

-2-6 =-8 = the y coordinate of B

(-2,-8) is the point B

A is 5 to the right of the midpoint, so B is 5 to the left of the midpoint

A is 6 up from the midpoint so B is 6 down from the midpoint

7 0

2 years ago

Please help !! [90 points if correct]

3241004551 [841]

Answer:

2201.8348 ; 3 ; x / (1 + 0.01)

Step-by-step explanation:

Final amount (A) = 2400 ; rate (r) = 6% = 0.06, time, t = 1.5 years

Sum = principal = p

Using the relation :

A = p(1 + rt)

2400 = p(1 + 0.06(1.5))

2400 = p(1 + 0.09)

2400 = p(1.09)

p = 2400 / 1.09

p = 2201.8348

2.)

12000 amount to 15600 at 10% simple interest

A = p(1 + rt)

15600 = 12000(1 + 0.1t)

15600 = 12000 + 1200t

15600 - 12000 = 1200t

3600 = 1200t

t = 3600 / 1200

t = 3 years

3.)

A = p(1 + rt)

x = p(1 + x/100 * 1/x)

x = p(1 + x /100x)

x = p(1 + 1 / 100)

x = p(1 + 0.01)

x = p(1.01)

x / 1.01 = p

x / (1 + 0.01)

7 0

3 years ago

1. (5pts) Find the derivatives of the function using the definition of derivative.

andreyandreev [35.5K]

2.8.1

$f(x) = \dfrac4{\sqrt{3-x}}$

By definition of the derivative,

$f'(x) = \displaystyle \lim_{h\to0} \frac{f(x+h)-f(x)}{h}$

We have

$f(x+h) = \dfrac4{\sqrt{3-(x+h)}}$

and

$f(x+h)-f(x) = \dfrac4{\sqrt{3-(x+h)}} - \dfrac4{\sqrt{3-x}}$

Combine these fractions into one with a common denominator:

$f(x+h)-f(x) = \dfrac{4\sqrt{3-x} - 4\sqrt{3-(x+h)}}{\sqrt{3-x}\sqrt{3-(x+h)}}$

Rationalize the numerator by multiplying uniformly by the conjugate of the numerator, and simplify the result:

$f(x+h) - f(x) = \dfrac{\left(4\sqrt{3-x} - 4\sqrt{3-(x+h)}\right)\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{\left(4\sqrt{3-x}\right)^2 - \left(4\sqrt{3-(x+h)}\right)^2}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{16(3-x) - 16(3-(x+h))}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{16h}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)}$

Now divide this by h and take the limit as h approaches 0 :

$\dfrac{f(x+h)-f(x)}h = \dfrac{16}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ \displaystyle \lim_{h\to0}\frac{f(x+h)-f(x)}h = \dfrac{16}{\sqrt{3-x}\sqrt{3-x}\left(4\sqrt{3-x} + 4\sqrt{3-x}\right)} \\\\ \implies f'(x) = \dfrac{16}{4\left(\sqrt{3-x}\right)^3} = \boxed{\dfrac4{(3-x)^{3/2}}}$