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3 years ago

(5x^{2}+3x-1)-(3x^{2}-4x-6) find the difference PLEASE HURRY IM TIMED

Mathematics

2 answers:

Andreyy893 years ago

6 0

Answer:

=2x2+7x+5

Step-by-step explanation:

Rus_ich [418]3 years ago

4 0

Answer:

2x^2 + 7 x + 5

Step-by-step explanation:

I can't waste time giving an explanation if you're timed ;) if this helped you, leave a brainliest! :)

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7.Mrs. Deaton’s 6th grade class is creating a mural for the school entryway. Each student is creating a square with a side len

saw5 [17]

2,500 inches squared

10 in.•5 squares= 50in. per side
The entire mural is a square. Therefore, all sides are equal.
50 in.•50 in.=2,500in. squared

Hope this helps!

6 0

3 years ago

A street light is mounted at the top of a 15-ft-tall pole. A man 6 feet tall walks away from the pole with a speed of 5 ft/s alo

harina [27]

Answer:

25/3 ft/s

Step-by-step explanation:

Height of pole , h=15 ft

Height of man, h'=6 ft

Let BD=x

BE=y

DE=BE-BD=y-x

All right triangles are similar

When two triangles are similar then the ratio of their corresponding sides are equal.

Therefore,

$\frac{AB}{CD}=\frac{BE}{DE}$

$\frac{15}{6}=\frac{y}{y-x}$

$\frac{5}{2}=\frac{y}{y-x}$

$5y-5x=2y$

$5y-2y=5x$

$3y=5x$

$y=\frac{5}{3}x$

Differentiate w.r.t t

$\frac{dy}{dt}=\frac{5}{3}\frac{dx}{dt}$

We have dx/dt=5ft/s

Using the value

$\frac{dy}{dt}=\frac{5}{3}(5)=\frac{25}{3}ft/s$

Hence, the tip of his shadow moving with a speed 25/3 ft/s when he is 45 feet from the pole.

8 0

3 years ago

Read 2 more answers

QUESTION IS STATED IN PICTURE DONT FORGET TO ROUND TO THE NEAREST HUNDRETH. USE 3.14 pi IF NEEDED

Anna007 [38]

The area of a circle is given by:

$A=\pi\cdot r^2$

Where r is the circle's radius. In this case we have a diameter of 8cm so the radius 4cm. Using pi=3.14 we have that the area of the circle is:

$A=3.14\cdot4^2=50.24$

So the area is 50.24 cm^2.

5 0

1 year ago

Urgent!

boyakko [2]

Step-by-step explanation:

18.

the last one.

-4.9t² + 24.5t + 117.6 = 0

the ball starts at 117.6 meters height. then, during the first seconds the thrust upwards is adding height, until finally, with more and more seconds passing, gravity will win with a vengeance and pulls the ball down faster and faster than any other force in any other direction.

19.

the general solution for a quadratic equation

y = ax² + bx + c

x = (-b ± sqrt(b² - 4ac))/(2a)

in our case

x = t

a = -4.9

b = 24.5

c = 117.6

t = (-24.5 ± sqrt(600.25 - -4×4.9×117.6))/-9.8 =

= (-24.5 ± sqrt(600.25 + 2304.96))/-9.8 =

= (-24.5 ± sqrt(2905.21))/-9.8 = (-24.5 ± 53.9)/-9.8

t1 = (-24.5 + 53.9)/-9.8 = -3

t2 = (-24.5 - 53.9)/-9.8 = 8

a negative time does not make any sense, so the real solution is : the ball reached the ground after 8 seconds.

20.

now the equation is

-4.9t² + 24.5t + 117.6 = 49

but this is quickly transformed again into a "= 0" equation :

-4.9t² + 24.5t + 68.6 = 0

t = (-24.5 ± sqrt(600.25 - -4×4.9×68.6))/-9.8 =

= (-24.5 ± sqrt(600.25 + 1344.56))/-9.8 =

= (-24.5 ± sqrt(1944.81))/-9.8 = (-24.5 ± 44.1)/-9.8

t1 = (-24.5 + 44.1)/-9.8 = -2

t2 = (-24.5 - 44.1)/-9.8 = 7

again, negative time does not make sense.

the ball will be at 49 m height after 7 seconds.

5 0

3 years ago

A big ship drops its anchor. E(t) models the anchor's elevation relative to the water's surface (in meters) as a function of tim

forsale [732]

Plug in 5 for t, while ignoring the +75.
$2.4(5) = 12$
Therefore, the anchor drops 12 m. every 5 seconds

3 0

4 years ago

Read 2 more answers