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maks197457 [2]
3 years ago
6

(5x2 + 2x + 11) − (7 + 4x − 2x2)

Mathematics
2 answers:
max2010maxim [7]3 years ago
5 0
(5-2) x^2 + ( 2-4) x + (11-7 )
= 3x^2 -2x + 4
ASHA 777 [7]3 years ago
4 0
The answer would end up being 12x+4
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I need help with this please
Colt1911 [192]

Answer:

56.55600 cm

Step-by-step explanation:

Radius ( r ) = 9 cm

Formula : -

Perimeter ( semi - circle ) = πr + 2r

Perimeter ( semi - circle )

= πr + 2r

= ( 3.142 x 9 ) + ( 2 x 9 )

= 9 ( 3.142 + 2 )

= 9 x 6.284

= 56.55600 cm

3 0
3 years ago
Read 2 more answers
How do you write the numerical expression for 1/2×(6×1)+11
Mademuasel [1]
I'm assuming they just want you to simplify the expression:
1/2 × 6 + 11
3 + 11
14
8 0
4 years ago
Can someone help me with this problem?
____ [38]

Answer:

88

Step-by-step explanation:

First find the area of the fountain using the area formula of a circle.

The formula for area is A = \pi r^2. Here the radius is 5 ft. So the area is A =\pi r^2 = \pi 5^ 2= 25\pi = 78.5.

We will subtract the area of 78.5 from the larger circle of the brick path and fountain to find only the area of the brick path.

The larger circle has an area of A =\pi r^2 = \pi 9^ 2= 81\pi = 254.34 where the radius is 9ft.

Subtract to find the area of the path: 254.34 - 78.5 = 175.84. Divide this area by 2 to find the number of bricks needed since each brick is 2 square feet.

175.84/2 = 87.92 = 88 bricks

5 0
4 years ago
a cheetah reaches a speed of 30 m/s as it chases its prey. What is the kinetic energy of the cheetah?
ZanzabumX [31]
Kinetic energy is in movement, so the answer is obviously the cheetah running :)
3 0
3 years ago
After heating up in a teapot, a cup of hot water is poured at a temperature of 210°F. The cup sits to cool in a room at a temper
Fiesta28 [93]

Newton's Law of Cooling:

T(t)=T_{s}+(T_{o}-T_{s})e^{-kt}

T(t) = Temperature given at a time

t = Time

T_{s} = Surrounding temperature

T_{o}= Initial temperature

e = Constant (Euler's number) ≈ 2.72

k = Constant

Using this information, find the value of k, to the nearest thousandth, then use the resulting equation to determine the temperature of the water cup after 4 minutes.

First, plug in the given values in the equation and solve for k:

T(t) = 197°, t = 1.5 minutes, T_{s} = 70° and T_{o}= 210°  

T(t)=T_{s}+(T_{o}-T_{s})e^{-kt}\\197=70+(210-70)e^{-1.5k} \\197 -70 = (140)e^{-1.5k} \\127 =(140)e^{-1.5k}\\\frac{127}{140}=e^{-1.5k} \\ln(\frac{127}{140})=-1.5k\\-0.097=-1.5k\\0.0649 = k

k ≈ 0.065

Let the temperature of the water cup after t = 4 minutes be T(t) = x

Now, let's plug the new time and k constant in the equation and solve for x:

T(t)=T_{s}+(T_{o}-T_{s})e^{-kt}\\\\\x=70+(210-70})e^{-0.065*4}\\\\x=70+(140})e^{-0.26}, -0.26=-\frac{26}{100}=-\frac{13}{50} \\

x=70+(140})e^{-\frac{13}{50}}\\\\

x=70+(140})e^{\frac{1}{\frac{13}{50}}\\\\\\

x=70+e^{\frac{140}{\frac{13}{50}}\\\\\\

x=70+{\frac{140}{\sqrt[50]{e^{13}}}\\

x = 70 +\frac{140}{1.3} \\x=70+107.947\\

x=177.95 ≈ 178

Temperature of water after 4 minutes is 178°

sorry if there's any misspelling or wrong step but I hope my answer is correct ':3

3 0
3 years ago
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