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shusha [124]
3 years ago
5

An owner wants to discount his pizza from $15 to $14.25, how much percent does he need to discount?

Mathematics
1 answer:
IrinaVladis [17]3 years ago
3 0

Answer:

5%

Step-by-step explanation:

Amount of money that is discounted

= $15 - $14.25

$0.75

Percentage that needed to be discount

=(0.75/15) * 100%

= 5%

Thus, owner needs to discount 5% to discount his pizza from $15 to $14.25.

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Half of 6900 and 8100
klio [65]

6900 + 8100 = 15000

15000 divided by 2 = 7500

So your answer is 7500

5 0
4 years ago
Read 2 more answers
-2x + 5y = -4<br> 3x + 4y = -17<br><br> Use elimination to solve
mario62 [17]

Answer:

(- 3, - 2 )

Step-by-step explanation:

Given the 2 equations

- 2x + 5y = - 4 → (1)

3x + 4y = - 17 → (2)

Multiplying (1) by 3 and (2) by 2 and adding the result will eliminate x

- 6x + 15y = - 12 → (3)

6x + 8y = - 34 → (4)

Add (3) and (4) term by term

(- 6x + 6x) + (15y + 8y) = (- 12 - 34), that is

23y = - 46 ( divide both sides by 23 )

y = - 2

Substitute y = - 2 in either of the 2 equations.

Substituting in (2)

3x - 8 = - 17 ( add 8 to both sides )

3x = - 9 ( divide both sides by 3 )

x = - 3

Solution is (- 3, - 2 )

5 0
3 years ago
Read 2 more answers
A contractor is given a scale drawing of a rectangular patio. The scale from the patio to the drawing is 4 ft to 1 in. What is t
iogann1982 [59]

Answer:

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Step-by-step explanation:

6 0
3 years ago
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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
Write as an algebraic expression <br> the quotient of 30 and 5
kirill [66]

Answer:

30÷5=6.

Step-by-step explanation:

Quotient means to divide, 30÷5 is 6 since 5x6 or 6x5 is 30.

4 0
3 years ago
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