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Karo-lina-s [1.5K]
3 years ago
10

A pizza place charges $5.00 for a cheese pizza and $1.75 for each additional topping. Which equation models the total price of a

cheese pizza, P, with t additional toppings?
A) P = 5.00t + 1.75

B) P = 6.75t

C) P = 1.75t + 5.00

D) P = 6.75 + t
Mathematics
1 answer:
inysia [295]3 years ago
4 0

Answer:

C

Step-by-step explanation: This is c because 1.75 t is right because they don't know how many toppings you are going to put on but they do know how much you will automatically have to pay for the pizza itself. That is why it is 1.75t(toppings) plus the 5 dollars for pizza.

Hope this helps!

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Solve for x:a(a²+b²)x²+b²x-a​
m_a_m_a [10]

Answer:

x = a/(a² + b²) or x = -1/a  

Step-by-step explanation:

a(a²+ b²)x² + b²x - a =0

Use the quadratic equation formula:

x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a} =\dfrac{-b\pm\sqrt{D}}{2a}

1. Evaluate the discriminant D

D = b² - 4ac = b⁴ - 4a(a² + b²)(-a) = b⁴ + 4a⁴ + 4a²b²  = (b² + 2a²)²

2. Solve for x

\begin{array}{rcl}x & = & \dfrac{-b\pm\sqrt{D}}{2a}\\\\ & = & \dfrac{-b^{2}\pm\sqrt{(b^{2}+2a^{2})^{2}}}{2a(a^{2} + b^{2})}\\\\ & = & \dfrac{-b^{2}\pm(b^{2}+ 2a^{2})}{2a(a^{2} + b^{2})}\\\\x = \dfrac{-b^{2}+(b^{2} + 2a^{2})}{2a(a^{2} + b^{2})}&\qquad& x =\dfrac{-b^{2}-(b^{2} + 2a^{2})}{2a(a^{2} + b^{2})}\\\\x =\dfrac{-b^{2}+(b^{2} + 2a^{2})}{2a(a^{2} + b^{2})}&\qquad& x =\dfrac{-b^{2}-(b^{2} +2a^{2})}{2a(a^{2} + b^{2})}\\\\\end{array}

\begin{array}{rcl}x = \large \boxed{\mathbf{\dfrac{a}{a^{2} + b^{2}}}}&\qquad& x =\dfrac{-b^{2}-(b^{2} +2a^{2})}{2a(a^{2} + b^{2})}\\\\&\qquad& x =\dfrac{-2b^{2}- 2a^{2}}{2a(a^{2} + b^{2})}\\\\&\qquad& x =\dfrac{-2(a^{2}+ b^{2})}{2a(a^{2} + b^{2})}\\\\&\qquad& x =\large \boxed{\mathbf{-\dfrac{1}{a}}}\\\\\end{array}

5 0
3 years ago
Convert the following decimals 1 / 8
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3 years ago
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HELP FAST ITS TO RPROVE MY TEACHER WRONG
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Answer

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Step-by-step explanation:

3 0
3 years ago
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Shureka Washburn has scores of 67​, 68​, 76​, and 63 on her algebra tests. a. Use an inequality to find the scores she must make
mash [69]
Average=(total number)/(number of items)
given that the final exam counts as two test, let the final exam be x. The weight of the final exams on the average is 2, thus the final exam can be written as 2x because any score Shureka gets will be doubled before the averaging.
Hence our inequality will be as follows:
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8 0
3 years ago
Hat is the solution to the equation 3/3(4c + 16)=2c+9 ?<br><br> c =
marusya05 [52]
If it is \frac{3}{3(4c+16)}=2c+9, go to AAAAAA
if it is (\frac{3}{3})(4c+16)=2c+9, go to BBBBB

AAAAAAAAA
\frac{3}{3(4c+16)}=2c+9
\frac{1}{4c+16}=2c+9
times 4c+16 to both sides
1=(2c+9)(4c+16)
distribute
1=8c^2+68c+144
minus 1 both sides
0=8c^2+68c+143
use quadratic formula
c=\frac{-17- \sqrt{3} }{4} or \frac{-17+ \sqrt{3} }{4}




BBBBBBBBBBBB
(3/3)(4c+16)=2c+9
1(4c+16)=2c+9
4c+16=2c+9
minus 2c both sides
2c+16=9
minus 16 both sides
2c=-7
divide both sides by 2
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c=-3.5





if it is \frac{3}{3(4c+16)}=2c+9,
c=\frac{-17- \sqrt{3} }{4} or \frac{-17+ \sqrt{3} }{4}


if it is (\frac{3}{3})(4c+16)=2c+9, c=-3.5



7 0
4 years ago
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