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olga55 [171]
3 years ago
10

An ice-cream shop uses the following ingredients to make one sundae.

Mathematics
2 answers:
vampirchik [111]3 years ago
6 0

Answer:

7

Step-by-step explanation:

If each sundae requires 2 Ice-cream scoops, and you have 14, then you can find the number of sundaes by dividing 14/2

14/2 = 7

Thepotemich [5.8K]3 years ago
5 0

Answer:

7 sundaes

Step-by-step explanation:

To make one sundae, 2 scoops of ice cream are used.

If 14 scoops of ice cream are used, then since 14/2 = 7, 7 times the amount of ice cream needed for 1 sundae was used, so 7 sundaes were made.

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4 years ago
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3 years ago
An expression is shown below: 2x3y + 18xy − 10x2y − 90y Part A: Rewrite the expression by factoring out the greatest common fact
SCORPION-xisa [38]

Answer:

Part A : 2y( x³ + 9x - 5x² - 45 ), Part B : 2y( x - 5 )( x² + 9 )

Step-by-step explanation:

Part A : Let's break every term down here to their " prime factors ", and see what is common among them,

2x³y + 18xy − 10x²y − 90y -

2x³y  = 2 * x³ * y,

18xy = 2 * 3 * 3 * x * y,

− 10x²y = 2 * - 5 * x² * y, - so as you can see for this example I purposely broke down - 10 into 2 and - 5. I could have placed the negative on the 2, but as that value was must likely common among all the terms, I decided to place it on the 5. The same goes for " − 90y. " I placed the negative there on the 5 once more.

− 90y = 2 * - 5 * 3 * 3 * y

The terms common among each term are 2 and y. Therefore, the GCF ( greatest common factor ) is 2x. Let's now factor the expression using this value.

2y( x³ + 9x - 5x² - 45 )

Part B : Let's simply factor this entire expression. Of course starting with the " factored " expression : 2y( x³ + 9x - 5x² - 45 ),

2y\left(x^3+9x-5x^2-45\right) - Factor out " (x^3+9x-5x^2-45\right)) " by grouping,

\left(x^3-5x^2\right)+\left(9x-45\right) - Factor 9 from 9x - 45 and x² from x³ - 5x²,

9\left(x-5\right)+x^2\left(x-5\right) - Factor out common term x - 5,

\left(x-5\right)\left(x^2+9\right) - And our solution is thus 2y( x - 5 )( x² + 9 )

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3 years ago
How can i differentiate this equation?
Dmitry_Shevchenko [17]

\bf y=\cfrac{2x^2-10x}{\sqrt{x}}\implies y=\cfrac{2x^2-10x}{x^{\frac{1}{2}}} \\\\\\ \cfrac{dy}{dx}=\stackrel{\textit{quotient rule}}{\cfrac{(4x-10)(\sqrt{x})~~-~~(2x^2-10x)\left( \frac{1}{2}x^{-\frac{1}{2}} \right)}{\left( x^{\frac{1}{2}} \right)^2}} \\\\\\ \cfrac{dy}{dx}=\cfrac{(4x-10)(\sqrt{x})~~-~~(2x^2-10x)\left( \frac{1}{2\sqrt{x}} \right)}{\left( x^{\frac{1}{2}} \right)^2} \\\\\\ \cfrac{dy}{dx}=\cfrac{(4x-10)(\sqrt{x})~~-~~\left( \frac{2x^2-10x}{2\sqrt{x}} \right)}{x}


\bf\cfrac{dy}{dx}=\cfrac{(4x-10)(\sqrt{x})~~-~~\left( \frac{2x^2-10x}{2\sqrt{x}} \right)}{x} \\\\\\ \cfrac{dy}{dx}=\cfrac{ \frac{(4x-10)(\sqrt{x})(2\sqrt{x})~~-~~(2x^2-10x)}{2\sqrt{x}}}{x} \\\\\\ \cfrac{dy}{dx}=\cfrac{(4x-10)(\sqrt{x})(2\sqrt{x})~~-~~(2x^2-10x)}{2x\sqrt{x}}


\bf \cfrac{dy}{dx}=\cfrac{(4x-10)2x~~-~~(2x^2-10x)}{2x\sqrt{x}}\implies \cfrac{dy}{dx}=\cfrac{8x^2-20x~~-~~(2x^2-10x)}{2x\sqrt{x}} \\\\\\ \cfrac{dy}{dx}=\cfrac{8x^2-20x~~-~~2x^2+10x}{2x\sqrt{x}} \implies \cfrac{dy}{dx}=\cfrac{6x^2-10x}{2x\sqrt{x}} \\\\\\ \cfrac{dy}{dx}=\cfrac{2x(3x-5)}{2x\sqrt{x}}\implies \cfrac{dy}{dx}=\cfrac{3x-5}{\sqrt{x}}

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3 years ago
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