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sleet_krkn [62]
3 years ago
11

50% of all the cakes jenny baked were party cakes, 1/5 were fruit cakes and the remainder were sponge cakes. What percentage of

cakes were sponge cakes?
Mathematics
1 answer:
anygoal [31]3 years ago
6 0
So basically you want to find the total percentage of cakes that weren't sponge cakes first.

You already have the 50% of party cakes. For the 1/5 percent of fruit cakes, you can multiply it by 20/20, to get 20/100, and then just take the 20 to get 20% that were fruit cakes.

Now you can just add the percentages together.

50% + 20% = 70%

So now you know 70% weren't sponge cakes, out of 100%.

So here you can just subtract 70% from 100% to figure out the remaining part of 100%, which must be sponge cakes.

100% - 70% = 30%

So 30% of the cakes were sponge cakes.
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Bingel [31]

Answer: It would be C

Step-by-step explanation:

because on Sat. there were 80 discs and on Mon. there were 55 discs, so 55-80=25 and C is your answer

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AnsweYou’ve seen two methods for finding the area of ΔABC—using coordinate algebra (by hand) and using geometry software. How are the two methods similar? How are they different? Why might coordinate algebra be important in making and using geometry software?

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Step-by-step explanation:

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1 year ago
What are the solutions of the inequality? -2(3x + 2) ≥ -6x - 4 Question 2 options: all real numbers x ≤ 6 no solution x ≥ 0
Tju [1.3M]

Solution, -2\left(3x+2\right)\ge \:-6x-4\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:True\quad \forall \:x\in \mathbb{R}\:\\ \:\mathrm{Interval\:Notation:}&\:\left(-\infty \:,\:\infty \:\right)\end{bmatrix}

Steps:

-2\left(3x+2\right)\ge \:-6x-4

\mathrm{Expand\:}-2\left(3x+2\right),\\\mathrm{Apply\:the\:distributive\:law}:\quad \:a\left(b+c\right)=ab+ac,\\a=-2,\:b=3x,\:c=2,\\-2\cdot \:3x+\left(-2\right)\cdot \:2,\\\mathrm{Apply\:minus-plus\:rules},\\+\left(-a\right)=-a,\\-2\cdot \:3x-2\cdot \:2,\\\mathrm{Simplify}\:-2\cdot \:3x-2\cdot \:2,\\\mathrm{Multiply\:the\:numbers:}\:2\cdot \:3=6,\\-6x-2\cdot \:2,\\\mathrm{Multiply\:the\:numbers:}\:2\cdot \:2=4,\\-6x-4,\\-6x-4\ge \:-6x-4

\mathrm{Add\:}4\mathrm{\:to\:both\:sides},\\-6x-4+4\ge \:-6x-4+4

\mathrm{Simplify},\\-6x\ge \:-6x

\mathrm{Add\:}6x\mathrm{\:to\:both\:sides},\\-6x+6x\ge \:-6x+6x

\mathrm{Simplify},\\0\ge \:0

\mathrm{Therefore,\:the\:final\:solution\:is},\\True\quad \forall \:x\in \mathbb{R}

Or\:No\:Solution\:x\geq 0


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