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harkovskaia [24]
3 years ago
9

Find the value of x when 6-3x= 5x-10x+20

Mathematics
1 answer:
icang [17]3 years ago
5 0

Answer:

7

Step-by-step explanation:

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a baker has 10 cups of sugar to make cookies each batch calls for 1 1/3 cups sugar how many batches can he make
Nata [24]
4/3 x 7=9.3333
4/3 x 8=10.666
Can't go over 10, so you can only make 7 batches.
Hope I've helped!

6 0
3 years ago
Is the ratio 375 miles to 15 gallons a rate? Explain.
vodka [1.7K]

Answer:

It's not correct

Because

Ratio is the comparison of two quantities having same units

So this ratio can't be taken

3 0
3 years ago
How to solve this equation (-14+3/2b)-(1+2/8b) step by step<br> please thank you
zepelin [54]
-14+3/2b-(1+2/8b), then take out the blanket become -14+3/2b-1-2/8b , get 5/4b-15=0 so 5/4b=15 ,get b=15×4÷5. b=12

5 0
3 years ago
What polynomial must be added to x^2-2x+6 so that the sum is 3x^2+7x?
slavikrds [6]
2x^2+9x-6 i found this by subtracting 3x^2+7x from the other equation
5 0
3 years ago
There are 10 true-false questions and 20 multiple choice questions from which to choose a five-question quiz how many ways can t
beks73 [17]

Answer:

In 68229 ways can the quiz be selected such that there is atleast three multiple choice questions

Step-by-step explanation:

Given:

Number of True or false questions= 10

Number of multiple choice questions= 20

To Find:

How many ways can 5 questions can be selected if there must be at least three multiple choice questions =?

Solution:

Combination

A combination is a mathematical technique that determines the number of possible arrangements in a collection of items where the order of the selection does not matter. In combinations, you can select the items in any order.  

The question States there sholud be ATLEAST 3 multiple choice question,

So, we may have

(3 Multiple choice question and 2 true or false question) or

(4 Multiple choice question and 1 true or false question) or

(5 Multiple choice question and 0 true or false question)

Required Number of ways = (20C3 X10C2) +(20C4 X10C1) + (20C5 X10C0)

Required Number of ways =(\frac{20!}{20!(20-3)!}\times\frac{10!}{10!(10-2)!})+(\frac{20!}{20!(20-4)!} \times \frac{10!}{10!(10-1)!}) +(\frac{20!}{20!(20-4)!} \times \frac{10!}{10!(10-0)!})

Required Number of ways = ( 1140 x 42) + (4845 x 10) +(15504 x 1)

Required Number of ways = 47880+48450+15504

Required Number of ways = 68229

3 0
3 years ago
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