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garri49 [273]
1 year ago
5

A grocer has 2 kinds of tea, one that sells for $8.00 per pound and the other that sells for $6.00 per pound. How many pounds of

each kind must he use to make 50 pounds of tea that will sell for $7.40 per pound?
Answer in _____ pounds of $8.00 tea and _____ pounds of $6.00 tea
ASAP-WILL MARK BRANLIEST!
Mathematics
2 answers:
Brilliant_brown [7]1 year ago
3 0

Answer:

35 pounds of $8.00 tea and 15 pounds of $6.00 tea

Step-by-step explanation:

Let x represent the pounds of tea of the $8 kind. 8x would represent the total cost of that type of tea. We can also say 50 - x is the pounds of tea of the $6 kind, so 6(50 - x) is the total cost of that type of tea. 50 * 7.4 represents the total cost of all the tea, and the expression 8x + 6(50 -x) would also represent the same thing. We can write this as an equation,

8x + 6(50 -x) = 50 * 7.4

simplify,

8x + 300 - 6x = 370

2x + 300 = 370

and solve.

2x = 70

x = 35

This means there is 35 pounds of the $8 kind.

We can subtract that from 50 to find,

50 - 35 = 15,

15 pounds is the amount of the $6 kind.

dsp731 year ago
3 0

Answer:

amount of $8 tea = 35 lb

amount of $6 tea = 15 lb

Step-by-step explanation:

Let x = amount of $8 tea

Then 50 - x =  amount of $6 tea

Total cost of $8 tea is = 8x

Total cost of $6 tea = 6(50 - x)

Total cost of $7.40 tea = 7.4(50)

Total cost of $8 tea + Total cost of $6 tea = Total cost of $7.40 tea

8x + 6(50 - x) = 7.4(50)

8x + 300 - 6x = 370

2x = 70

x = 35

50 - x = 15

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Answer:

0.4

Step-by-step explanation:

Given

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This question can be Solved by using Venn diagram

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P\left ( wear \ ring\ or \ necklace )+P\left ( neither\ ring\ or\ necklace)=1

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The sum of probabilty is equal to 1 because it completes the set

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7 0
1 year ago
Which of the following is the expansion of (3c + d2)6?
vovangra [49]

Answer: Option A) 729c^6 + 1,458c^5d^2 + 1,215c^4d^4 + 540c^3d^6 + 135c^2d^8 + 18cd^{10} + d^{12} is the correct expansion.

Explanation:

on applying binomial theorem,  (a+b)^n=\sum_{r=0}^{n} \frac{n!}{r!(n-r)!} a^{n-r} b^r

Here a=3c, b=d^2 and n=6,

Thus, (3c+d^2)^6=\sum_{r=0}^{6} \frac{6!}{r!(6-r)!} (3c)^{n-r} (d^2)^r

⇒ (3c+d^2)^6= \frac{6!}{(6-0)!0!} (3c)^{6-0}.(d^2)^0+\frac{6!}{(6-1)!1!} (3c)^{6-1}.(d^2)^1+\frac{6!}{(6-2)!2!} (3c)^{6-2}.(d^2)^2+\frac{6!}{(6-3)!3!} (3c)^{6-3}.(d^2)^3+\frac{6!}{(6-4)!4!} (3c)^{6-4}.(d^2)^4+\frac{6!}{(6-5)!5!} (3c)^{6-5}.(d^2)^5+\frac{6!}{(6-6)!6!} (3c)^{6-6}.(d^2)^6

⇒(3c+d^2)^6= \frac{6!}{(6-)!0!} (3c)^6.d^0+\frac{6!}{(5)!1!} (3c)^5.d^2+\frac{6!}{(4)!2!} (3c)^4.d^4+\frac{6!}{(6-3)!3!} (3c)^3.d^6+\frac{6!}{(2)!4!} (3c)^2.d^8+\frac{6!}{(1)!5!} (3c).d^{10}+\frac{6!}{(0)!6!} (3c)^0.d^{12}

⇒(3c+d^2)^6=(3c)^6.d^0+\frac{720}{120} (3c)^5.d^2+\frac{720}{48} (3c)^4.d^4+\frac{720}{36} (3c)^3.d^6+\frac{720}{48} (3c)^2.d^8+\frac{720}{120} (3c).d^{10}+.d^{12}

⇒(3c+d^2)^6=729c^6 + 1,458c^5d^2 + 1,215c^4d^4 + 540c^3d^6 + 135c^2d^8 + 18cd^{10} + d^{12}

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iragen [17]

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