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antoniya [11.8K]
3 years ago
15

Patty is baking for the holidays. 3/5of her treats are cookies and 1/6of her treats are cupcakes. The rest are brownies. What fr

action of her treats are brownies?
Mathematics
1 answer:
BartSMP [9]3 years ago
7 0
So the answer would be 7/30 because 3/5 + 1/6 = 23/30 and you add 7 to that and get 30 well that’s what I believe hope this helped
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Van and Polina have each drawn a conclusion about multiplying negative fractions.
Leto [7]
Ivan only, because an odd number of negatives does not always make negatives while an even number does.
7 0
3 years ago
Read 2 more answers
Question Help Find the original price of a pair of shoes if the sale price is $11 after a 75% discount.
fenix001 [56]
If it was a 75% discount, then u r actually paying 25%

so....25% of the original price is $ 11

0.25x = 11
x = 11 / 0.25
x = 44 <=== the original price
4 0
3 years ago
The units’ digit of a two-digit number is 7 more than the tens’ digit. If 26 is added to the number, the result obtained is five
Vikki [24]
Let the digit in the tens place be x

tens place = x (therefore the value is 10x)
ones place = x + 7
The number is 10x + x + 7

Add 26 to it
10x + x + 7 + 26 = 11x + 33

It is 5 times the sum of the digit
11x + 33  = 5(x + x + 7)
11x 33 = 10x + 35
x = 2
x+ 7 = 2 + 7 = 9

So the number is 29

3 0
3 years ago
To describe a specific arithmetic sequence, Elijah wrote the recursive formula:
Rufina [12.5K]

Answer:

Y(n) = 7n + 23

Step-by-step explanation:

Given:

f(0) = 30

f(n+1) = f(n) + 7

For n=0 : f(1) = f(0) + 7

For n=1 :  f(2) = f(1) + 7

For n=2 : f(3) = f(2) + 7 and so on.

Hence the sequence is an arithmetic progression with common difference 7 and first term 30.

We have to find a general equation representing the terms of the sequence.

General term of an arithmetic progression is:

T(n) = a + (n-1)d

Here a = 30 and d = 7

Y(n) = 30 + 7(n-1) = 7n + 23

7 0
3 years ago
Plz..... Solve this questions for me...
faust18 [17]

6.\\\text{We know}\ 9=3^2.\ \text{Therefore}\\\\9^2=(3^2)^2\\\\\text{use}\ (a^n)^m=a^{nm}\\\\9^2=(3^2)^2=3^{2\cdot2}=3^4\\\\\boxed{3^4=9^2}\\=================

7.\\METHOD\ 1:\\\\\sqrt{7^4}=\sqrt{7^{2\cdot2}}\\\\\text{use}\ (a^n)^m=a^{nm}\\\\=\sqrt{(7^2)^2}\\\\\text{use}\ \sqrt{a^2}=a\ \text{for}\ a\geq0\\\\=\boxed{7^2}\\\\METHOD\ 2:\\\\\text{use}\ a^\frac{m}{n}=\sqrt[n]{a^m}\\\\\sqrt{7^4}=7^\frac{4}{2}=\boxed{7^2}\\=================

8.\\5^a\times5^b=5^{11}\\\\a)\\\text{use}\ a^n\times a^m=a^{n+m}\\\\5^a\times5^b=5^{a+b}\\\\5^{a+b}=5^{11}\Rightarrow a+b=11\\\\11\ \text{is odd. The sum of even and odd is odd. Therefore}\ a\ \text{and}\ b\ \text{can't be even.}\\--------------------\\\\b)\\5\ \text{solutions}\\\\11=5+6\\11=4+7\\11=3+8\\11=2+9\\11=1+10

3 0
3 years ago
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