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vladimir1956 [14]
3 years ago
8

your teacher ask you to find a recipe that includes two ingredients with a ratio of 1/2 cup and 1/8 cups

Mathematics
1 answer:
klemol [59]3 years ago
8 0
The answer is 0.625 I am not so sure
You might be interested in
Answer for the points<br> 6 − (9 − 2) + 3 × 6
AfilCa [17]

Answer:

17

Step-by-step explanation:

USING BIDMAS (Brackets Indices Division Multiplication Addition Subtraction)

6 − (9 − 2) + 3 × 6

9-2=7

3x6=18

6-7+18=17

8 0
3 years ago
Write an equation for the graph below.
topjm [15]

Answer:

y= 25x

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
mia hired a moving company, it charged 500$ for service mia give them a 16% tip.How much did she tip the movers,
snow_tiger [21]

Answer: 80 dollars

Step-by-step explanation:

You can find a percentage with one of two ways.

For hundreds like this, oftentimes it's quicker and more simple to think of it as just 100. The proportion from 100 to 500 is just 5.  16% of 100 is 16. So, multiply 16 * 5. You would get 80.

The second method is you can use 0.16 * 500. It's the more conventional method, and you get 80, like the last method.

3 0
3 years ago
What is the area model for 821 x 54
Kay [80]
4,334 Is The Answer


821
x
54
———
3284
1050
————
4,334

7 0
3 years ago
How do you solve this equation?
Travka [436]
Exponentail thingies

easy, look at all them them, see that they have 5 in common?
rremember how esay it was to factor
ax^2+bx+c=0

now we have
5^(2x)-6(5^x)+5=0
remember that 5^(2x)=(5^2)^x or (5^x)^2
in other words, we can rewrite it as
1(5^x)^2-6(5^x)+5=0
if yo want, replace 5^x with a and factor
1a^2-6a+5=0
(a-1)(a-5)=0
a=5^x
(5^x-1)(5^x-5)=0
set each to zero

5^x-1=0
5^x=1
take the log₅ of both sides
x=log₅1


5^x-4=0
5^x=4
take the log₅ of both sides
x=log₅4

x=log₅1 and/or log₅4







second quesiton

same thing

1(2^x)-10(2^x)+16=0
factor
(2^x-8)(2^x-2)=0
set each to zero

2^x-8=9
2^x=8
x=3

2^x-2=0
2^x=2
x=1

x=3 or 1







first one
x=log₅1 and/or log₅4
second one
x=1 and/or 3


7 0
3 years ago
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