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shtirl [24]
3 years ago
12

A recipe that serves 6 uses 1 1/4 cups of flour. How many cups are needed to serve 8?

Mathematics
1 answer:
pickupchik [31]3 years ago
7 0

Answer:1 2/3

Step-by-step explanation:

6 uses 1 1/4. To improper fraction 6 uses 5/4. Thus 1 will use 5/4 ×1/6= 5/24. Then 8 will use 8×5/24=5/3=1 2/3

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Rewrite 32/35 and 9/10
hram777 [196]

The question was incorrect. Please find the correct content below.

Compare 32/35 and 9/10.

Comparing 32/35 and 9/10 we have 32/35 is greater than 9/10, that is 32/35 > 9/10.

Fraction is the ratio of two numbers. The upper number is called Numerator and the Lower number is called the Denominator.

We know that if the denominators are the same for two fractions then which has the greatest numerator is a greater fraction than the other.

Given the fractions are 32/35, 9/10

To compare this two fractions we have to make denominators equal first.

LCM of 10,35 = 70

Calculating the fractions,

32/35 = (32*2)/(35*2) = 64/70

9/10 = (9*7)/(10*7) = 63/70

Since 64 > 63

So 64/70 > 63/70

Therefore, 32/35 > 9/10

Hence fraction 32/35 is greater than the other fraction 9/10.

Learn more about Fraction here -

brainly.com/question/78672

#SPJ10

6 0
2 years ago
How many times does 8go into14
eimsori [14]

Answer: 1

Step-by-step explanation: it only goes into 14 once because 8 • 2 =16

4 0
3 years ago
Read 2 more answers
Find the missing term of 3b2 − = -9b2 ?<br> -6b^2<br> -9b^2<br> -12b^2
gladu [14]
3b^2 - x = -9b^2 
x = 3b^2 +9b^2

x = 12b^2

3b^2 -12b^2 = -9b^2

hope this will help you 
7 0
3 years ago
A sphere with a diameter of 16mm has the same surface area as the total surface area of a right cylinder with the base diameter
OLEGan [10]
4πrsquare= 2πrsquare+2πrh
2πrsquare=2πrh
R=h
So h=16/2=8mm
5 0
3 years ago
Simplify the expression using trigonometric identities: sec (–θ) – cos θ.
ololo11 [35]

Answer:

sin θ . tan θ

Step-by-step explanation:

Note : -

sec ( - θ ) = sec θ

Formula / Identity : -

sec θ = 1 / cos θ

sec ( - θ ) - cos θ

= [ 1 / cos θ ] - cos θ

{ LCM = cos θ }

= [ 1 / cos θ ] - [ cos²θ / cos θ ]

= [ 1 - cos²θ ] / cos θ

{ 1 - cos²θ = sin²θ }

= sin²θ / cos θ

{ sin²θ = sin θ . sin θ }

= sin θ . sin θ / cos θ

{ sin θ / cos θ = tan θ }

= sin θ . tan θ

Hence, simplified.

4 0
2 years ago
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