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babymother [125]
3 years ago
7

Jeff's cookie recipe calls for 3 cups of flour. If Jeff wants to cook only 2/3

Mathematics
2 answers:
kramer3 years ago
7 0

Answer:

2 cups

Step-by-step explanation:

he would need 2 cups of flour because of the recipe calls for 3 cups 2/3 of 3 is 2

oee [108]3 years ago
6 0

Answer: 2 cups of flour

Step-by-step explanation:

2/3 of 3 is 2

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A company that manufactures storage bins for grains made a drawing of a silo. The silo has a conical base, as shown below:
sasho [114]
To find the volume of this storage bin, you will use the formula for volume of a cylinder and of a cone.

V = Bh (cylinder)
Pi x r^2 x h
Pi x 2^2 x 10

V = 1/3 x pi x r^2 x h
1/3 x pi x 2^2 x 3

All of these choices show correct answers!
4 0
3 years ago
here are the dimension of the oven tray jenny bakes the cake on.600mm and 500mm.a cake tin is 25 cm in diameter .how many cake t
castortr0y [4]
First we have 2 types of units of measurement so let’s make all the same type.
600mm= 60 cm
500mm= 50 cm
Tray dimensions are 60 by 50 cm and is a rectangle

Cake tin is round and has diameter 25 cm

We can fit 2 rows and 2 columns of tins
Since (25+25) is less or equal to 50
And (25+25) is less or equal to 60

Answer 4 cake tins

7 0
3 years ago
Please I need help on this
vovangra [49]

Answer:the third one

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7 0
3 years ago
[4-2]-10=1 solve the equation
a_sh-v [17]
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3 0
3 years ago
Simplify. Your answer should contain only positive exponents with no fractional exponents in the denominator.
mart [117]

Answer:

\dfrac{x^{\frac{2}{3}}y^{\frac{1}{4}}}{4y^{2}}.

Step-by-step explanation:

The given expression is  

\dfrac{3y^{\frac{1}{4}}}{4x^{-\frac{2}{3}}y^{\frac{3}{2}}\cdot 3y^{\frac{1}{2}}}

We need to simplify the expression such that answer should contain only positive exponents with no fractional exponents in the denominator.

Using properties of exponents, we get

\dfrac{3}{4\cdot 3}\cdot \dfrac{y^{\frac{1}{4}}}{x^{-\frac{2}{3}}y^{\frac{3}{2}+\frac{1}{2}}}    [\because a^ma^n=a^{m+n}]

\dfrac{1}{4}\cdot \dfrac{y^{\frac{1}{4}}}{x^{-\frac{2}{3}}y^{2}}

\dfrac{1}{4}\cdot \dfrac{x^{\frac{2}{3}}y^{\frac{1}{4}}}{y^{2}}         [\because a^{-n}=\dfrac{1}{a^n}]

\dfrac{x^{\frac{2}{3}}y^{\frac{1}{4}}}{4y^{2}}

We can not simplify further because on further simplification we get negative exponents in numerator or fractional exponents in the denominator.

Therefore, the required expression is \dfrac{x^{\frac{2}{3}}y^{\frac{1}{4}}}{4y^{2}}.

5 0
3 years ago
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