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bulgar [2K]
3 years ago
14

A bread recipe calls for 1/3 cup of butter. You want to make 1/2 of a batch. How much butter do you need?

Mathematics
2 answers:
algol [13]3 years ago
7 0

If you want to make \dfrac{1}{2} of a batch and a bread recipe calls for \dfrac{1}{3} cup of butter, then you should multiply the fraction needed to make the whole batch by the fraction of a batch you want tot make:

\dfrac{1}{3}\cdot \dfrac{1}{2}=\dfrac{1}{6}.

Answer: you need \dfrac{1}{6} cup od butter.

Sergio [31]3 years ago
6 0
Do 1/3 x 1/2 to get 1/6, or you could think, 1/3 divided by 2, which would equal 1/3 x 1/2
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bagirrra123 [75]
This means that X in the equation equals -1
7 0
2 years ago
Help due now pleaseee
DENIUS [597]

Answer:

d

Step-by-step explanation:

base² + Altitude² = Hypotenuse²

 x² + 9² = 13²

x² + 81 = 169

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      x = \sqrt{88}

     x = \sqrt{2*2*2*11}=2\sqrt{2*11}\\\\x=2\sqrt{22}

6 0
2 years ago
Read 2 more answers
<img src="https://tex.z-dn.net/?f=tan20%2B4sin20%3D%5Csqrt%7B3%7D" id="TexFormula1" title="tan20+4sin20=\sqrt{3}" alt="tan20+4si
zepelin [54]

tan( 20 ) + 4 Sin( 20 ) =

( Sin( 20 ) / Cos( 20 ) ) + 4 Sin( 20 ) =

Sin( 20 ) + 4 Sin( 20 ).Cos( 20 ) / Cos( 20 ) =

Sin( 20 ) + 2 × 2 Sin(20).Cos(20)/ Cos(20) =

Sin( 20 ) + 2 × Sin( 40 ) / Cos( 20 ) =

Sin( 20 ) + 2Sin( 40 ) / Cos( 20 ) =

Sin( 20 ) + 2Cos( 50 ) / Cos ( 20 ) =

Sin( 20 ) + 2Cos( 20 + 30 ) / Cos( 20 ) =

________________________________

2 × Cos( 30 + 20 ) =

2 × [ Cos(30).Cos(20) - Sin(30).Sin(20) ] =

2 × [ √3/2 × Cos(20) - 1/2 × Sin(20) ] =

√3 Cos(20) - Sin(20)

_________________________________

Sin( 20 ) + 2Cos ( 20 + 30 ) / Cos( 20 ) =

Sin( 20 ) + √3 Cos(20) - Sin(20) / Cos(20) =

Sin(20) - Sin(20) + √3 Cos(20) / Cos(20) =

0 + √3 Cos(20) / Cos(20) =

√3 Cos(20) / Cos(20) =

Cos(20) simplifies from the numerator and denominator of fraction

√3 × 1 / 1 =

√3

And we're done ....

5 0
2 years ago
Can u help fast please
kotegsom [21]
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5 0
3 years ago
Read 2 more answers
5. Mis the midpoint of CD. C has coordinates (-1,-1) and
lana66690 [7]

Answer/Step-by-step Explanation:

4. Midpoint (M) of AB, for A(-2, -3) and B(1, 2) is given as:

M(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2})

Let A(-2, -3) = (x_1, y_1)

B(1, 2) = (x_2, y_2)

Thus:

M(\frac{-2 + 1}{2}, \frac{-3 + 2}{2})

M(\frac{-1}{2}, \frac{-1}{2})

5. Given M(3, 5) as midpoint of CD, and C(-1, -1),

let C(-1, -1) = (x_2, y_2)

D(?, ?) = (x_1, y_1)

M(3, 5) = (\frac{x_1 +(-1)}{2}, \frac{y_1 +(-1)}{2})

Rewrite the equation to find the coordinates of D

3 = \frac{x_1 - 1}{2} and 5 = \frac{y_1 - 1}{2}

Solve for each:

3 = \frac{x_1 - 1}{2}

3*2 = \frac{x_1 - 1}{2}*2

6 = x_1 - 1

6 + 1= x_1 - 1 + 1

7 = x_1

x_1 = 7

5 = \frac{y_1 - 1}{2}

5*2 = \frac{y_1 - 1}{2}*2

10 = y_1 - 1

10 + 1= y_1 - 1 + 1

11 = y_1

y_1 = 11

Coordinates of D is (7, 11)

7 0
3 years ago
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