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wlad13 [49]
3 years ago
15

Hi, a recipe calls for 1/2 cups of flour. How much flour should be used if only half of the recipe is to be made. Thank You.

Mathematics
2 answers:
Luden [163]3 years ago
4 0

Answer:

¼ cup of flour

Step-by-step explanation:

½ of ½ cup = ½×½ cup = ¼ cup

Allushta [10]3 years ago
3 0

Answer:

1/4 cup of flour

Step-by-step explanation:

If you are doing half the recipe, and the recipe calls for 1/4 cup of flour, you divide the recipe (And portions), in two. So, half of 1/2 is 1/4,

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2 years ago
Cos^2x+cos^2(120°+x)+cos^2(120°-x)<br>i need this asap. pls help me​
o-na [289]

Answer:

\frac{3}{2}

Step-by-step explanation:

Using the addition formulae for cosine

cos(x ± y) = cosxcosy ∓ sinxsiny

---------------------------------------------------------------

cos(120 + x) = cos120cosx - sin120sinx

                   = - cos60cosx - sin60sinx

                   = - \frac{1}{2} cosx - \frac{\sqrt{3} }{2} sinx

squaring to obtain cos² (120 + x)

= \frac{1}{4}cos²x + \frac{\sqrt{3} }{2}sinxcosx + \frac{3}{4}sin²x

--------------------------------------------------------------------

cos(120 - x) = cos120cosx + sin120sinx

                   = -cos60cosx + sin60sinx

                   = - \frac{1}{2}cosx + \frac{\sqrt{3} }{2}sinx

squaring to obtain cos²(120 - x)

= \frac{1}{4}cos²x - \frac{\sqrt{3} }{2}sinxcosx + \frac{3}{4}sin²x

--------------------------------------------------------------------------

Putting it all together

cos²x + \frac{1}{4}cos²x + \frac{\sqrt{3} }{2}sinxcosx + \frac{3}{4}sin²x + \frac{1}{4}cos²x - \frac{\sqrt{3} }{2}sinxcosx + \frac{3}{4}sin²x

= cos²x + \frac{1}{2}cos²x + \frac{3}{2}sin²x

= \frac{3}{2}cos²x + \frac{3}{2}sin²x

= \frac{3}{2}(cos²x + sin²x) = \frac{3}{2}

                 

5 0
2 years ago
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ELEN [110]

Answer:

(- 3, 37) and (- \frac{5}{2}, \frac{63}{2} )

Step-by-step explanation:

Given the 2 equations

2x² - y + 19 = 0 → (1)

y + 11x = 4 → (2) ← subtract 11x from both sides

y = 4 - 11x → (3)

Substitute y = 4 - 11x into (1)

2x² - (4 - 11x) + 19 = 0

2x² - 4 + 11x + 19 = 0

2x² + 11x + 15 = 0 ← in standard form

(2x + 5)(x + 3) = 0 ← in factored form

Equate each factor to zero and solve for x

2x + 5 = 0 ⇒ 2x = - 5 ⇒ x = - \frac{5}{2}

x + 3 = 0 ⇒ x = - 3

Substitute these values into (3) for corresponding values of y

x = - \frac{5}{2} : y = 4 + \frac{55}{2} = \frac{63}{2} ⇒ (- \frac{5}{2}, \frac{63}{2} )

x = - 3 : y = 4 + 33 = 37 ⇒ (- 3, 37 )

4 0
2 years ago
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