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nika2105 [10]
3 years ago
6

How many 1/4 pound servings of walnuts, s, are in 3/4 pound ofwalnuts?​

Mathematics
2 answers:
andrew-mc [135]3 years ago
4 0

Answer:

3 servings

Step-by-step explanation:

Take the number of pounds and divide by the size of the servings

3/4 ÷ 1/4

Copy dot flip

3/4 * 4/1

3/1 * 4/4

3 servings

Lady_Fox [76]3 years ago
3 0

Answer:

3

Step-by-step explanation:

To find this out, you need to use division. Divide 3/4 by 1/4.

3/4 divided by 1/4 equals 3.

So there are 3 1/4 pound servings in 3/4 pounds of walnuts.

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Let f(x)=16^x, find f(1/4)
sergejj [24]

Answer:

2

Step-by-step explanation:

f( \frac{1}{4} ) =  {16}^{ \frac{1}{4} }  =   \sqrt[4 ]{16}  = 2

That's it, hope you enjoyed it.

6 0
2 years ago
Which are solutions of the equation x2 - 16 = 0? Check all that apply.
Maurinko [17]
Last one
Because you have to substitute it
So you do X times 2 - 16 = 0
And the only X that works is the last one cause 8 X 2 = 16 and 16 -16=0
5 0
2 years ago
Question 1<br> Determine the midpoint of the segment with endpoint (-9,3) and (7,-8).<br> Question 2
mario62 [17]

Answer:

(-1, -2.5)

Step-by-step explanation:

the midpoint of a segment is (\frac{x1+x2}{2} ,\frac{y1+y2}{2}), given points (x1, y1) and (x2, y2)

midpoint=(\frac{-9+7}{2}, \frac{3-8}{2} ) = (\frac{-2}{2}, \frac{-5}{2}  ) = (-1, -2.5)

6 0
3 years ago
Read 2 more answers
The maximum afternoon temperature in Granderson is modeled by t=60-30 cos (x<img src="https://tex.z-dn.net/?f=%20%5Cpi%20" id="T
kicyunya [14]
The equation is t=60-30 cos (x \frac{\pi}{6})

the -30 expression, is subtracting, if negative, from the 60
if the cosine returned is negative, you'd end up with a +30,
negative * negative = positive

and then the 30 expression will ADD to the 60 amount
so the temperature "t" is highest, when the 30 expression, is
positive and and it's highest

when does that happen, when cosine is negative and at its highest,
well, cosine range is -1\le cos(\theta ) \le 1
so the lowest value cosine can provide is -1, when is cosine -1?
well, at \pi

so...  let's find a value that makes that expression to cos(\pi)

\bf t=60-30 cos (x \frac{\pi}{6}) \qquad x=6&#10;\\\\&#10;thus&#10;\\\\&#10;t=60-30 cos (6 \frac{\pi}{6})\implies t=60-30 cos (\pi )&#10;\\\\&#10;t=60-30[-1]\implies t=60+30\implies t=90

----------------------------------------------------------------------------------------
so...for August, that'll mean


\bf t=60-30 cos (x \frac{\pi}{6}) \qquad x=7&#10;\\\\&#10;t=60-30 cos (\frac{7\pi}{6})\implies t=60-30\left( -\cfrac{\sqrt{3}}{2} \right)&#10;\\\\&#10;t=60+(15\cdot \sqrt{3})\implies t\approx85.98^o


4 0
3 years ago
Which is the graph of g(x) = [X + 3]?
Vanyuwa [196]

Answer

D or the fourth one because i took the quiz and got it right

3 0
3 years ago
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