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Hoochie [10]
2 years ago
12

A chocolate chip cookie recipe ask for one and one half times as much flour as chocolate chips. If three and three quarters cups

of flour is used what quantity of chocolate chips would then be needed according to the recipe?
Mathematics
1 answer:
babymother [125]2 years ago
6 0

Answer:

1/2 a cup.

Step-by-step explanation:

3.5 times as much flour as chocolate chips....

f = 3.5c

f = 1 3/4 (or 1.75)

1.75 = 3.5c

1.75 / 3.5 = c

0.5 = c......chocolate chips needed would be 1/2 a cup

Hope that helps!

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The altitude of a helicopter increases 1000 feet every 4 minutes .what is the unit rate change (feet per minute) of the altitude
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Step-by-step explanation:

divide 1000 by 4 and you will get the unit rate

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A 4 pack of toy cars cost 3.64. What is the unit price
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5 0
1 year ago
Use the definition of logarithm to simplify each expression.
konstantin123 [22]
1)\ \ \ log 10 ^{10}=x\ \ \ \Leftrightarrow\ \ \ 10^x=10^{10}\ \ \ \Leftrightarrow\ \ \ \ \ \ x=10\\\\2) \ \ \ log 10 ^{100}=x\ \ \ \Leftrightarrow\ \ \ 10^x=10^{100}\ \ \ \Leftrightarrow\ \ \ x=100\\ \\ 3)\ \ \ log 10 ^ {10-4}=x\ \ \ \Leftrightarrow\ \ \ 10^x=10^{10-4}\ \ \ \Leftrightarrow\ \ \ x=6
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3 years ago
If two angels are supplementary and one of them is 30 degrees what is the measurement for the second angle
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6 0
3 years ago
Calculus help with the graph of a integral and finding the following components.
lys-0071 [83]
A)

\bf \displaystyle F(x)=\int\limits_{4}^{\sqrt{x}}\cfrac{2t-1}{t+2}\cdot dt\qquad x=16\implies \displaystyle F(x)=\int\limits_{4}^{\sqrt{16}}\cfrac{2t-1}{t+2}\cdot dt
\\\\\\
\displaystyle F(x)=\int\limits_{4}^{4}\cfrac{2t-1}{t+2}\cdot dt\implies 0


why is 0?  well, the bounds are the same.


b)

let's use the second fundamental theorem of calculus, where F'(x) = dF/du * du/dx

\bf \displaystyle F(x)=\int\limits_{4}^{\sqrt{x}}\cfrac{2t-1}{t+2}\cdot dt\implies F(x)=\int\limits_{4}^{x^{\frac{1}{2}}}\cfrac{2t-1}{t+2}\cdot dt\\\\
-------------------------------\\\\
u=x^{\frac{1}{2}}\implies \cfrac{du}{dx}=\cfrac{1}{2}\cdot x^{-\frac{1}{2}}\implies \cfrac{du}{dx}=\cfrac{1}{2\sqrt{x}}\\\\
-------------------------------\\\\

\bf \displaystyle F(x)=\int\limits_{4}^{u}\cfrac{2t-1}{t+2}\cdot dt\qquad F'(x)=\cfrac{dF}{du}\cdot \cfrac{du}{dx}
\\\\\\
\displaystyle\cfrac{d}{du}\left[ \int\limits_{4}^{u}\cfrac{2t-1}{t+2}\cdot dt \right]\cdot \cfrac{du}{dx}\implies \cfrac{2u-1}{u+2}\cdot \cfrac{1}{2\sqrt{x}}
\\\\\\
\cfrac{2\sqrt{x}}{\sqrt{x}+2}\cdot \cfrac{1}{2\sqrt{x}}\implies \cfrac{2\sqrt{x}-1}{2x+4\sqrt{x}}


c)

we know x = 16, we also know from section a) that at that point f(x) = y = 0, so the point is at (16, 0), using section b) let's get the slope,

\bf \left. \cfrac{2\sqrt{x}-1}{2x+4\sqrt{x}} \right|_{x=16}\implies \cfrac{2\sqrt{16}-1}{2(16)+4\sqrt{16}} \implies \cfrac{7}{48}\\\\
-------------------------------\\\\
\begin{cases}
x=16\\
y=0\\
m=\frac{7}{48}
\end{cases}\implies \stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-0=\cfrac{7}{48}(x-16)
\\\\\\
y=\cfrac{7}{48}x-\cfrac{7}{3}\implies y=\cfrac{7}{48}x-2\frac{1}{3}


d)

\bf 0=\cfrac{2\sqrt{x}-1}{2x+4\sqrt{x}}\implies 0=\cfrac{2\sqrt{x}-1}{(\sqrt{x}+2)(2\sqrt{x})}

now, we can get critical points from zeroing out the derivative, we also get critical points from zeroing out the denominator, however, the ones from the denominator are points where the function is not differentiable, namely, is not a smooth curve, is a sharp jump, a cusp, or a spike, and therefore those points are usually asymptotic, however, they're valid critical points, let's check both,

\bf 0=\cfrac{2\sqrt{x}-1}{2x+4\sqrt{x}}\implies 0=\cfrac{2\sqrt{x}-1}{(\sqrt{x}+2)(2\sqrt{x})}\\\\
-------------------------------\\\\
0=2\sqrt{x}-1\implies 1=2\sqrt{x}\implies \cfrac{1}{2}=\sqrt{x}\implies \left(\cfrac{1}{2}  \right)^2=x
\\\\\\
\boxed{\cfrac{1}{4}=x}\\\\
-------------------------------\\\\
0=(\sqrt{x}+2)(2\sqrt{x})\implies 0=2\sqrt{x}\implies \boxed{0=x}
\\\\\\
0=\sqrt{x}+2\implies -2=\sqrt{x}\implies (-2)^2=x\implies \boxed{4=x}

now, doing a first-derivative test on those regions, we get the values as in the picture below.

so, you can see where is increasing and decreasing.

5 0
3 years ago
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