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4 years ago

jaylen is making biscuits using the recipe. How many. cups of flour will he need to make 30 biscuits ? pls help

Mathematics

2 answers:

finlep [7]4 years ago

7 0

Answer:

$7\frac{1}{2}$ cups

Step-by-step explanation:

Jaylen is making biscuits using the recipe.

The recipe calls for 8 biscuits and need to make 30 biscuits.

∵ For 8 biscuits Jaylen needs flour = 2 cups

∴ for 1 biscuits Jaylen needs flour = $\frac{2}{8}$

∴ for 30 biscuits Jaylen needs flour = $\frac{2}{8}\times 30$

= $\frac{60}{8}$

= $7\frac{1}{2}$ cups

Jaylen needs $7\frac{1}{2}$ cups flour for 30 biscuits.

valentina_108 [34]4 years ago

5 0

30*2=60
so 60 cups of flour he will need

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4 years ago

Read 2 more answers

Fit a trigonometric function of the form f(t)=c0+c1sin(t)+c2cos(t)f(t)=c0+c1sin⁡(t)+c2cos⁡(t) to the data points (0,5.5)(0,5.5),

larisa86 [58]

Answer

$f(t)=-0.2+4.1sin(t)+4cos(t)$

Step-By-Step Explanation

Given the function $f(t)=c_0+c_1sin(t)+c_2cos(t)$ .

For each pair (t, f(t)) in the data points (0,5.5), (π/2,0.5), (π,−2.5), (3π/2,−7.5)

$f(0)=c_0+0c_1+c_2=5.5$ .

$f(\pi /2)=c_0+c_1+0c_2=0.5$ .

$f(\pi)=c_0+0sin(t)-c_2=-2.5$ .

$f(3\pi /2)=c_0-c_1+0c_2=-7.5$ .

Expressing this as a system of linear equations in matrix form AX=B

$\left(\begin{array}{ccc} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & -1 \\ 0 & -1 & 0 \end{array} \right)\left( \begin{array}{c} c_{0} \\ c_{1} \\ c_{2}\\ \end{array} \right)=\left(\begin{array}{c} 5.5 \\ 0.5 \\ -2.5 \\ -7.5 \end{array} \right)$

Where

$A=\left(\begin{array}{ccc} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & -1 \\ 0 & -1 & 0 \end{array} \right)$ ,

$B=\left(\begin{array}{c}5.5\\0.5\\-2.5\\-7.5\end{array} \right)$

$X=\left(\begin{array}{c}c_0\\c_1\\c_2\end{array}\right)$

To determine the values of X, we use the expression

$X=(A^{T}A)^{-1}A^{T}B$

$A^{T}A= \left(\begin{array}{ccc} 3 & 1 & 0 \\ 1 & 2 & 0 \\ 0 & 0 & 2 \end{array} \right)$

$(A^{T}A)^{-1}= \left(\begin{array}{ccc} 0.4 & -0.2 & 0 \\ -0.2 & 0.6 & 0 \\ 0 & 0 & 0.5 \end{array} \right)$

$A^{T}B=\left(\begin{array}{c} 3.5 \\ 8 \\ 8 \end{array} \right)$

Therefore:

$X=\left(\begin{array}{ccc} 0.4 & -0.2 & 0 \\ -0.2 & 0.6 & 0 \\ 0 & 0 & 0.5 \end{array} \right)\left( \begin{array}{c} 3.5 \\ 8 \\ 8 \end{array} \right)$

$X=\left(\begin{array}{c}c_0\\c_1\\c_2\end{array}\right)=\left(\begin{array}{c} -0.2 \\4.1\\4\end{array}\right)$

Therefore, the trigonometric function which fits to the given data is:

$f(t)=-0.2+4.1sin(t)+4cos(t)$

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3 years ago

Answer the following

Amanda [17]

The set A satisfying the given inequality is A = (- $\infty$ , -10].

<h3>What are some properties of an inequality relation? </h3>

Following are some facts which are true for an inequality relation:

Equal numbers can be added or subtracted from both sides of an inequality without affecting the inequality sign.
The Inequality sign is unchanged if both sides are multiplied or divided by a positive number, but when multiplied or divided by a negative number the inequality sign is reversed.

$\frac{5x-2}{8} - \frac{3x-5}{10} &\ge& x+y\\\\\Rightarrow\;\; \frac{13}{40}x + \frac{1}{4}&\ge& x+y\\\\\Rightarrow\;\;\;\;\;\; -\frac{27}{40}x &\ge & y - \frac{1}{4}\\\\\Rightarrow\;\;\;\;\;\;\;\;\;\; -x &\ge & \frac{40}{27}\left( y-\frac{1}{4} \right).\hspace{1cm}(1)$

Since y ∈ B, -2 ≤ y ≤ 7. So,

$\;\;\;\;\;\;\;\,-2 - \frac{1}{4}\; \le\; y - \frac{1}{4} \;\le\; 7 - \frac{1}{4}\\\\\Rightarrow\;\;\;\;\;\;\;\;\; -\frac{9}{4}\; \le\; y - \frac{1}{4} \;\le\; \frac{27}{4}\\\\\Rightarrow\;\;\; -\frac{9}{4}\cdot \frac{40}{27} \;\le\; \frac{40}{27} \left( y-\frac{1}{4} \right) \;\le \;\frac{27}{4}\cdot \frac{40}{27}\\\\\Rightarrow\;\;\;\;\;\;\;\; -\frac{10}{3} \;\le\; \frac{40}{27}\left( y - \frac{1}{4} \right)\; \le\; 10.$