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3 years ago

A basket of laundry is being separated. Divide 48 pieces of clothing into 2 groups so that the ratio is 1:3

Mathematics

1 answer:

Alla [95]3 years ago

5 0

Ratio is 1:3.....added = 4

1/4(48) = 48/4 = 12
3/4(48) = 144/4 = 36

the 2 groups will be 12 and 36

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If 2/5 of the macaroni and cheese provides 16 pieces to the table,how many pieces were in the pan when it was full? Support you

astra-53 [7]

Bearing in mind that a whole will be 5/5, so the full pan is 5/5.

$\bf \begin{array}{ccll}
macaroni&pieces\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\
\frac{2}{5}&16\\\\
\frac{5}{5}&p
\end{array}\implies \cfrac{\quad\frac{2}{5} \quad }{\frac{5}{5}}=\cfrac{16}{p}\implies \cfrac{2}{5}\cdot \cfrac{5}{5}=\cfrac{16}{p}
\\\\\\
\cfrac{2}{5}=\cfrac{16}{p}\implies 2p=80\implies p=\cfrac{80}{2}\implies p=40$

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3 years ago

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Mamont248 [21]

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3 years ago

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Travon had a balance of $345.28 in his checking account. During the week amounts: $65.08, $24.50, and $118.95. He then made a de

Vladimir79 [104]

Select Is a Function or Is not a Function to correctly classify each relation.

Title Is a Function Is not a Function{(3, 7),(3, 6),(5, 4),(4, 7)} {(1, 5),(3, 5),(4, 6),(6, 4)} {(2, 3),(4, 2),(4, 6),(5, 8)} {(0, 4),(3, 2),(4, 2),(6, 5)}

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3 years ago

Factor the expression using the GCF 12 + 20

Rus_ich [418]

Answer:

3 + 5

Step-by-step explanation:

GCF = 4

12 / 4 = 3

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6 0

2 years ago

in exercises 15-20 find the vector component of u along a and the vecomponent of u orthogonal to a u=(2,1,1,2) a=(4,-4,2,-2)

Nimfa-mama [501]

Answer:

The component of $\vec u$ orthogonal to $\vec a$ is $\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)$ .

Step-by-step explanation:

Let $\vec u$ and $\vec a$ , from Linear Algebra we get that component of $\vec u$ parallel to $\vec a$ by using this formula:

$\vec u_{\parallel \,\vec a} = \frac{\vec u \bullet\vec a}{\|\vec a\|^{2}} \cdot \vec a$ (Eq. 1)

Where $\|\vec a\|$ is the norm of $\vec a$ , which is equal to $\|\vec a\| = \sqrt{\vec a\bullet \vec a}$ . (Eq. 2)

If we know that $\vec u =(2,1,1,2)$ and $\vec a=(4,-4,2,-2)$ , then we get that vector component of $\vec u$ parallel to $\vec a$ is:

$\vec u_{\parallel\,\vec a} = \left[\frac{(2)\cdot (4)+(1)\cdot (-4)+(1)\cdot (2)+(2)\cdot (-2)}{4^{2}+(-4)^{2}+2^{2}+(-2)^{2}} \right]\cdot (4,-4,2,-2)$

$\vec u_{\parallel\,\vec a} =\frac{1}{20}\cdot (4,-4,2,-2)$

$\vec u_{\parallel\,\vec a} =\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)$

Lastly, we find the vector component of $\vec u$ orthogonal to $\vec a$ by applying this vector sum identity:

$\vec u_{\perp\,\vec a} = \vec u - \vec u_{\parallel\,\vec a}$ (Eq. 3)

If we get that $\vec u =(2,1,1,2)$ and $\vec u_{\parallel\,\vec a} =\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)$ , the vector component of $\vec u$ is:

$\vec u_{\perp\,\vec a} = (2,1,1,2)-\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)$

$\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)$

The component of $\vec u$ orthogonal to $\vec a$ is $\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)$ .

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3 years ago