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Debora [2.8K]
3 years ago
5

2 women and 7 girls can do a peice of work in 4 days.4 women and 4 girls can do the same work in 3 days. How long would it take

for 1 women and 1 girl to do the work.
Mathematics
1 answer:
svetlana [45]3 years ago
3 0

Answer:

x: work a woman can do in 1 day

y: work a girl can do in 1  day

Then

2x + 7y =1/4

4x + 4y =1/3

4x + 14y =2/4 (1)

4x + 4y  =1/3 (2)

=> Let (1) - (2), 10y =2/4-1/3 <=>10y = 1/6 <=> y = 1/(6x10) = 1/60 (work)

=> From (2),4x = 1/3 - 4x1/60 => 4x = 16/60 => x = 4/60 (work)

=> 1 day, a woman and a girl can do: 1/60+4/60 = 5/60 =1/12 (work)

Then, the day required for a woman and a girl to complete work: 12 days

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5 is multiplied by the difference of 7 and a number
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5(7 - x).

Step-by-step explanation:

That is 5 multiplied by 7 - x   where x is the number.

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The answer is C. (1, -3)

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A map is drawn with a scale of 1/2 inch to 5 actual. miles. If the map distance is 8 inches, whats the actual. distance
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3 years ago
compute the projection of → a onto → b and the vector component of → a orthogonal to → b . give exact answers.
Nina [5.8K]

\text { Saclar projection } \frac{1}{\sqrt{3}} \text { and Vector projection } \frac{1}{3}(\hat{i}+\hat{j}+\hat{k})

We have been given two vectors $\vec{a}$ and $\vec{b}$, we are to find out the scalar and vector projection of $\vec{b}$ onto $\vec{a}$

we have $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{b}=\hat{i}-\hat{j}+\hat{k}$

The scalar projection of$\vec{b}$onto $\vec{a}$means the magnitude of the resolved component of $\vec{b}$ the direction of $\vec{a}$ and is given by

The scalar projection of $\vec{b}$onto

$\vec{a}=\frac{\vec{b} \cdot \vec{a}}{|\vec{a}|}$

$$\begin{aligned}&=\frac{(\hat{i}+\hat{j}+\hat{k}) \cdot(\hat{i}-\hat{j}+\hat{k})}{\sqrt{1^2+1^1+1^2}} \\&=\frac{1^2-1^2+1^2}{\sqrt{3}}=\frac{1}{\sqrt{3}}\end{aligned}$$

The Vector projection of $\vec{b}$ onto $\vec{a}$ means the resolved component of $\vec{b}$ in the direction of $\vec{a}$ and is given by

The vector projection of $\vec{b}$ onto

$\vec{a}=\frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2} \cdot(\hat{i}+\hat{j}+\hat{k})$

$$\begin{aligned}&=\frac{(\hat{i}+\hat{j}+\hat{k}) \cdot(\hat{i}-\hat{j}+\hat{k})}{\left(\sqrt{1^2+1^1+1^2}\right)^2} \cdot(\hat{i}+\hat{j}+\hat{k}) \\&=\frac{1^2-1^2+1^2}{3} \cdot(\hat{i}+\hat{j}+\hat{k})=\frac{1}{3}(\hat{i}+\hat{j}+\hat{k})\end{aligned}$$

To learn more about scalar and vector projection visit:brainly.com/question/21925479

#SPJ4

3 0
1 year ago
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