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3 years ago

7

Working together, two pumps can drain a certain pool in 6 hours. If it takes the older pump 15 hours to drain the pool by itself

, how pump to drain the pool on its own?

1 answer:

stellarik [79]3 years ago

5 0

Answer:

10

Step-by-step explanation:

Rate of drainage for pump A and B = [1 pool drained]/[6 hours]

Rate of drainage for pump A (older pump) = [1 pool drained]/[15 hours]

Rate for pump B (newer pump) = [1 pool drained]/[t hours]

Thus, adding the rate of drainage of A and B,

1/t + 1/15 = 1/6

t=10

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Find the right quotient. 12x a y b ÷ (-6x a y)

Elena-2011 [213]

Answer: - 2 b

Step-by-step explanation:

Here the given expression,

$\frac{12xayb}{-6xay}$

= $\frac{12\times x\times a\times y\times b\times x^{-1}\times a^{-1}\times y^{-1}}{-6}$ $( \frac{1}{a^x} = a^{-x} )$

= $-2 x^{1-1}\times a^{1-1}\times y^{1-1}\times b$

$(a^m\times a^n = a^{m+n})$

= $-2\times x^{0}\times a^{0}\times y^{0}\times b$

= $-2 \times b$ $( a^0 = 1 )$

= $- 2b$

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3 years ago

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Jasmine earns $1200 per month as a base salary, plus 3% of her total sales. Write a linear model of her monthly earnings. Then u

BartSMP [9]

Answer:

The linear model of her monthly earnings is; Y =1200 +0.03*X. The amount earned is 1350; choice A

Step-by-step explanation:

Let Y denote her monthly earnings and X her total sales. The linear model can thus be formulated as; Y =1200 +0.03*X. We are then to find the value of Y given X =5000. We simply substitute X =5000 into the linear model which yields; Y =1350

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3 years ago

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Let L be the line with parametric equations x=5+t,y=6,z=−2−3t. Find the vector equation for a line that passes through the point

scZoUnD [109]

Answer:

The required equations are

$(-5 \hat i + 7 \hat j + 8 \hat k )+\lambda \left((10+\frac {3}{\sqrt {10}})\hat i -\hat j +(6- \frac {9}{\sqrt {10}})\hat k\right)=0$ and

$(-5 \hat i + 7 \hat j + 8 \hat k )+\lambda \left((10-\frac {3}{\sqrt {10}})\hat i -\hat j +(6+ \frac {9}{\sqrt {10}})\hat k\right)=0$ .

Step-by-step explanation:

The given parametric equation of the line, $L$ , is $x=5+t, y=6, z=-2-3t,$ so, an arbitrary point on the line is $R(x,y,z)=R(5+t, 6, -2-3t)$

The vector equation of the line passing through the points $P(-5,7,-8)$ and $R(5+t, 6, -2-3t)$ is

$\vec P + \lambda \vec{(PR)}=0$

$\Rightarrow (-5 \hat i + 7 \hat j - 8 \hat k )+\lambda \left((5+t-(-5))\hat i + (6-7)\hat j +(-2-3t-8)\hat k\right)=0$

$\Rightarrow (-5 \hat i + 7 \hat j - 8 \hat k )+\lambda \left((10+t)\hat i -\hat j +(6-3t)\hat k\right)=0\;\cdots (i)$

The given equation can also be written as

$\frac {x-5}{1}=\frac {v-6}{0}=\frac{z+2}{-3}=t \; \cdots (ii)$

The standard equation of the line passes through the point $P_0(x_0,y_0,z_0)$ and having direction $\vec v= a_1 \hat i +a_2 \hat j +a_3 \hat k$ is

$\frac {x-x_0}{a_1}=\frac {y-y_0}{a_2}=\frac{z-z_0}{a_3}=t \;\cdots (iii)$

Here, The value of the parameter, $t$ , of any point $R$ at a distance $d$ from the point, $P_0$ , can be determined by

$|t \vec v|=d\;\cdots (iv)$

Comparing equations $(ii)$ and $(iii)$

The line is passing through the point $P_0 (5,6,-2)$ having direction $\vec v=\hat i -3 \hat k$ .

Note that the point $Q(5,6,-2)$ is the same as $P_0$ obtained above.

Now, the value of the parameter, $t$ , for point $R$ at distance $d=3$ from the point $Q(5,6,-2)$ can be determined by equation $(iv)$ , we have

$|t(\hat i -3 \hat k)|=3$

$\Rightarrow t^2|(\hat i -3 \hat k)|^2=9$

$\Rightarrow 10t^2=9$

$\Rightarrow t^2=\frac {9}{10}$

$\Rightarrow t=\pm \frac {3}{\sqrt {10}}$

Put the value of $t$ in equation $(i)$ , the required equations are as follows:

For $t= \frac {3}{\sqrt {10}}$

$(-5 \hat i + 7 \hat j - 8 \hat k )+\lambda \left((10+\frac {3}{\sqrt {10}})\hat i -\hat j +\left(6-3\times \frac {3}{\sqrt {10}})\hat k\right)=0$

$\Rightarrow (-5 \hat i + 7 \hat j - 8 \hat k )+\lambda \left((10+\frac {3}{\sqrt {10}})\hat i -\hat j +(6- \frac {9}{\sqrt {10}})\hat k\right)=0$

and for $t= -\frac {3}{\sqrt {10}}$ ,

$(-5 \hat i + 7 \hat j - 8 \hat k )+\lambda \left((10+\left (-\frac {3}{\sqrt {10}}\right))\hat i -\hat j +(6-3\times \left(-\frac {3}{\sqrt {10}}\right)\hat k\right)=0$

$\Rightarrow (-5 \hat i + 7 \hat j - 8 \hat k )+\lambda \left((10-\frac {3}{\sqrt {10}})\hat i -\hat j +(6+ \frac {9}{\sqrt {10}})\hat k\right)=0$

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Construct the graph of the direct proportion y = kx for k = -1

Sergio [31]

Answer:

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