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3 years ago

Find all the values of k so that the given points are 29 square units apart. (-5, 5) and (k,0).

1 answer:

7 0

Distance formula between 2 points (x1,y1) and (x2,y2)

D= $\sqrt{(x1-x2)^{2}+(y1-y2)^{2}}$

D=29 and one point is (-5,5) and the other is (k,0)
sub

29= $\sqrt{(-5-k)^{2}+(5-0)^{2}}$
29= $\sqrt{(-5-k)^{2}+(5)^{2}}$
square both sides
841= $(-5-k)^{2}+(5)^{2}$
841= $(-5-k)^{2}+25$
minus 25 both sides
816= $(-5-k)^{2}$
square root both sides don't forget positive and negative roots
+/-4√51=-5-k
add 5 to both sides
5+/-4√51=-k
times -1 both sides
-5+/-4√51=k

the possible values for k are
k=-5+4√51 or k=-5-4√51

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A small theater had 6 rows of 26 chairs each. An extra 9 chairs have just been brought in. How many chairs are in the theater no

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Answer:165

Step-by-step explanation:

6•26 = 156. 156+9=165

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Write an equation of the line with the given properties. Your answer should be written in standard form.

Vanyuwa [196]

Given:

The slope of the line is m=0.

The line passes through the point P(-9,-3).

To find:

The equation of the line in standard form.

Solution:

Standard form of a line is:

$Ax+By=C$

The slope intercept form of the line is

$y-y_1=m(x-x_1)$

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It is given that the slope of the line is 3 and it passes through the point (-9,-3), so the equation of the line is

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$y+3=0$

$y=-3$

Therefore, the standard form of the given line is $y=-3$ .

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3 years ago

Suppose that a large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved. Another br

Lina20 [59]

Answer:

The differential equation for the amount of salt A(t) in the tank at a time t > 0 is $\frac{dA}{dt}=12 - \frac{2A(t)}{500+t}$ .

Step-by-step explanation:

We are given that a large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved. Another brine solution is pumped into the tank at a rate of 3 gal/min, and when the solution is well stirred, it is then pumped out at a slower rate of 2 gal/min.

The concentration of the solution entering is 4 lb/gal.

Firstly, as we know that the rate of change in the amount of salt with respect to time is given by;

$\frac{dA}{dt}= \text{R}_i_n - \text{R}_o_u_t$

where, $\text{R}_i_n$ = concentration of salt in the inflow $\times$ input rate of brine solution

and $\text{R}_o_u_t$ = concentration of salt in the outflow $\times$ outflow rate of brine solution

So, $\text{R}_i_n$ = 4 lb/gal $\times$ 3 gal/min = 12 lb/gal

Now, the rate of accumulation = Rate of input of solution - Rate of output of solution

= 3 gal/min - 2 gal/min

= 1 gal/min.

It is stated that a large mixing tank initially holds 500 gallons of water, so after t minutes it will hold (500 + t) gallons in the tank.

So, $\text{R}_o_u_t$ = concentration of salt in the outflow $\times$ outflow rate of brine solution

= $\frac{A(t)}{500+t} \text{ lb/gal } \times 2 \text{ gal/min}$ = $\frac{2A(t)}{500+t} \text{ lb/min }$

Now, the differential equation for the amount of salt A(t) in the tank at a time t > 0 is given by;

= $\frac{dA}{dt}=12\text{ lb/min } - \frac{2A(t)}{500+t} \text{ lb/min }$

or $\frac{dA}{dt}=12 - \frac{2A(t)}{500+t}$ .

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3 years ago

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Answer:

Step-by-step explanation:

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bogdanovich [222]

Answer:

Step-by-step explanation:

f(x)=17x

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f(x)-g(x)

51-12= 39

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