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Anon25 [30]
3 years ago
11

Find the value of −b8 when b=20

Mathematics
1 answer:
Viefleur [7K]3 years ago
6 0
-160

Since -(20x8) the answer would be -160
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What is your wire value in the X value is five? 18.6, 16.6, 17.6, 19.6
Delvig [45]

Answer:

18.6 i think

Step-by-step explanation:

6 0
3 years ago
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Two numbers are in the ratio 7:5 and their difference is 22.Find the numbers​
omeli [17]

Answer:

Let x and y be the two numbers. Then

yx=57  and x−y=20

Cross multiplying the proportion

5x=7y  or y=75x

So, x−y=20

x−75x=20⇒77x−5x=72x=20

⇒x=2140=70

y=75x=75×70=50

So smaller number is 50.

Check :yx=5070=57 and 70−50=20

4 0
3 years ago
A line passes through the points (7, 10) and (7, 20). Which statement is true about the line? It has a slope of zero because x 2
yaroslaw [1]

Answer:

It has no slope because x 2 minus x 1 in the formula m = StartFraction y 2 minus y 1 Over x 2 minus x 1 EndFraction is zero, and the denominator of a fraction cannot be zero.

Step-by-step explanation:

Points:

  • (7, 10) and (7, 20)

Slope-intercept form:

  • y= mx+b

slope is:

  • m= (y2-y1)/(x2-x1)= (20-10)/(7-7)= 10/0,

denominator of the fraction is zero, we can't divide by zero, so this line has no slope

----------

It has a slope of zero because x 2 minus x 1 in the formula m = StartFraction y 2 minus y 1 Over x 2 minus x 1 EndFraction is zero, and the numerator of a fraction cannot be zero.

  • no, numerator is not zero

It has a slope of zero because x 2 minus x 1 in the formula m = StartFraction y 2 minus y 1 Over x 2 minus x 1 EndFraction is zero, and the denominator of a fraction cannot be zero.

  • no, it has no slope

It has no slope because x 2 minus x 1 in the formula m = StartFraction y 2 minus y 1 Over x 2 minus x 1 EndFraction is zero, and the numerator of a fraction cannot be zero.

  • no, numerator is not zero

It has no slope because x 2 minus x 1 in the formula m = StartFraction y 2 minus y 1 Over x 2 minus x 1 EndFraction is zero, and the denominator of a fraction cannot be zero.

  • yes
3 0
4 years ago
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What is the perimeter of an Olympic size swimming pool
Sauron [17]
150 is the perimeter of the pool
4 0
4 years ago
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The population of mosquitoes in a certain area increases at a rate proportional to the current population, and in the absence of
Alexandra [31]

Answer:

Population of mosquitoes in the area at any time t is:

P(t) =504,943.26  -104,943.26e^{0.693t}

Step-by-step explanation:

assume population at any time t = P(t)

population increases at a rate proportional to the current population:

⇒dP/dt ∝ P

 \implies \frac{dP}{dt} =kP----(1)

where k is constant rate at which population is doubled

solving (1)

ln|P(t)|=kt +C\\P(t)= e^{kt+C}\\P(t)=Ce^{kt}

t=0\\P(0) = P_{o}\\\implies C= P_{o}\\P(t) =P_{o}e^{kt}\\ ---- (2)

initial population = 400,000

population is doubled every week

                                                 ⇒P(1)=2P(0)

Using (2)

                                 P_{o}e^{k(1)} = 2P_{o} e^{k(0)}\\

                                            e^{k} =2\\k=ln|2|\\

In presence of predators amount is decreased by 50,000 per day

Then amount decreased per week = 350,000

In this case (1) becomes

\frac{dP}{dt}=kP-350,000\\\frac{dP}{dt} - kP=-350,000\\ ---(3)

solving (3) by calculating integrating factor

                                          I.F=e^{\int-k dt}

Multiplying I.F with all terms of (3)

e^{-kt}\frac{dP}{dt} - ke^{-kt}P =-350,000 e^{-kt}\\\frac{d}{dt}(e^{-kt}P) =  -350,000 e^{-kt}

Integrating w.r.to t

                         e^{-kt}P(t)= \frac{350,000e^{-kt}}{k} +C

                         P(t) =\frac{350,000}{k} +Ce^{kt}\\

                                          k=ln|2| =0.693

                          P(t) =504,943.26 + Ce^{0.693t}\\

at t=0

                        P(0) =504,943.26 + Ce^{0.693(0)}

                        400,000 =504,943.26 + C

                           C = -104,943.26

So, population of mosquitoes in the area at any time t is

                  P(t) =504,943.26  -104,943.26e^{0.693t}

6 0
3 years ago
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