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stellarik [79]
3 years ago
7

PLEASE HELP! URGENT Help with 6 & 7 please! Leave a explanation. Thanks :)

Mathematics
2 answers:
krok68 [10]3 years ago
8 0

Answer:

3,000,000

Step-by-step explanation:


timurjin [86]3 years ago
8 0

#6) He gets 1.5 percent commission on total contracts and then an additional 2.5 percent on anything over 1.5 million.

Since his total contracts is 2,555,500,

He gets 1.5 percent of that, plus 2.5 percent of "2,555,500 - 1,500,000" which is 1,055,500

So his total pay is:

0.015*2,555,500 + 0.025*1,055,500 = $64,720


#7) Tobias earns $1,959 per month plus 6.3% commission on sales.  So let "s" be sales. His monthly earnings can be calculated by:

1,959 + 0.063s

In order for this amount to be $2,784.84, we need to solve:

1,959 + 0.063s = 2,784.84

Solve:

0.063s = 825.84

s = $13,108.57143

or $13,108.57

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Over the interval 2 to 4 by how much does the function increse
gulaghasi [49]

The answer is 2 because two plus two equals 4 and 4-2=2 so the answer is 2. Hope this could help a lot and this was a useful answer


3 0
3 years ago
How you does this you how to fine what equal one
miskamm [114]
7/16 + 3/8 + Blank = 1
7/16 + 6/16 + 3/16 = 1
convert 3/8 to 6/16 then add 7+6 which = 13
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6 0
3 years ago
"Suppose an object falling in the atmosphere has mass m=15kg and the drag coefficient is γ=9kg/s. Recall that the differential e
Art [367]

Answer:

a. v(t)= -6.78e^{-16.33t} + 16.33 b. 16.33 m/s

Step-by-step explanation:

The differential equation for the motion is given by mv' = mg - γv. We re-write as mv' + γv = mg ⇒ v' + γv/m = g. ⇒ v' + kv = g. where k = γ/m.Since this is a linear first order differential equation, We find the integrating factor μ(t)=e^{\int\limits^  {}k \, dt } =e^{kt}. We now multiply both sides of the equation by the integrating factor.

μv' + μkv = μg ⇒ e^{kt}v' + ke^{kt}v = ge^{kt} ⇒ [ve^{kt}]' = ge^{kt}. Integrating, we have

∫ [ve^{kt}]' = ∫ge^{kt}

    ve^{kt} = \frac{g}{k}e^{kt} + c

    v(t)=   \frac{g}{k} + ce^{-kt}.

From our initial conditions, v(0) = 9.55 m/s, t = 0 , g = 9.8 m/s², γ = 9 kg/s , m = 15 kg. k = y/m. Substituting these values, we have

9.55 = 9.8 × 15/9 + ce^{-16.33 * 0} = 16.33 + c

       c = 9.55 -16.33 = -6.78.

So, v(t)=   16.33 - 6.78e^{-16.33t}. m/s = - 6.78e^{-16.33t} + 16.33 m/s

b. Velocity of object at time t = 0.5

At t = 0.5, v = - 6.78e^{-16.33 x 0.5} + 16.33 m/s = 16.328 m/s ≅ 16.33 m/s

6 0
3 years ago
Solve with interval notation Solve: 5 | x − 2 | + 4 > 8
Gnesinka [82]

Answer:

x<6/5, x>14/5

Step-by-step explanation:

Steps

$5\left|x-2\right|+4>8$

$\mathrm{Subtract\:}4\mathrm{\:from\:both\:sides}$

$5\left|x-2\right|+4-4>8-4$

$\mathrm{Simplify}$

$5\left|x-2\right|>4$

$\mathrm{Divide\:both\:sides\:by\:}5$

$\frac{5\left|x-2\right|}{5}>\frac{4}{5}$

$\mathrm{Simplify}$

$\left|x-2\right|>\frac{4}{5}$

$\mathrm{Apply\:absolute\:rule}:\quad\mathrm{If}\:|u|\:>\:a,\:a>0\:\mathrm{then}\:u\:<\:-a\:\quad\mathrm{or}\quad\:u\:>\:a$

$x-2<-\frac{4}{5}\quad\mathrm{or}\quad\:x-2>\frac{4}{5}$

Show Steps

$x-2<-\frac{4}{5}\quad:\quad x<\frac{6}{5}$

Show Steps

$x-2>\frac{4}{5}\quad:\quad x>\frac{14}{5}$

$\mathrm{Combine\:the\:intervals}$

$x<\frac{6}{5}\quad\mathrm{or}\quad\:x>\frac{14}{5}$

5 0
3 years ago
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mrs_skeptik [129]
855=95d

855 minutes is equal to 95 minutes times the number of days run
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