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4 years ago

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Last month, Belinda worked 160 hours at a rate of $10.00 per hour. If her employer retains 9% of her gross salary, is $1,500 a r

easonable estimate of her net salary?

1 answer:

sleet_krkn [62]4 years ago

8 0

We have been given that last month, Belinda worked 160 hours at a rate of $10.00 per hour.

Let us find amount earned by Belinda last month as:

$\text{Amount earned in 160 hours}=\$10\times 160$

$\text{Amount earned in 160 hours}=\$1600$

We are also told that Belinda's employer retains 9% of her gross salary. This means that Belinda's net income will be 91% of $1600. $(100\%-9\%=91\%)$ .

$\text{Belinda's net income would be}=\$1600 \times \frac{91}{100}$

$\text{Belinda's net income would be}=\$16\times 91$

$\text{Belinda's net income would be}=\$1456$

Upon rounding $1456 to nearest hundred, we will get:

$\text{Belinda's net income would be}\approx \$1500$

Therefore, $1500 is a reasonable estimate of her net salary.

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Two chemicals A and B are combined to form a chemical C. The rate, or velocity, of the reaction is proportional to the product o

Orlov [11]

Answer:

25.35 grams of C is formed in 14 minutes

after a long time , the limiting amount of C = 60g ,

A = 0 gram

and B = 30 grams; will remain.

Step-by-step explanation:

From the information given;

Let consider x(t) to represent the number of grams of compound C present at time (t)

It is obvious that x(0) = 0 and x(5) = 10 g;

And for x gram of C;

$\dfrac{2}{3}x$ grams of A is used ;

also $\dfrac{1}{3} x$ grams of B is used

Similarly; The amounts of A and B remaining at time (t) are;

$40 - \dfrac{2}{3}x$ and $50 - \dfrac{1}{3}x$

Therefore ; rate of formation of compound C can be said to be illustrated as ;

$\dfrac{dx}{dt }\propto (40 - \dfrac{2}{3}x)(50-\dfrac{1}{3}x)$

= $k \dfrac{2}{3}( 60-x) \dfrac{1}{3}(150-x)$

where;

k = proportionality constant.

= $\dfrac{2}{9}k (60-x)(150-x)$

By applying the separation of variable;

$\dfrac{1}{(60-x)(150-x)}dx= \dfrac{2}{9}k dt \\ \\ \\$

Solving by applying partial fraction method; we have:

$\{ \dfrac{1}{90(60-x)} - \dfrac{1}{90(150-x)} \}dx = \dfrac{2}{9}kdt$

$\dfrac{1}{90}(\dfrac{1}{x-150}-\dfrac{1}{x-60})dx =\dfrac{2}{9}kdt$

Taking the integral of both sides ; we have:

$\dfrac{1}{90}\int\limits(\dfrac{1}{x-150}- \dfrac{1}{x-60})dx= \dfrac{2}{9}\int\limits kdx$

$\dfrac{1}{90}(In(x-150)-In(x-60)) = \dfrac{2}{9}kt+C$

$\dfrac{1}{90}(In(\dfrac{x-150}{x-60})) = \dfrac{2}{9}kt+C$

$In( \dfrac{x-150}{x-60})= 20 kt + C_1 \ \ \ \ \ where \ \ C_1 = 90 C$

$\dfrac{x-150}{x-60}= Pe ^{20 kt} \ \ \ \ \ where \ \ P= e^{C_1}$

Applying the initial condition x(0) =0 to determine the value of P

Replace x= 0 and t =0 in the above equation.

$\dfrac{0-150}{0-60}= Pe ^{0}$

$\dfrac{5}{2}=P$

Thus;

$\dfrac{x-150}{x-60}=Pe^{20kt} \\ \\ \\ \dfrac{x-150}{x-60}=\dfrac{5}{2}e^{20kt} \\ \\ \\ 2x -300 =5e^{20kt}(x-60)$

$2x - 300 = 5xe^{20kt} - 300 e^{20kt} \\ \\ 5xe^{20kt} -2x = 300 e^{20kt} -300 \\ \\ x(5e^{20kt} -2) = 300 e^{20kt} -300 \\ \\ x= \dfrac{300 e^{20kt}-300}{5e^{20kt}-2}$

Thus;

$x(t)= \dfrac{300 e^{20kt}-300}{5e^{20kt}-2}$

Applying the initial condition for x(7) = 15 , to find the value of k

Replace t = 7 into $x(t)= \dfrac{300 e^{20kt}-300}{5e^{20kt}-2}$

$x(7)= \dfrac{300 e^{20k(7)}-300}{5e^{20k(7)}-2}$

$15= \dfrac{300 e^{140k}-300}{5e^{140k}-2}$

$75e^{140k}-30 ={300 e^{140k}-300}$

$225e^{140k}=270$

$e^{140k}=\dfrac{270}{225}$

$e^{140k}=\dfrac{6}{5}$

$140 k = In (\dfrac{6}{5})$

$k = \dfrac{1}{140}In (\dfrac{6}{5})$

k = 0.0013

Thus;

$x(t)= \dfrac{300 e^{20kt}-300}{5e^{20kt}-2}$

$x(t)= \dfrac{300 e^{20(0.0013)t}-300}{5e^{20(0.0013)t}-2}$

$x(t)= \dfrac{300 e^{(0.026)t}-300}{5e^{(0.026)t}-2}$

The amount of C formed in 14 minutes is ;

$x(14)= \dfrac{300 e^{(0.026)14}-300}{5e^{(0.026)14}-2}$

x(14) = 25.35 grams

Thus 25.35 grams of C is formed in 14 minutes

NOW; The limiting amount of C after a long time is:

$\lim_{t \to \infty} = \lim_{t \to \infty} (\dfrac{300 e^{(0.026)t}-300}{5e^{(0.026)t}-2})$

$\lim_{t \to \infty} (\dfrac{300- 300 e^{(0.026)t}}{2-5e^{(0.026)t}})$

As; $\lim_{t \to \infty} e^{-20kt} = 0$

⇒ $\dfrac{300}{5}$

= 60 grams

Therefore as t → $\infty$ ; x = 60

and the amount of A that remain = $40 - \dfrac{2}{3}x$

= $40 - \dfrac{2}{3}(60)$

= 40 -40

=0 grams

The amount of B that remains = $50 - \dfrac{1}{3}x$

= $50 - \dfrac{1}{3}(60)$

= 50 - 20

= 30 grams

Hence; after a long time ; the limiting amount of C = 60g , and 0 g of A , and 30 grams of B will remain.

I Hope That Helps You Alot!.

5 0

4 years ago

PLEASE NO LINKS DONT WASTE YOUR TIME IM NOT GONNA CLICK. What are the zeros of this function?

Artyom0805 [142]

Answer: C. x = 4 and x = 6

Step-by-step explanation:

Zeros of a function would be the value of x when f(x) is 0.

Based on the given graph, the parabola has 2 locations where the y-value is 0 and the graph hits the x-axis.

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3 years ago

A solid right circular cone of diameter 14cm and height 8cm is melted to form a hollow sphere. If the external diameter of the s

pshichka [43]

Answer:

Step-by-step explanation:

Diameter of cone=14 cm

so, Radius of cone= 7cm

Height of cone= 8cm

∴ Volume of cone= 1/3*pi*r^2*h

=1/3*pi*49*8 cm^3

In Sphere, External Radius R=5cm

Let, Internal Radius = r

Volume of Sphere = External Volume- Internal Volume

= 4/3*pi*R^3-4/3*pi*r^3

=4/3*pi[R^3-r^3]

=4/3*pi[5^3-r^3]

But During Conversion Volume Remains Same

So, Volume of Sphere=Volume of Cone

4/3*pi[125-r^3]= 1/3*pi*49*8

4[125-r^3]=49*8

[125-r^3]=49*2

r^3=125-98

r^3=27

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3 years ago

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Answer:

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Step-by-step explanation:

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1500 + 45 = 1545

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Molodets [167]

Answer:

See verification below

Step-by-step explanation:

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Now, to apply the theorem, we have that f(0)=0²+0-1=-1, f(5)=5²+5-1=29, then f(0)=-1<11<29=f(5). Additionally, f is continous since it is a polynomial. Then the IVT applies, and such c exists.

To find, c, solve the quadratic equation f(c)=11. This equation is c²+c-1=11. Rearranging, c²+c-12=0. Factor the expression to get (c+4)(c-3)=0. Then c=-4 or c=3. -4 is not in the interval, then we take c=3. Indeed, f(3)=3²+3-1=9+3-1=11.

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