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madam [21]
3 years ago
9

Martin is saving for a gaming system. The total cost of the gaming system and three games is $325.49. About how much money shoul

d he save per week to puchased the gaming system and games in 20 weeks?
Mathematics
1 answer:
cestrela7 [59]3 years ago
4 0
First you would do $325.49 divided by 20 and you would get $16.27. so you would need $16.27 a week.
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Que es el teorema del coseno
kakasveta [241]

Answer:

La regla del coseno establece que el cuadrado de la longitud de cualquier lado de un triángulo es igual a la suma de los cuadrados de la longitud de los otros lados menos el doble de su producto multiplicado por el coseno de su ángulo incluido.

Step-by-step explanation:

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3 years ago
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nadezda [96]

Answer:

.

Step-by-step explanation:

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2 years ago
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The following function represents the value of a car, in dollars, after x years:
Assoli18 [71]

Answer:

Option D.The decrease in the value of the car, which is 8%

Step-by-step explanation:

we have a exponential function of the form

f(x)=a(b)^{x}

where

y is the value of the car

x is the time in years

a is the initial value

b is the base

r is the rate of decrease

b=1+r

In this problem we have

a=$24,000 initial value of the car

b=0.92

so

0.92=1+r

r=0.92-1=-0.08=-8%-----> is negative because is a rate of decrease

8 0
3 years ago
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Justin spends $21.75 on 6 rides and three sodas for his friends. Jerry spends $39.50 on 10 rides and 6 sodas for his friends. Ho
SIZIF [17.4K]
R = rides
S = sodas

6R + 3S = $21.75 —> -12R - 6S = -43.5
10R + 6S = $39.50–>10R + 6S = 39.5

Multiplying Justin’s whole equation by -2 will bring out the 6S’, so we can focus on the cost of one ride.

-2R = -4
Divide both sides by -2
So for one ride, it would cost $2.

To find the cost for one soda, we plug in the cost for a ride.

6(2) + 3S = $21.75
12 + 3S = $21.75
3S = $9.75
So for one soda, it would cost $3.25.
8 0
3 years ago
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Surface integrals using an explicit description. Evaluate the surface integral \iint_{S}^{}f(x,y,z)dS using an explicit represen
Jobisdone [24]

Parameterize S by the vector function

\vec r(x,y)=x\,\vec\imath+y\,\vec\jmath+f(x,y)\,\vec k

so that the normal vector to S is given by

\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}=\left(\vec\imath+\dfrac{\partial f}{\partial x}\,\vec k\right)\times\left(\vec\jmath+\dfrac{\partial f}{\partial y}\,\vec k\right)=-\dfrac{\partial f}{\partial x}\vec\imath-\dfrac{\partial f}{\partial y}\vec\jmath+\vec k

with magnitude

\left\|\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}\right\|=\sqrt{\left(\dfrac{\partial f}{\partial x}\right)^2+\left(\dfrac{\partial f}{\partial y}\right)^2+1}

In this case, the normal vector is

\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}=-\dfrac{\partial(8-x-2y)}{\partial x}\,\vec\imath-\dfrac{\partial(8-x-2y)}{\partial y}\,\vec\jmath+\vec k=\vec\imath+2\,\vec\jmath+\vec k

with magnitude \sqrt{1^2+2^2+1^2}=\sqrt6. The integral of f(x,y,z)=e^z over S is then

\displaystyle\iint_Se^z\,\mathrm d\Sigma=\sqrt6\iint_Te^{8-x-2y}\,\mathrm dy\,\mathrm dx

where T is the region in the x,y plane over which S is defined. In this case, it's the triangle in the plane z=0 which we can capture with 0\le x\le8 and 0\le y\le\frac{8-x}2, so that we have

\displaystyle\sqrt6\iint_Te^{8-x-2y}\,\mathrm dx\,\mathrm dy=\sqrt6\int_0^8\int_0^{(8-x)/2}e^{8-x-2y}\,\mathrm dy\,\mathrm dx=\boxed{\sqrt{\frac32}(e^8-9)}

5 0
3 years ago
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