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3 years ago

In Oscars monthly budget each category is Assigned a certain percentage of his monthly income Oscars monthly income is $2250

Mathematics

2 answers:

ella [17]3 years ago

8 0

Answer:

Step-by-step explanation:

Marizza181 [45]3 years ago

3 0

I'm 100% sure of my answer

C,oscar budgets $485 of his monthly income for telephone,utilities, and emergencies.

hope this helps

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If the diagonal of a square is 12 to the square root of two then what is the area of the square

erastovalidia [21]

Answer:

Let there be a square ABCD with diagonal AC=12 root 2 cm

Since angles of a square are all 90 degrees

pythagoras theoream

AB^2+BC^2=AC^2

Since all sides of square are equal

2AB^2=(12 root 2)^2

2AB ^2=144*2

AB^2=144

AB= root 144=12 cm

area of square = s^2 =12^2 = 144cm^2

hope it helps

Step-by-step explanation:

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3 years ago

Mr. Sanchez is reading a biography. On Day 1 he had 225 pages left to

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4 years ago

Find the extreme values of f subject to both constraints f(x,y) = 2x^2+3y^2-4x-5, x^2 + y^2 <=16

Softa [21]

$f(x,y)=2x^2+3y^2-4x-5$
$f_x=4x-4=0\implies x=1$
$f_y=6y=0\implies y=0$

$f(x,y)$ has only one critical point at $(x,y)=(1,0)$ . The function has Hessian

$\mathbf H(x,y)=\begin{bmatrix}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\end{bmatrix}=\begin{bmatrix}4&0\\0&6\end{bmatrix}$

which is positive definite for all $(x,y)$ , which means $f(x,y)$ attains a minimum at the critical point with a value of $f(1,0)=-7$ .

To find the extrema (if any) along the boundary, parameterize it by $x=4\cos t$ and $y=4\sin t$ , with $0\le t$ . On the boundary, we have

$f(x(t),y(t))=F(t)=2(4\cos t)^2+3(4\sin t)^2-4(4\cos t)-5=32\cos^2t+48\sin^2t-16\cos t-5$
$F(t)=35-16\cos t-8\cos2t$

Find the critical points along the boundary:

$F'(t)=16\sin t+16\sin2t=16\sin t+32\sin t\cos t=16\sin t(1+2\cos t)=0$
$\implies t=0,\dfrac{2\pi}3,\pi,\dfrac{4\pi}3$

Respectively, plugging these values into $F(t)$ gives 11, 47, 43, and 47. We omit the first and third, as we can see the absolute extrema occur when $F(t)=47$ .

Now, solve for $x,y$ for both cases:

$t=\dfrac{2\pi}3\implies\begin{cases}x=4\cos t=-2\\y=4\sin t=2\sqrt3\end{cases}$

$t=\dfrac{4\pi}3\implies\begin{cases}x=4\cos t=-2\\y=4\sin t=-2\sqrt3\end{cases}$

so $f(x,y)$ has two absolute maxima at $(x,y)=(-2,\pm2\sqrt3)$ with the same value of 47.

5 0

3 years ago

Find the solutions of each equation on the interval [0, 2π). si (3pi/2+x)+ sin (3pi/2+x)=-2

ANEK [815]

Answer:

The solutions are 0° and 3π

Step-by-step explanation:

On solving the equation given;

$sin(\frac{3\pi}{2}+x )+ sin(\frac{3\pi}{2}+x ) = -2\\2sin(\frac{3\pi}{2}+x ) = -2\\sin(\frac{3\pi}{2}+x ) = -1\\\frac{3\pi}{2}+x = sin^{-1}-1\\ \frac{3\pi}{2}+x = -\frac{\pi}{2} \\x = -\frac{\pi}{2} -\frac{3\pi}{2} \\x = \frac{-4\pi}{2} \\x = -2\pi\\$

<u>Since sin is negative in the 3rd and 4th quadrant, </u>

In the 3rd quadrant;

x =180°+2π

x = π + 2π

x = 3π

In the 4th quadrant;

x = 360°-2π

x = 2π-2π

x = 0°

6 0

4 years ago

Plz help I'll give brainlest no links Determine the volume of the composite figure...

Verizon [17]

Answer:

36π + (18√29)π

Step-by-step explanation:

π × r² + π × r × √(r² + h²)

π × 6² + π × 6 × √(6² + 15²)

36π + (18√29)π ≈417.621

7 0

3 years ago