When is a quadrilateral a parallelogram?

1 1/2 times a number equals 6

olga nikolaevna [1]

$1\frac{1}{2}x=6\\x=4$

⭐ Please consider brainliest! ⭐

✉️ If any further questions, inbox me! ✉️

8 0

4 years ago

Read 2 more answers

Solve for x: two thirds plus one third times x equals two times x

Salsk061 [2.6K]

Answer: x = 2/5 in fraction form

In decimal form, the answer is x = 0.4

====================================================

Work Shown:

2/3 = two thirds

2/3 + (1/3)x = 2x is the equation you're given

Multiply every term by 3 so that we clear out the fractions

3*(2/3) + 3*(1/3)x = 3*2x

2 + x = 6x

From here, we solve for x

2 + x = 6x

2+x-x = 6x-x .... subtract x from both sides

2 = 5x

5x = 2

5x/5 = 2/5 ... divide both sides by 5

x = 2/5

x = 0.4

7 0

3 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 slope to y=8/9x-9 and the y intercept

Liula [17]

$The \ general \ form \ of \ a \ straight-line \ equation \ is: \\\\ y=mx+b\\\\where \ "m " \ is \ the \ slope \ of \ the \ line \ and \ "b" \ is \ y -intercept \\ \\ y= \frac{8}{9}x-9 \\\\m=\frac{8}{9}\\b=-9 \\ \\Answer: \ Slope \ is \ equal \ to \ 6 \\ y-intercept \ is \ equal \ ( -9).$

7 0

3 years ago

Read 2 more answers

A right rectangular prism is shown. The two dimensional net is also shown. Find the surface area of this prism.

Natasha_Volkova [10]

Hey!

------------------------------------------------

Surface Area Formula: 2LW * 2LH * 2WH

------------------------------------------------

2LW:

5 x 1 = 5

5 x 2 = 10

------------------------------------------------

2LH:

4 x 1 = 4

4 x 2 = 8

------------------------------------------------

2WH:

5 x 4 = 20

20 x 2 = 40

------------------------------------------------

Surface Area:

40 + 8 + 10= 58

------------------------------------------------

Answer:

D) 58 cm2

------------------------------------------------

Hope This Helped! Good Luck!

5 0

3 years ago