Ask question

Ask question

All categories

SVETLANKA909090 [29]

2 years ago

11

PLEASEEEEEEEEEE HELPPPPPPPPPPPPp It is in the picture.

2 answers:

ki77a [65]2 years ago

8 0

Answer:

D is the answer

Step-by-step explanation:

In-s [12.5K]2 years ago

6 0

Answer: D

Step-by-step explanation:

You might be interested in

2x-5y+5z=-10<br> 5x-4y+3z=-19<br> X-y+5z=17

yarga [219]

Answer:

x = -125/71 , y = 448/71 , z = 356/71

Step-by-step explanation:

Solve the following system:

{2 x - 5 y + 5 z = -10 | (equation 1)

5 x - 4 y + 3 z = -19 | (equation 2)

x - y + 5 z = 17 | (equation 3)

Swap equation 1 with equation 2:

{5 x - 4 y + 3 z = -19 | (equation 1)

2 x - 5 y + 5 z = -10 | (equation 2)

x - y + 5 z = 17 | (equation 3)

Subtract 2/5 × (equation 1) from equation 2:

{5 x - 4 y + 3 z = -19 | (equation 1)

0 x - (17 y)/5 + (19 z)/5 = (-12)/5 | (equation 2)

x - y + 5 z = 17 | (equation 3)

Multiply equation 2 by 5:

{5 x - 4 y + 3 z = -19 | (equation 1)

0 x - 17 y + 19 z = -12 | (equation 2)

x - y + 5 z = 17 | (equation 3)

Subtract 1/5 × (equation 1) from equation 3:

{5 x - 4 y + 3 z = -19 | (equation 1)

0 x - 17 y + 19 z = -12 | (equation 2)

0 x - y/5 + (22 z)/5 = 104/5 | (equation 3)

Multiply equation 3 by 5:

{5 x - 4 y + 3 z = -19 | (equation 1)

0 x - 17 y + 19 z = -12 | (equation 2)

0 x - y + 22 z = 104 | (equation 3)

Subtract 1/17 × (equation 2) from equation 3:

{5 x - 4 y + 3 z = -19 | (equation 1)

0 x - 17 y + 19 z = -12 | (equation 2)

0 x+0 y+(355 z)/17 = 1780/17 | (equation 3)

Multiply equation 3 by 17/5:

{5 x - 4 y + 3 z = -19 | (equation 1)

0 x - 17 y + 19 z = -12 | (equation 2)

0 x+0 y+71 z = 356 | (equation 3)

Divide equation 3 by 71:

{5 x - 4 y + 3 z = -19 | (equation 1)

0 x - 17 y + 19 z = -12 | (equation 2)

0 x+0 y+z = 356/71 | (equation 3)

Subtract 19 × (equation 3) from equation 2:

{5 x - 4 y + 3 z = -19 | (equation 1)

0 x - 17 y+0 z = (-7616)/71 | (equation 2)

0 x+0 y+z = 356/71 | (equation 3)

Divide equation 2 by -17:

{5 x - 4 y + 3 z = -19 | (equation 1)

0 x+y+0 z = 448/71 | (equation 2)

0 x+0 y+z = 356/71 | (equation 3)

Add 4 × (equation 2) to equation 1:

{5 x + 0 y+3 z = 443/71 | (equation 1)

0 x+y+0 z = 448/71 | (equation 2)

0 x+0 y+z = 356/71 | (equation 3)

Subtract 3 × (equation 3) from equation 1:

{5 x+0 y+0 z = (-625)/71 | (equation 1)

0 x+y+0 z = 448/71 | (equation 2)

0 x+0 y+z = 356/71 | (equation 3)

Divide equation 1 by 5:

{x+0 y+0 z = (-125)/71 | (equation 1)

0 x+y+0 z = 448/71 | (equation 2)

0 x+0 y+z = 356/71 | (equation 3)

Collect results:

Answer: {x = -125/71 , y = 448/71 , z = 356/71

5 0

3 years ago

in a laboratory, solution A will be mixed with solution B in a ratio of 3 to 2. what amount of solution A is needed to mix with

Ede4ka [16]

1,125 ML of solution A will be mixed

5 0

2 years ago

Given that the measure of ∠x is 60°, and the measure of ∠y is 90°, find the measure of ∠z.

UkoKoshka [18]

The answer is 30 because u need to add 60+90=150 and then subtract 180-150 which gives you 30

7 0

3 years ago

Read 2 more answers

Find the absolute extrema for f(x,y)=4-x^2-y^4+1/2y^2 over the closed disk D:x^2+y^2 is less than or equal to 1

algol [13]

Find the critical points of $f(x,y)$ :

$\dfrac{\partial f}{\partial x}=-2x=0\implies x=0$

$\dfrac{\partial f}{\partial y}=y-4y^3=y(1-4y^2)=0\implies y=0\text{ or }y=\pm\dfrac12$

All three points lie within $D$ , and $f$ takes on values of

$\begin{cases}f(0,0)=4\\f\left(0,-\frac12\right)=\frac{65}{16}\\f\left(0,\frac12\right)=\frac{65}{16}\end{cases}$

Now check for extrema on the boundary of $D$ . Convert to polar coordinates:

$f(x,y)=f(\cos t,\sin t)=g(t)=4-\cos^2-\sin^4t+\dfrac12\sin^2t=3+\dfrac32\sin^2t-\sin^4t$

Find the critical points of $g(t)$ :

$\dfrac{\mathrm dg}{\mathrm dt}=3\sin t\cos t-4\sin^3t\cos t=\sin t\cos t(3-4\sin^2t)=0$

$\implies\sin t=0\text{ or }\cos t=0\text{ or }\sin t=\pm\dfrac{\sqrt3}2$

$\implies t=n\pi\text{ or }t=\dfrac{(2n+1)\pi}2\text{ or }\pm\dfrac\pi3+2n\pi$

where $n$ is any integer. There are some redundant critical points, so we'll just consider $0\le t< 2\pi$ , which gives

$t=0\text{ or }t=\dfrac\pi3\text{ or }t=\dfrac\pi2\text{ or }t=\pi\text{ or }t=\dfrac{3\pi}2\text{ or }t=\dfrac{5\pi}3$

which gives values of

$\begin{cases}g(0)=3\\g\left(\frac\pi3\right)=\frac{57}{16}\\g\left(\frac\pi2\right)=\frac72\\g(\pi)=3\\g\left(\frac{3\pi}2\right)=\frac72\\g\left(\frac{5\pi}3\right)=\frac{57}{16}\end{cases}$

So altogether, $f(x,y)$ has an absolute maximum of 65/16 at the points (0, -1/2) and (0, 1/2), and an absolute minimum of 3 at (-1, 0).

5 0

2 years ago

Plzz Help ASAP!! i posted the picture?

Andre45 [30]

Hi!

So the original formula for a problem likes this or a quadratic is:

ax^2 + bx+ c

so a is the accelaration; b is the initial velocity; and c is the initial height

so none of those are the zeros, which means c is out

zeros are the x values when y is 0, that also means when the parabola crosses the x axis

so d is out too because the maximum height is the vertex which mean that there would probably be a y value. in some cases the vertex is the zero, but not here

B is out because x isn't time it is distance

your answer is A!

Hope this helps!

8 0

2 years ago

Read 2 more answers