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4 years ago

Will give Brainliest! if right!

Mathematics

2 answers:

umka2103 [35]4 years ago

6 0

It would be 16 i think because theres two cases so 8 multiply by 2 is 16

Jet001 [13]4 years ago

4 0

There are 7 pairs per poster and u have 8 posters so, I think u should do 7 times 8 and u get 56 so its B.56

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treveon paddled 35 miles downstream in his boat. the speed of the water current was 4 miles per hour. mark paddled 10 miles upst

Nataly [62]

Answer: 6622

Step-by-step explanation:

3 0

3 years ago

You want to save $500 for a trip to Legoland you have planned in 6 months. Which inequality shows the least amount you must save

sladkih [1.3K]

Answer:

6x ≥ 500

Step-by-step explanation:

I took the test but also it means that in six months, you have to save that amount of money to get exactly or more than 500, so this would be the correct answer!

7 0

3 years ago

In the coordinate plane, the point X(-4, 0) is translated to the point X'(1, -3). Under the same translation, the points

topjm [15]

sorry i don't know the answer

6 0

3 years ago

Use words to write a comparison statement for the problem above.

pantera1 [17]

Looking at the question we can see that it says:

$41253>35214$

We should know what these symbols mean:

stands for something less than the other, for example:

We can say that 2 is less than 5 so in a symbolic form it can be written as,

$2$

Similarly,

$>$ stands for something greater than the other, for example:

We can say that 7 is greater than 5, in a symbolic form it can be written as,

$7>5$

So here, $42153>35214$ means that, 42153 is greater than 35214.

6 0

3 years ago

Read 2 more answers

The plane x + y + 2z = 12 intersects the paraboloid z = x2 + y2 in an ellipse. Find the points on the ellipse that are nearest t

Soloha48 [4]

The distance between a point $(x,y,z)$ and the origin is $\sqrt{x^2+y^2+z^2}$ . But since $(\sqrt{f(x)})'=\frac{f'(x)}{2\sqrt{f(x)}}$ , both $f(x)$ and $\sqrt{f(x)}$ have the same critical points, so we can consider instead the squared distance, $x^2+y^2+z^2$ .

We're looking for the extrema of $x^2+y^2+z^2$ subject to $x+y+2z=12$ and $z=x^2+y^2$ . The Lagrangian is

$L(x,y,z,\lambda,\mu)=x^2+y^2+z^2+\lambda(x+y+2z-12)+\mu(z-x^2-y^2)$

with critical points where the partial derivatives vanish:

$L_x=2x+\lambda-2\mu x=0\implies\lambda=2x(\mu-1)$

$L_y=2y+\lambda-2\mu y=0\implies\lambda=2y(\mu-1)$

$L_z=2z+2\lambda+\mu=0$

$L_\lambda=x+y+2z-12=0$

$L_\mu=z-x^2-y^2=0$

From the first two equations, it follows that $x=y$ .

Then in the last two equations,

$x+y+2z-12=0\implies x+z=6$

$z-x^2-y^2=0\implies z=2x^2$

$\implies x+2x^2=6\implies2x^2+x-6=(2x-3)(x+2)=0\implies x=\dfrac32\text{ or }x=-2$

If $x=\frac32$ , then $z=6-\frac32=\frac92$ ; if $x=-2$ , then $z=8$ .

So there are two critical points, $\left(\frac32,\frac32,\frac92\right)$ and $(-2,-2,8)$ .

Let $f(x,y,z)=\sqrt{x^2+y^2+z^2}$ . We have a minimum distance of $f\left(\frac32,\frac32,\frac92\right)=\boxed{\frac{3\sqrt{11}}2}$ and maximum distance of $f(-2,-2,8)=\boxed{6\sqrt2}$ .

8 0

4 years ago