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3 years ago

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Mathematics

1 answer:

Zanzabum3 years ago

4 0

Pic is kinda blurry so I can’t really answer sorry

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Find x <br><br> H,I x+12<br> I,J 16+x<br> H,J 20

AlladinOne [14]

Answer:

x=-4

Step-by-step explanation:

x+12+x+16=20

2x=20-28

x=-8/2

x=-4

plz mark me brainliest

5 0

3 years ago

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What numbers fall between 4.7 and 4.8 on a number line

SVEN [57.7K]

Answer:

4.71, 4.72, 4.73, 4.74, 4.75, 4.76,4.77, 4.78, and 4.79

Step-by-step explanation:

all you need to do is move into hundredths and you need to know that there are 10 hundredths between 2 tenths.

3 0

3 years ago

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a cylinder has a diameter of 24 inches. if its height is half its radius , what is the volume of the cylinder on cubic inches?

maksim [4K]

Answer:

approx. 905 in³

Step-by-step explanation:

The height is half its radius, and therefore the height is (1/2)(12 in), where 12 in represents half the diameter (24 in).

The formula for the volume of a cylinder of radius r and height h is

V = (1/3)(πr²)(h). Using the values r = 12 in, h = (1/2)(12 in), we get:

V= (1/3)π(12 in)²(6 in), or

V = (1/3)(144 in²)(6 in), or

V = 288π in³, or approx. 905 in³

8 0

3 years ago

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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

If $400 is invested at the rate of 6% for one year, what will bet the total value of the investment (the amount)?

Olenka [21]

Find 6% of $400 and add to $400.

6% of $400 =

= 6% * $400

= 0.06 * $400

= $24

6% of $400 is $24. That is the interest earned on $400 at 6% for 1 year.

Now we add it to the original amount, $400.

$400 + $24 = $424

Answer: $424

3 0

4 years ago

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