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2 years ago

What is the definition of a mechanical wave?

Mathematics

1 answer:

solmaris [256]2 years ago

3 0

A.Requires a medium
A machismo wave is not capable of transmitting it’s energy through its vaccum so it needs a medium.

You might be interested in

How many solutions does the equation have?

SSSSS [86.1K]

Let's solve and find out.

4(2n - 4) + 3 = 8n - 19
8n - 16 + 3 = 8n - 19
8n - 13 = 8n - 19
-13 = -19

-13 and -19 are not equal, so the equation has no solutions.

Answer:
A. 0

6 0

3 years ago

Read 2 more answers

A candle in the shape of a circular cone has a base of radius r and a height of h that is the same length as the radius. which e

ladessa [460]

Answer:

$\frac{r(1-\sqrt{2})}{-3}$

Step-by-step explanation:

Volume of cone = $\frac{1}{3} \pi r^{2} h$

Since we are given that a circular cone has a base of radius r and a height of h that is the same length as the radius

= $\frac{1}{3} \pi r^{2} \times r$

= $\frac{1}{3} \pi r^{3}$

Surface area of cone including 1 base = $\pi r^{2} +\pi\times r \times \sqrt{r^2+h^2}$

Since r = h

So, area = $\pi r^{2} +\pi\times r \times \sqrt{r^2+r^2}$

= $\pi r^{2} +\pi\times r \times \sqrt{2r^2}$

= $\pi r^{2} +\pi\times r^2 \times \sqrt{2}$

Ratio of volume of cone to its surface area including base :

$\frac{\frac{1}{3} \pi r^{3}}{\pi r^{2} +\pi\times r^2 \times \sqrt{2}}$

$\frac{\frac{1}{3}r}{1+\sqrt{2}}$

$\frac{r}{3(1+\sqrt{2})}$

Rationalizing

$\frac{r}{3(1+\sqrt{2})} \times \frac{1-\sqrt{2}}{1-\sqrt{2}}$

$\frac{r(1-\sqrt{2})}{-3}$

Hence the ratio the ratio of the volume of the candle to its surface area(including the base) is $\frac{r(1-\sqrt{2})}{-3}$

8 0

3 years ago

Read 2 more answers

A. Region C <br> B. Region D <br> C. Region A<br> D. Region B

Fittoniya [83]

Answer:

dont know

Step-by-step explanation:

8 0

3 years ago

Estimate. then ,divide using the standard algorithm and check<br> 45.15/21

insens350 [35]

I estimated to 45 then divided 21 and got about 2.1

7 0

3 years ago

What is the answer please.... need help

mina [271]

Answer:

$f\left(x\right)=x^3-6x^2+3x+10$ is the function of the least degree has the real coefficients and the leading coefficients of 1 and with the zeros -1, 5, and 2.

Step-by-step explanation:

Given the function

$f\left(x\right)=x^3-6x^2+3x+10$

As the highest power of the x-variable is 3 with the leading coefficients of 1.

So, it is clear that the polynomial function of the least degree has the real coefficients and the leading coefficients of 1.

solving to get the zeros

$f\left(x\right)=x^3-6x^2+3x+10$

$0=x^3-6x^2+3x+10$ ∵ $f(x)=0$

$Factor\:x^3-6x^2+3x+10\::\:\left(x+1\right)\left(x-2\right)\left(x-5\right)=0$

$\left(x+1\right)\left(x-2\right)\left(x-5\right)=0$

Using the zero factor principle

if $ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)$

$x+1=0\quad \mathrm{or}\quad \:x-2=0\quad \mathrm{or}\quad \:x-5=0$

$x=-1,\:x=2,\:x=5$

Therefore, the zeros of the function are:

$x=-1,\:x=2,\:x=5$

$f\left(x\right)=x^3-6x^2+3x+10$ is the function of the least degree has the real coefficients and the leading coefficients of 1 and with the zeros -1, 5, and 2.

Therefore, the last option is true.

8 0

3 years ago