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dimulka [17.4K]
2 years ago
11

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.
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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

as

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

so

\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
2 years ago
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