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1 year ago

A. B. C. D. E. Which expressions listed on the left are equivalent to ? Check all that apply. (Assume that a ≠ 0.) A B C D E

Mathematics

2 answers:

Neko [114]1 year ago

4 0

The Answer Is .A,B,AND E

netineya [11]1 year ago

4 0

Answer:

the answer is A, B, and E

Step-by-step explanattion

it's asking for the square root of 36a^8/225a^2 on E d gen u ity .

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When a baseball is hit by a batter, the height of the ball, h(t), at time t, t=0, is determined by the equation h(t)=-16t^2+64t+

Natali [406]

T over x is your starting debut
add 40 from your original answer and divide by two
triplemente the second number and switch your trinomial by three

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1 year ago

Read 2 more answers

A pipe that is 120 cm long resonates to produce sound of wavelengths 480 cm, 160 cm, and 96 cm but does not resonate at any wave

erma4kov [3.2K]

Answer:

A Pipe that is 120 cm long resonates to produce sound of wavelengths 480 cm, 160 cm and 96 cm but does not resonate at any wavelengths longer than these. This pipe is:

A. closed at both ends

B. open at one end and closed at one end

C. open at both ends.

D. we cannot tell because we do not know the frequency of the sound.

The right choice is:

B. open at one end and closed at one end .

Step-by-step explanation:

Given:

Length of the pipe, $L$ = 120 cm

Its wavelength $\lambda_1$ = 480 cm

$\lambda_2$ = 160 cm and $\lambda_3$ = 96 cm

We have to find whether the pipe is open,closed or open-closed or none.

Note:

The fundamental wavelength of a pipe which is open at both ends is 2L.
The fundamental wavelength of a pipe which is closed at one end and open at another end is 4L.

So,

The fundamental wavelength:

⇒ $4L=4(120)=480\ cm$

It seems that the pipe is open at one end and closed at one end.

Now lets check with the subsequent wavelengths.

For one side open and one side closed pipe:

An odd-integer number of quarter wavelength have to fit into the tube of length L.

⇒ $\lambda_2=\frac{4L}{3}$ ⇒ $\lambda_3=\frac{4L}{5}$

⇒ $\lambda_2=\frac{4(120)}{3}$ ⇒ $\lambda_3=\frac{4(120)}{5}$

⇒ $\lambda_2=\frac{480}{3}$ ⇒ $\lambda_3=\frac{480}{5}$

⇒ $\lambda_2=160\ cm$ ⇒ $\lambda_3=96\ cm$

So the pipe is open at one end and closed at one end .

6 0

1 year ago

Write an integer whose absolute value is greater than itself.

raketka [301]

The integer -1 has an absolute value of 1, which is greater than itself. Since all negative integers are by definition integers, their respective absolute values will be greater than themselves.

3 0

1 year ago

For the binomial expansion of (x + y)^10, the value of k in the term 210x 6y k is a) 6 b) 4 c) 5 d) 7

Sloan [31]

Answer:

a) 6

Step-by-step explanation:

Expanding the polynomial using the formula:

$(x+y)^n=\sum_{k=0}^n \binom{n}{k} x^{n-k} y^k$

Also

$\binom{n}{k}=\frac{n!}{(n-k)!k!}$

I think you mean $210x^6y^4$

We can deduce that this term will be located somewhere in the middle. So I will calculate $k= 5; k=6 \text{ and } k =7$ .

For $k=5$

$\binom{10}{5} (y)^{10-5} (x)^{5}=\frac{10!}{(10-5)! 5!}(y)^{5} (x)^{5}= \frac{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5! }{5! \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 } \\ =\frac{30240}{120} =252 x^{5} y^{5}$

Note that we actually don't need to do all this process. There's no necessity to calculate the binomial, just $x^{n-k} y^k$

For $k=6$

$\binom{10}{6} \left(y\right)^{10-6} \left(x\right)^{6}=\frac{10!}{(10-6)! 6!}\left(y\right)^{4} \left(x\right)^{6}=210 x^{6} y^{4}$

5 0

2 years ago

if f(x)= 2x^2+5sqrt(x-2), complete the following statement: The domain for f(x) is all real numbers _____ than or equal to 2. (f

SCORPION-xisa [38]

Answer:

Greater.

Step-by-step explanation:

The in the function $f(x)=2x^2+5\sqrt{x-2}$ , the square root term $\sqrt{x-2}$ has to give a real number if $f(x)$ is to be real. This can only happen if $x\geq 2$ because if $x$ then $\sqrt{x-2}$ will give a complex number and therefore $f(x)$ will not be real.

Thus, the domain for f(x) is all real numbers greater than or equal to 2.

7 0

2 years ago