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3 years ago

Help Plz Will mark brainliest

Mathematics

1 answer:

Karolina [17]3 years ago

8 0

Answer:

b. $(x - 2) ^{2} - 36$

d. $(x - 2) ^{2} - 36$

Step-by-step explanation:

$(x + 4)(x - 8) \\ \\ = {x}^{2} - 8x + 4x + 4( - 8) \\ \\ = {x}^{2} - 4x - 32 \\ \\ = {x}^{2} - 4x + 4 - 4 - 32 \\ \\ = {x}^{2} - 4x + {2}^{2} - 36 \\ \\ = (x - 2) ^{2} - 36$

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Which is equivalent to the expression below?<br><br> 3p+4+p+12+3q

slega [8]

3p+16+p+3q

because 12+4=16

3 0

3 years ago

How to you write 1/25 using exponents

Montano1993 [528]

Answer:

$5^{-2}$

Step-by-step explanation:

Using the rule of exponents

$a^{-m}$ ⇔ $\frac{1}{a^{m} }$ , then

$\frac{1}{25}$ = $\frac{1}{5^{2} }$ = $5^{-2}$

5 0

3 years ago

Read 2 more answers

Layla wrote a report on insects. She listed the lengths of a tiger beetles, carpenter ants, and aphids. The lengths were 1/2 inc

ahrayia [7]

First we would change the fractions into decimals and compare them. That way it would be easy to solve when needed.

1/2=0.50
5/8=0.625
1/8=0.125

Now we can see that 1/8 (0.125) is the smallest, followed by 1/2 (0.50), and 5/8 the largest. Now we can match the sizes with the insects.

Since the tiger beetle is the largest we will place that with the length of 5/8 since that is the largest.

The carpenter ant is second largest so it would be matched with 1/2

The aphid is the smallest so that would be with 1/8 as the smallest

*Hope that helped :D*

3 0

3 years ago

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:

⇒ $4L=4(120)=480\ cm$