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earnstyle [38]
3 years ago
9

Serena is writing an expression that is equivalent to 12ab – 6b, but she has not completed it.

Mathematics
2 answers:
Veseljchak [2.6K]3 years ago
7 0

Answer:

the person above me is correct!!!!!!!

Step-by-step explanation:

azamat3 years ago
4 0

Given:

The expression is

12ab-6b

The equivalent expression of this expression is

\_\_\_?\_\_\_(2a-1)

To find:

The missing value.

Solution:

We have,

12ab-6b

The factor form of 12ab is

12ab=2\times 2\times 3\times a\times b

The factor form 6b is

6b=2\times 3\times b

The common factors of 12ab and 6b are 2,3,b.

HCF(12ab,6b)=2\times 3\times b

HCF(12ab,6b)=6b

Taking out the highest common factors(HCF) from the given expression, we get

12ab-6b=6b(2a-1)

On comparing this with \_\_\_?\_\_\_(2a-1), we get the missing value which is equal to 6b.

Therefore, the correct option is A.

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3 years ago
Can someone help me with this ASAP please I’m being timed !
Paladinen [302]

Answer:

A

Step-by-step explanation:

8 0
3 years ago
The ΔPQR is right-angled at P, and PN is an altitude. If QN = 12 in and NR = 6 in, find PN, PQ, PR.
zaharov [31]

Answer:

PR = 6√2, PN = 6, PQ = 12√2

Step-by-step explanation:

Find the unknowns through Sin, Cos, Tan formulae.

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4 0
4 years ago
The length of a 1-inch paperclip is about 1.6 x 10−5 miles. It takes about 1.5 x 1010 paperclips linked together to reach the mo
son4ous [18]

Answer:

Distance from the moon is 2.4 × 10⁵ miles

Step-by-step explanation:

Length of 1-inch paperclip = 1.6\times 10^{-5} miles

Number of paperclips used to reach the moon = 1.5 × 10¹⁰

Total distance from the moon = Number of paper clips used × Length of one clip

= (1.5 × 10¹⁰) × (1.6 × 10⁻5)

= (1.5 × 1.6) (10¹⁰× 10⁻⁵)

= 2.4 × 10⁽¹⁰⁻⁵⁾ miles

= 2.4 × 10⁵ miles

Distance from the moon is 2.4 × 10⁵ miles.

5 0
3 years ago
He loudness of a pile driver is 112 dB. About how many times the sound intensity of the jackhammer is the sound intensity of a p
chubhunter [2.5K]

Complete question :

The loudness of a jack hammer is 96 dB. Its sound intensity is about 0.004.

The loudness of a compactor is 94 dB. Its sound intensity is about 0.0025.

The sound intensity of the jack hammer is about 1.6 times the sound intensity of the compactor.

The loudness of a pile driver is 112 dB. About how many times the sound intensity of the jackhammer is the sound intensity of a pile driver? Round to the nearest ten.

Answer:

40

Step-by-step explanation:

Given that:

Sound intensity of jackhammer = 0.004

Sound intensity of compactor = 0.0025

Sound intensity of jackhammer is about 1.6 times the sound intensity of a compactor

Each factor of 10 in intensity corresponds to 10dB

Pile driver with loudness of 112dB ; 112 / 10 = 11.2

Hence, 10^11.2 = 1.58489 * 10^11

Hence, sound intensity (I) in watts per m²

I = 1.58489 * 10^11 * 10^-12

I = 1.58489 * 10-1

I = 0.158489

Comparing the intensity of pile driver and jackhammer :

Intensity of piledriver / Intensity of jackhammer

0.158489 / 0.004

= 39.622

= 40 (nearest ten)

3 0
3 years ago
Read 2 more answers
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