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olasank [31]
3 years ago
6

The Greatest Common Factor (GCF) of 24, 36, and 48 is O A. 2 O B. 24 O C. 12 O D. 6​

Mathematics
2 answers:
Karolina [17]3 years ago
7 0

Answer: C. 12

Step-by-step explanation:

GREYUIT [131]3 years ago
6 0

Answer:

Greatest common factor (GCF) of 24, 36, 48 is 12.

Step-by-step explanation:

i searched it up;)

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In a G.P the 3rd term is 4 times the 1st term and the sum of the 2nd term and the 4th term is 30.find the common ratio,find the
STALIN [3.7K]

Answer:

the common ratio is either 2 or -2.

the sum of the first 7 terms is then either 765 or 255

Step-by-step explanation:

a geometric sequence or series of progression (these are the most common names for the same thing) means that every new term of the sequence is created by multiplying the previous term by a constant factor which is called the common ratio.

so,

a1

a2 = a1×f

a3 = a2×f = a1×f²

a4 = a3×f = a1×f³

the problem description here tells us

a3 = 4×a1

and from above we know a3 = a1×f².

so, f² = 4

and therefore the common ratio = f = 2 or -2 (we need to keep that in mind).

again, the problem description tells us

a2 + a4 = 30

a1×f + a1×f³ = 30

for f = 2

a1×2 + a1×2³ = 30

2a1 + 8a1 = 30

10a1 = 30

a1 = 3

for f = -2

a1×-2 + a1×(-2)³ = 30

-10a1 = 30

a1 = -3

the sum of the first n terms of a geometric sequence is

sn = a1×(1 - f^(n+1))/(1-f) for f <>1

so, for f = 2

s7 = 3×(1 - 2⁸)/(1-2) = 3×-255/-1 = 3×255 = 765

for f = -2

s7 = -3×(1 - (-2)⁸)/(1 - -2) = -3×(1-256)/3 = -3×-255/3 =

= -1×-255 = 255

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2 years ago
I need project on thales theorem
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3 years ago
BONUS QUESTION: Is it true or false that each system of linear equations
Anastaziya [24]

Answer:

true because I know hdhdnndnd

Step-by-step explanation:

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5 0
3 years ago
Read 2 more answers
Describe and correct the error(s) made in each of the problems below.
ladessa [460]

Answer:

\displaystyle \frac{1-x}{(5-x)(-x)} =-\frac{x-1 }{ x(x-5)}

\displaystyle \frac{5}{s}\times \frac{2}{5} =\frac{2}{s}

Step-by-step explanation:

<u>Errors in Algebraic Operations </u>

It's usual that students make mistakes when misunderstanding the application of algebra's basic rules. Here we have two of them

  • When we change the signs of all the terms of a polynomial, the expression must be preceded by a negative sign
  • When multiplying negative and positive quantities, if the number of negatives is odd, the result is negative. If the number of negatives is even, the result is positive.
  • Not to confuse product of fractions with the sum of fractions. Rules are quite different

The first expression is

1-x / (5-x)(-x)=x-1 / x(x-5)

Let's arrange into format:

\displaystyle \frac{1-x}{(5-x)(-x)} =\frac{x-1 }{ x(x-5)}

We can clearly see in all of the factors in the expression the signs were changed correctly, but the result should have been preceeded with a negative sign, because it makes 3 (odd number) negatives, resulting in a negative expression. The correct form is

\displaystyle \frac{1-x}{(5-x)(-x)} =-\frac{x-1 }{ x(x-5)}

Now for the second expression

5/s+2/5=2/s

Let's arrange into format

\displaystyle \frac{5}{s}+\frac{2}{5} =\frac{2}{s}

It's a clear mistake because it was asssumed a product of fractions instead of a SUM of fractions. If the result was correct, then the expression should have been

\displaystyle \frac{5}{s}\times \frac{2}{5} =\frac{2}{s}

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3 years ago
How to isolate scalars from a matrix
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The technique of matrix isolation involves condensing the substance to be studied with a large excess of inert gas (usually argon or nitrogen) at low temperature to form a rigid solid (the matrix). The early development of matrix isolation spectroscopy was directed primarily to the study of unstable molecules and free radicals. The ability to stabilise reactive species by trapping them in a rigid cage, thus inhibiting intermolecular interaction, is an important feature of matrix isolation. The low temperatures (typically 4-20K) also prevent the occurrence of any process with an activation energy of more than a few kJ mol-1. Apart from the stabilisation of reactive species, matrix isolation affords a number of advantages over more conventional spectroscopic techniques. The isolation of monomelic solute molecules in an inert environment reduces intermolecular interactions, resulting in a sharpening of the solute absorption compared with other condensed phases. The effect is, of course, particularly dramatic for substances that engage in hydrogen bonding. Although the technique was developed to inhibit intermolecular interactions, it has also proved of great value in studying these interactions in molecular complexes formed in matrices at higher concentrations than those required for true isolation.
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3 years ago
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