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lana [24]
3 years ago
12

Which expression represents the greatest common factor (GCF) of 48 and 136?

Mathematics
2 answers:
Romashka-Z-Leto [24]3 years ago
7 0

Answer:

B

Step-by-step explanation:

C and D dont go into 48 AND 136 evenly. B is the largest one there that goes into both 48 AND 136.

ta-da! :P

Lina20 [59]3 years ago
3 0

The answer is B

2 x 2 = 4

4 x 2 = 8

8 x 6 = 48

8 x 17 = 136

why isn't it C or D?

For C, 8 x 3 = 24 (throwing the 8 in there because the expression is the same as B).

24 x 2 = 48, but 24 x 5 = 120 and 24 x 6 = 144.

It's also the same reason why it's not D, except that the expression goes WAY into the 400's

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Answer:

109 hope this helps you

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2 years ago
The parallelogram shown below has an area of ​​135 units2 Find the missing base.
Nostrana [21]
<h3>Answer:  9</h3>

===================================================

Explanation:

The area of the parallelogram is equal to the base times height. The formula to use is A = b*h. The height and base are always perpendicular to one another.

The base is some unknown variable b. Multiply the base by the height 15 to get 15*b which is equal to the area 135

Therefore, 15*b = 135

Divide both sides by 15 and we isolate b

15b = 135

15b/15 = 135/15

b = 9

So the base is 9

Check:

area = base*height = 9*15 = 135

so the answer is confirmed

side note: we never use the value 17 at all. It is likely put in there as a distraction.

3 0
3 years ago
SOMEONE PLEASE HELP I will give you brainliest it’s due in 11:59 pm TODAY!! Please
Snezhnost [94]
Shorter side: 130ft
longer side: 260ft
greatest possible area: 33800ft^2

basically get the equations for area and perimeter which will be

A: lw (length x width)
P: l + 2w

Substitute them into eachother and you get the equation:
-2w^2+520w = A

find the vertex of this parabola which will give you the greatest width which is 130m
Then u can find length and area from this width
4 0
2 years ago
Read 2 more answers
<img src="https://tex.z-dn.net/?f=%24a%2Ba%20r%2Ba%20r%5E%7B2%7D%2B%5Cldots%20%5Cinfty%3D15%24%24a%5E%7B2%7D%2B%28a%20r%29%5E%7B
riadik2000 [5.3K]

Let

S_n = \displaystyle \sum_{k=0}^n r^k = 1 + r + r^2 + \cdots + r^n

where we assume |r| < 1. Multiplying on both sides by r gives

r S_n = \displaystyle \sum_{k=0}^n r^{k+1} = r + r^2 + r^3 + \cdots + r^{n+1}

and subtracting this from S_n gives

(1 - r) S_n = 1 - r^{n+1} \implies S_n = \dfrac{1 - r^{n+1}}{1 - r}

As n → ∞, the exponential term will converge to 0, and the partial sums S_n will converge to

\displaystyle \lim_{n\to\infty} S_n = \dfrac1{1-r}

Now, we're given

a + ar + ar^2 + \cdots = 15 \implies 1 + r + r^2 + \cdots = \dfrac{15}a

a^2 + a^2r^2 + a^2r^4 + \cdots = 150 \implies 1 + r^2 + r^4 + \cdots = \dfrac{150}{a^2}

We must have |r| < 1 since both sums converge, so

\dfrac{15}a = \dfrac1{1-r}

\dfrac{150}{a^2} = \dfrac1{1-r^2}

Solving for r by substitution, we have

\dfrac{15}a = \dfrac1{1-r} \implies a = 15(1-r)

\dfrac{150}{225(1-r)^2} = \dfrac1{1-r^2}

Recalling the difference of squares identity, we have

\dfrac2{3(1-r)^2} = \dfrac1{(1-r)(1+r)}

We've already confirmed r ≠ 1, so we can simplify this to

\dfrac2{3(1-r)} = \dfrac1{1+r} \implies \dfrac{1-r}{1+r} = \dfrac23 \implies r = \dfrac15

It follows that

\dfrac a{1-r} = \dfrac a{1-\frac15} = 15 \implies a = 12

and so the sum we want is

ar^3 + ar^4 + ar^6 + \cdots = 15 - a - ar - ar^2 = \boxed{\dfrac3{25}}

which doesn't appear to be either of the given answer choices. Are you sure there isn't a typo somewhere?

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