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3 years ago

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When Jeremy made 8 black and white copies and 2 color copies at a copy shop, the cost was $1.20. When he made 10 black and white

copies and 10 color copies, the cost was $3.00. How much do black and white copies cost?
A. $0.10
B. $0.15
C. $0.20
D. $0.30

1 answer:

marshall27 [118]3 years ago

5 0

I think it’s A and it’s B at the most but still...A

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Is 1.99 a rational number?

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Yes, an irrational number cannot be counted, for example, pi is an irrational number.

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2 years ago

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A frog sits at one end of a table which is 2m long. In its first jump the frog goes a distance of 1m along the table, with its s

Phantasy [73]

Ah, this is an infinite sum question, or a sum of geometric sequence
when will the sum reach 1cm of the edge or about 1.99m
so

2m=200cm
within 1cm means at least 1.99m
so
we will use m and not cm for consitancy

sum of geometric sequence is
$S_n= \frac{a_1(1-r^n)}{1-r}$
a1=first term=initial jjump=1
r=common ratio=1/2
n=?, we ar solving for that

so
we want it to equal 1.99 so

$1.99= \frac{1(1- (\frac{1}{2})^n)}{1-\frac{1}{2}}$
$1.99= \frac{(1- (\frac{1}{2})^n)}{\frac{1}{2}}$
$1.99= 2(1- (\frac{1}{2})^n)$
divide both sides by 2
$\frac{1.99}{2} = 1- (\frac{1}{2})^n$
times -1
$\frac{-1.99}{2} = (\frac{1}{2})^n-1$
add 1 or 2/2 to both sides
$\frac{0.01}{2} = (\frac{1}{2})^n$
take the ln of both sides
$ln(\frac{0.01}{2}) = ln((\frac{1}{2})^n)$
$ln(\frac{0.01}{2}) = n ln(\frac{1}{2})$
divide both sides by ln(1/2)
$\frac{ln(\frac{0.01}{2})}{ln(\frac{1}{2})} =n$
use your calculatro to find that n≈7.64386
so on 7th jump, it is not yet at 1cm to the edge but at 8th jump, it is past
so 8th jump

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3 years ago

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Answer:

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kolezko [41]

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3 years ago

Which of the following are solutions to the equation below?

Luden [163]

Answer:

c) -4/5

d) 4/5

Step-by-step explanation:

The given equation is 25x^2 -16 = 0

Adding 16 on both sides, we get

25x^2 = 16

Dividing both sides by 25, we get

x^2 = 16/25

Taking square root on both sides, we get

x = √(16/25)

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Therefore, x = 4/5 and x = -4/5

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Hope this will helpful to understand the concept.

Thank you.

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