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jarptica [38.1K]
3 years ago
14

Determine if the lines going through the given points are parallel.

Mathematics
1 answer:
iVinArrow [24]3 years ago
4 0

Answer:

Step-by-step explanation:

Line A: Neither

Line B: Neither

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Bill wants to see how many times he needs to roll a standard number cube before getting a 5. What is the probability that Bill w
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5/6 • 5/6 • 1/6 = 25/216 the answer is b
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Please help, been struggling with this for almost a week now.
nikitadnepr [17]

Answer:

A \approx 1140.3\ units^2

Step-by-step explanation:

To solve this problem we first need to find the diameter of the circle. We are told the point N is the circle, so we can assume it is the center of the circle. To find the diameter we can use the Pythagorean Theorem:

a^2 + b^2 = c^2\\40^2 + 9^2 = c^2\\1600 + 81 = c^2\\1681 = c^2\\\sqrt{1681} = c\\41 = c

The diameter is 41 units, which means the radius is \frac{41}{2}. From here we can calculate the area of the circle:

A_c = \pi r^2\\A_c = \pi (\frac{41}{2})^2\\A_c = \pi 420.25\\A_c \approx 1320.25

Now we need to subtract the area of the triangle from the area of the circle. Calculate the area of the triangle:

A_t = \frac{bh}{2}\\A_t = \frac{40 \times 9}{2}\\A_t = \frac{360}{2}\\A_t = 180

Subtract the triangle area from the circle area to get the area of the shaded area:

A = A_c - A_t\\A = \pi420.25 - 180\\A = 1140.25431\\A \approx 1140.3\ units^2

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How many eighths are in one fourth
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For p = $3000, r = 3.5%,and t = 5 years, Find the balance in an account When interest is compounded (a) quarterly, (b) monthly,a
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Answer:

a) $3,571.02

b) $3,572.9

c) $3,573.74

Step-by-step explanation:

Data provided in the question:

p = $3000,

r = 3.5%,

t = 5 years

a) quarterly

number of periods in a year, n = 4

Interest rate per period = 3.5% ÷ 4 = 0.875%

Now,

A = p\times \left( 1 + \frac{r}{n} \right)^{\Large{n \cdot t}}

A = total amount

n = number of times compounded per year

on substituting the respective values, we get

A = 3000 × \left( 1 + \frac{ 0.035 }{ 4 } \right)^{\Large{ 4 \cdot 5 }}

A = 3000 × [/tex]\cdot { 1.00875 } ^ { 20 }[/tex]

A = 3000 × 1.19034

A = $3,571.02

b) monthly

number of periods in a year, n = 12

Now,

A = p\times \left( 1 + \frac{r}{n} \right)^{\Large{n \cdot t}}

on substituting the respective values, we get

A = 3000 × \left( 1 + \frac{ 0.035 }{ 12 } \right)^{\Large{ 12 \cdot 5 }}

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A = 3000 × 1.190967

A = $3,572.9

c) continuously

A = pe^{r\times t}

on substituting the respective values, we get

A = 3,000 × e^{0.035\times 5}

or

A = 3,000 × e^{0.175}

or

A = 3,000 × 1.1912

or

A = $3,573.74

5 0
3 years ago
Read 2 more answers
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