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3 years ago

Given the equation: = 4, which property should be applied first?

Mathematics

1 answer:

Ghella [55]3 years ago

3 0

This is = 4 therefore no property is applied

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Leonard owns a business earning $37,800 in profits. These profits are appreciating at about 2.3% each year. What are Leonard's p

satela [25.4K]

Answer:

B. $45341.60

Step-by-step explanation:

Given:

Initial profit = $37800
Rate of appreciation = 2.3% per year

The equation for this:

$P(t) = 37800*(1 + 0.023)^t = 37800*1.023^t$

Find the value of P(8):

$P(8) = 37800*1.023^8 = 45341.60$

Correct choice is B

3 0

2 years ago

USPS charges a flate rate of $5.50 for the 1st pound of shipping, plus $1.35 for each additional pound. FedEx charges $8.00 for

Rasek [7]

usps is cheaper

Step-by-step explanation:

usps equals 6 dollars 85 cents and fedex charges 9 dollars and 10 cents

5 0

3 years ago

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4x+7=64 i need the answer

irina [24]

Answer:

jniancun

Step-by-step explanation:

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7 0

2 years ago

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A group of students went on a field trip. The students were divided into 6 buses and each bus had less than 52 students. How man

Vikki [24]

Answer:

A maximum of 306

Step-by-step explanation:

You can estimate that since each bus had less than 52 students, that 51 were on each bus. 51 times 6=306

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2 years ago

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An acute triangle has sides measuring 10 cm and 16 cm. The length of the third side is unknown.

saveliy_v [14]

Answer: Choice B

12.5 < x < 18.9

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Explanation:

We have a triangle with these side lengths:

a = 10
b = 16
c = x = unknown

Let's assume that b = 16 is the largest side of this triangle.

By the converse of the pythagorean theorem, we need $b^2 < a^2+c^2$ to be true in order for an acute triangle to happen.

So,

$b^2 < a^2 + c^2\\\\c^2 > b^2 - a^2\\\\c > \sqrt{b^2-a^2}\\\\x > \sqrt{16^2-10^2}\\\\x > \sqrt{156}\\\\x > 12.4899959967968 \ \text{(approximate)}\\\\x > 12.5$

Now let's consider the possibility that the missing side x is actually the longest side.

Using the same theorem as before, we would say,

$c^2 < a^2 + b^2\\\\c < \sqrt{a^2 + b^2}\\\\x < \sqrt{10^2 + 16^2}\\\\x < \sqrt{356}\\\\x < 18.8679622641132 \ \text{(approximate)}\\\\x < 18.9\\\\$

We found that x > 12.5 and x < 18.9

This is the same as saying 12.5 < x and x < 18.9

Put together, they form the approximate answer of 12.5 < x < 18.9

6 0

2 years ago