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2 years ago

INTEREST Bianca invested $5000 in an account that pays 5% annual interest.

Mathematics

1 answer:

gavmur [86]2 years ago

4 0

Answer:

y=5000(1.05)^x

Step-by-step explanation:

Account balance = Principal(1+rate)^time

Account balance = y

rate is 5% annual interest

time = x years

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Find the cosine of each acute angle

Lemur [1.5K]

Answer:

cos P = 53 degrees

cos Q = 37 degrees

Step-by-step explanation:

cos P = $\frac{adjacent}{hypothenuse}$

cos P = $\frac{24}{40}$

cos P = 0.60

Cos-1 of 0.60 is 53.13 (round to the nearest whole number)

cos P = 53 degrees

cos Q = $\frac{adjacent}{hypothenuse}$

cos Q = $\frac{32}{40}$

cos Q = 0.80

Cos of 0.80 is 36.86 (round to the nearest whole number)

cos Q = 37 degrees

To check anser:

90 + 53 + 37 = 180

180 = 180

4 0

2 years ago

A pair of opposite rays would be.

vodomira [7]

Answer:

Option (4)

Step-by-step explanation:

Definition of a ray states that a segments which connects two points but extending only in one direction infinitely is called as Ray.

Opposite rays will have a common end but the direction of the arrows will be different.

Therefore, from the picture attached,

Opposite rays will be PA and PB.

Option (4) will be the correct option.

7 0

3 years ago

A comuter has $245 in his commuter savings account b. How much must he add to his account if he wants to have 20 weeks worth of

Arisa [49]

I guess it (5) I really not sure

4 0

3 years ago

Robert ran a race that was 2/3 of a mile in length. If he has completed 5/6 of the race, how far has he run in miles?

Darina [25.2K]

Answer:

5/9

Step-by-step explanation:

Let's begin by determining the factors of the problem. We know that Robert ran a race that is 2/3 of a mile long. Therefore, we know that 2/3 is a factor in the equation. We also know that, of the 2/3 mile, he ran 5/6 of it. Now the tricky part is deciding how the two fractions are related.

To make this easier, let's substitute a different number for 5/6. We'll say 2.

If he has completed TWICE the length of the race, how would you determine how far he ran?

You would multiply 2/3 by 2! This same principle can be applied to the problem.

To determine the total distance run in miles, we can write it as 2/3 * 5/6

(NOTE that * is known as "times" or "multiplied by")

With this, you multiply the numerators (2 * 5 = 10) and the denominators (3 * 6 = 18) and then make your fraction... 10/18!

But you're not done yet. Always remember to simplify when possible. Both terms are divisible by 2. Therefore, it can be written as 5/9.

Hope this helped!

7 0

2 years ago

If A, B, and C are integers between 1 and 10 (inclusive), how many different combinations of A, B, and C exist such that A

Marianna [84]

$\fontsize{18}{10}{\textup{\textbf{The number of different combinations is 120.}}}$

Step-by-step explanation:

A, B and C are integers between 1 and 10 such that A<B<C.

The value of A can be minimum 1 and maximum 8.

If A = 1, B = 2, then C can be one of 3, 4, 5, 6, 7, 8, 9, 10 (8 options).

If A = 1, B = 3, then C has 7 options (4, 5, 6, 7, 8, 9, 10).

If A = 1, B = 4, then C has 6 options (5, 6, 7, 8, 9, 10).

If A = 1, B = 5, then C has 5 options (6, 7, 8, 10).

If A = 1, B = 6, then C has 4 options (7, 8, 9, 10).

If A = 1, B = 7, then C has 3 options (8, 9, 10).

If A = 1, B = 8, then C has 2 options (9, 10).

If A = 1, B = 9, then C has 1 option (10).

So, if A = 1, then the number of combinations is

$n_1=1+2+3+4+5+6+7+8=\dfrac{8(8+1)}{2}=36.$

Similarly, if A = 2, then the number of combinations is

$n_2=1+2+3+4+5+6+7=\dfrac{7(7+1)}{2}=28.$

If A = 3, then the number of combinations is

$n_3=1+2+3+4+5+6=\dfrac{6(6+1)}{2}=21.$

If A = 4, then the number of combinations is

$n_4=1+2+3+4+5=\dfrac{5(5+1)}{2}=15.$

If A = 5, then the number of combinations is

$n_5=1+2+3+4=\dfrac{4(4+1)}{2}=10.$

If A = 6, then the number of combinations is

$n_6=1+2+3=\dfrac{3(3+1)}{2}=6.$

If A = 7, then the number of combinations is

$n_7=1+2=\dfrac{2(2+1)}{2}=3.$

If A = 3, then the number of combinations is

$n_8=1.$

Therefore, the total number of combinations is

$n\\\\=n_1+n_2+n_3+n_4+n_5+n_6+n_7+n_8\\\\=36+28+21+15+10+6+3+1\\\\=120.$

Thus, the required number of different combinations is 120.

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Link : https://brainly.in/question/4909567.

8 0

3 years ago