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Furkat [3]
3 years ago
11

What is ​ 27/36 ​ as a decimal? Enter your answer in the box.

Mathematics
2 answers:
Darya [45]3 years ago
7 0

Answer: Correction it's 0.75

Step-by-step explanation:

tamaranim1 [39]3 years ago
5 0

Awnser=0.75

You can first reduce this fraction by dividing both the numerator and denominator by the greatest common Factor of 27 and 36 using GFC(27,36)=9.

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kandy is financing a $335,000 mortgage for 30 years at a fixed rate of 7.5% what is the total cost of the principal and interest
-BARSIC- [3]
Given:
<span>F= $335,000
n = 30 years at a fixed rate of i = 7.5%

Required:
the total cost of the principal

Solution:
F = P(1+i)^n
P = F/(1+i)^n
P = 335,000 / (1.0.075)^30
P = 38,264.05</span>
8 0
2 years ago
H(n) = 3^n+3; Find h (3n)
olya-2409 [2.1K]

Answer:

The option which you chose is absolutely right.

Step-by-step explanation:

h(n) = 3^ n + 3

h(3n) = 3 ^3n + 3

Hope it will help :)

5 0
2 years ago
Someone please help!!!!!
Nadya [2.5K]

Answer:

144 people were originally at the party.

Step-by-step explanation:

18 boys is one fifth of the second boys population, so the second population is 18x9 = 162. The original population would be 162-18, which equals 144.

6 0
3 years ago
Yesterday, Brian had 97 baseball cards. Today, he got more. Using
Ket [755]

Answer:

97+x

Step-by-step explanation:

Let x represent how many cards he got.

3 0
3 years ago
Suppose you solved a second-order equation by rewriting it as a system and found two scalar solutions: y = e^5x and z = e^2x. Th
xenn [34]

Answer:

The solutions are linearly independent because the Wronskian is not equal to 0 for all x.

The value of the Wronskian is \bold{W=-3e^{7x}}

Step-by-step explanation:

We can calculate the Wronskian using the fundamental solutions that we are provided and their corresponding the derivatives, since the Wroskian is defined as the following determinant.

W = \left|\begin{array}{cc}y&z\\y'&z'\end{array}\right|

Thus replacing the functions of the exercise we get:

W = \left|\begin{array}{cc}e^{5x}&e^{2x}\\5e^{5x}&2e^{2x}\end{array}\right|

Working with the determinant we get

W = 2e^{7x}-5e^{7x}\\W=-3e^{7x}

Thus we have found that the Wronskian is not 0, so the solutions are linearly independent.

3 0
3 years ago
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