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2 years ago

Assuming log, 4.4 = 1.4816 and log 7.7 = 2.0142, then whats the value of log, 7/4

Mathematics

2 answers:

Leni [432]2 years ago

6 0

7/4 = 7.7 / 4.4 , so:-

log 7/4 = log 7.7 / 4.4 = log 7.7 - log 4.4 = 2.0142 - 1.4816 = 0.5326

denis23 [38]2 years ago

4 0

Log (7/4) is equivalent to log (7.7/4.4), therefore it's log(7.7) - log(4.4), or...

2.0142 - 1.4816 = 0.5326

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The school chorus has 52 students which represents 26% of the seventh graders. How many students are in the seventh grade

barxatty [35]

Answer:

There are 200 students in the seventh grade.

Step-by-step explanation:

For equations like these, you have to use the is/of formula, which is:

Percent/100 = is/of.

Since you are trying to find how many students are in the seventh grade, this means that 52 is 26% of the total amount of students in the seventh grade. So, the formula would look like so:

26/100 = 52/n.

Next, you have to cross multiply, which means you would multiply 26 by n, and 52 by 100.

Then, you are going to have this left over from the cross multiplying:

5200 = 26n.

Divide 26 on both sides, leaving you with:

1n = 200, or n = 200.

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3 years ago

Each number in the sequence is 8 less than the previous number which list of number can be described by the following rule

Ganezh [65]

Answer: a

Step-by-step explanation:

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2 years ago

What is the The square root of 9 + 36

Blizzard [7]

Answer:

Step-by-step explanation:

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2 years ago

Read 2 more answers

36 - 40. In order to save for her high school graduation, Marie decided to save P200 pesos at the end of each month. If the

PtichkaEL [24]

Answer:

200.4 at 0.25%

Step-by-step explanation:

Given data

P= P200

r= 0.25%

t= 1 year

n= 12

A= P(1+ r/n)^nt

substitute

A= 200(1+ 0.0025/12)^12*1

A= 200(1+ 0.00020833333)^12

A= 200(1.0002)^12

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2 years ago

Find constants a and b such that the function y = a sin(x) + b cos(x) satisfies the differential equation y'' + y' − 5y = sin(x)

vichka [17]

Answers:

a = -6/37

b = -1/37

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Explanation:

Let's start things off by computing the derivatives we'll need

$y = a\sin(x) + b\cos(x)\\\\y' = a\cos(x) - b\sin(x)\\\\y'' = -a\sin(x) - b\cos(x)\\\\$

Apply substitution to get

$y'' + y' - 5y = \sin(x)\\\\\left(-a\sin(x) - b\cos(x)\right) + \left(a\cos(x) - b\sin(x)\right) - 5\left(a\sin(x) + b\cos(x)\right) = \sin(x)\\\\-a\sin(x) - b\cos(x) + a\cos(x) - b\sin(x) - 5a\sin(x) - 5b\cos(x) = \sin(x)\\\\\left(-a\sin(x) - b\sin(x) - 5a\sin(x)\right) + \left(- b\cos(x) + a\cos(x) - 5b\cos(x)\right) = \sin(x)\\\\\left(-a - b - 5a\right)\sin(x) + \left(- b + a - 5b\right)\cos(x) = \sin(x)\\\\\left(-6a - b\right)\sin(x) + \left(a - 6b\right)\cos(x) = \sin(x)\\\\$

I've factored things in such a way that we have something in the form Msin(x) + Ncos(x), where M and N are coefficients based on the constants a,b.

The right hand side is simply sin(x). So we want that cos(x) term to go away. To do so, we need the coefficient (a-6b) in front of that cosine to be zero

a-6b = 0

a = 6b

At the same time, we want the (-6a-b)sin(x) term to have its coefficient be 1. That way we simplify the left hand side to sin(x)

-6a -b = 1

-6(6b) - b = 1 .... plug in a = 6b

-36b - b = 1

-37b = 1

b = -1/37

Use this to find 'a'

a = 6b

a = 6(-1/37)

a = -6/37

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2 years ago