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Alborosie
3 years ago
5

Find the sum of the whole numbers from 1 to 740?

Mathematics
1 answer:
goldenfox [79]3 years ago
7 0

Answer:

126,170

Step-by-step explanation:

370 * 741 = 126170

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Express f(x) = 3 cos x + 5 sin x as Rsin(x + o), where e<br>is an acute angle in radians.​
Andrews [41]

Answer:

<h2>√34sin(x + 0.33π)</h2>

Step-by-step explanation:

The general form of the equation acosx + bsinx = Rsin(x + e) where R is the resultant of the constants 'a' and 'b' and e is the angle between them.

R = √a²+b²

e = tan^{-1}\frac{b}{a}

Given the function f(x) = 3 cos x + 5 sin x, comparing with the general equation;

a = 3, b = 5

R = √3²+5²

R = √9+25

R =√34

e = tan^{-1} \frac{5}{3} \\e = 59.09^{0}

in radians;

e =\frac{\pi }{180}*59.09\\ e = 0.33\pi rad

3 cos x + 5 sin x = √34sin(x + 0.33π)

4 0
3 years ago
(-(-)(-)(-10x))=-5 solve for x
kirza4 [7]
A negative times a negative is a positive so...

( - ) ( - ) ( - ) ( -10x ) = -5

10x = -5

x = -5/10

x = -1/2
4 0
4 years ago
If we used the points (1,8) and (2,16), what is the slope?
Naddik [55]

Answer:  8

Step-by-step explanation:  

See the image

7 0
3 years ago
Could somebody pls help me with this question
natulia [17]
simplify to 1 hour which will be 6 minutes in 1 hour

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5 0
3 years ago
Read 2 more answers
There are three grades in the school. One grade has 1/3 of the students, one grade has 1/4 of the students. What fraction of stu
AlladinOne [14]

Answer:

Step-by-step explanation:

Let x represent the total number of stdents that has all grades in the school.

There are three grades in the school. One grade has 1/3 of the students, this means that number of students that belongs tho this grade is 1/3 × x = x/3

One grade has 1/4 of the students, this means that number of students that belongs to this grade is 1/4 × x = x/4

Total number of students in both grades would be x/3 +/x/4 = 7x/12

The number of students in the remaining grade would be

x - 7x/12 = 5x/12

fraction of students in the remaining grade would be

(5x/12)/x = 5/12

4 0
4 years ago
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