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NikAS [45]
3 years ago
11

the number of boys and girls in a school are 480 and 576 respectively express the ratio of the number of boys to that of girls

Mathematics
2 answers:
Kaylis [27]3 years ago
7 0

Answer:

480:576 or 1:1.2

Step-by-step explanation:

Leni [432]3 years ago
3 0

Answer:

480 to 576

Step-by-step explanation:

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creativ13 [48]

(3x(a)-6x)- (6a-12)       Multiply the outside with the inside

3 0
3 years ago
The number of subscribers to a streaming video service doubles each week.
Ilya [14]

Answer:

Step-by-step explanation:

This is an exponential equation that is solved by taking the natural log of both sides.  The equation is

f(x)=2^x

If we are looking for x when f(x) = 64, then

64=2^x

Take the natural log of both sides:

ln(64)=ln(2^x)

The rules of logs allows us to bring the x down in front:

ln(64)=xln(2)

Divide both sides by ln(2) to get:

\frac{ln(64)}{ln(2)}=x

Do this on your calculator to get that x = 6.

You could also have just gone right to your calculator and started raising 2 to consecutive powers starting at like 3 or 4 to eventually get that 2 to the 6th power is equal to 64, but for the basics of solving log equations, you need to know how to do this.

5 0
3 years ago
Read 2 more answers
What is an equation of the line that passes through the point (4,-3)(4,−3) and is perpendicular to the line 4x+y=34x+y=3?
zlopas [31]

Step-by-step explanation:

Perpendicular slopes must be opposite reciprocals of each other:  m1 * m2 = –1

4x+ y= 3

y= -4x + 3

slope = -4

the new slope = 1/4

the equation formula y= mx+b

m = new slope

y = (x/4) + b

From the point given (4,-3)

y = -3. x = 4

-3 = 1 + b

b = -4

the equation =

y =  \frac{1}{4} x - 4

8 0
3 years ago
Consider the function f(x)=xln(x). Let Tn be the nth degree Taylor approximation of f(2) about x=1. Find: T1, T2, T3. find |R3|
Fynjy0 [20]

Answer:

R3 <= 0.083

Step-by-step explanation:

f(x)=xlnx,

The derivatives are as follows:

f'(x)=1+lnx,

f"(x)=1/x,

f"'(x)=-1/x²

f^(4)(x)=2/x³

Simialrly;

f(1) = 0,

f'(1) = 1,

f"(1) = 1,

f"'(1) = -1,

f^(4)(1) = 2

As such;

T1 = f(1) + f'(1)(x-1)

T1 = 0+1(x-1)

T1 = x - 1

T2 = f(1)+f'(1)(x-1)+f"(1)/2(x-1)^2

T2 = 0+1(x-1)+1(x-1)^2

T2 = x-1+(x²-2x+1)/2

T2 = x²/2 - 1/2

T3 = f(1)+f'(1)(x-1)+f"(1)/2(x-1)^2+f"'(1)/6(x-1)^3

T3 = 0+1(x-1)+1/2(x-1)^2-1/6(x-1)^3

T3 = 1/6 (-x^3 + 6 x^2 - 3 x - 2)

Thus, T1(2) = 2 - 1

T1(2) = 1

T2 (2) = 2²/2 - 1/2

T2 (2) = 3/2

T2 (2) = 1.5

T3(2) = 1/6 (-2^3 + 6 *2^2 - 3 *2 - 2)

T3(2) = 4/3

T3(2) = 1.333

Since;

f(2) = 2 × ln(2)

f(2) = 2×0.693147 =

f(2) = 1.386294

Since;

f(2) >T3; it is significant to posit that T3 is an underestimate of f(2).

Then; we have, R3 <= | f^(4)(c)/(4!)(x-1)^4 |,

Since;

f^(4)(x)=2/x^3, we have, |f^(4)(c)| <= 2

Finally;

R3 <= |2/(4!)(2-1)^4|

R3 <= | 2 / 24× 1 |

R3 <= 1/12

R3 <= 0.083

5 0
3 years ago
Simplify the expression?! :) please!
Arlecino [84]
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