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4 years ago

What is an equivalent fraction for 8/9

Mathematics

2 answers:

svetoff [14.1K]4 years ago

6 0

16/18
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Gelneren [198K]4 years ago

4 0

$\frac{16}{18}$

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The larger number is 18 more than twice the smaller. If the sum of the two numbers is 93, find both numbers

Ganezh [65]

Answer:

larger number=18+(2x)

smaller number=x

(18+2x)+x=93

18 +3x=93

3x=93-18

3x=75

x=25

18+2(25)

18+50

larger number =68

smaller number=25

6 0

3 years ago

Find the radian measure of the central angle of a circle of radius r = 60 inches that intercepts an arc of length s = 70 inche

Finger [1]

Step-by-step explanation:

s = rθ

70 in = (60 in) θ

θ = 7/6 radians

θ ≈ 1.167 radians

7 0

4 years ago

The sum of a certain number and 3 times the number is 40. what is the number?

xxTIMURxx [149]

3n + n = 40
4n = 40
n = 40/4
n = 10 <==

6 0

3 years ago

Find the absolute maximum and absolute minimum values of f on the given interval.

anyanavicka [17]

The question is missing parts. Here is the complete question.

Find the absolute maximum and absolute minimum values of f on the given interval.

$f(x)=xe^{-\frac{x^{2}}{32} }$ , [ -2,8]

Answer: Absolute maximum: f(4) = 2.42;

Absolute minimum: f(-2) = -1.76;

Step-by-step explanation: Some functions have absolute extrema: maxima and/or minima.

Absolute maximum is a point where the function has its greatest possible value.

Absolute minimum is a point where the function has its least possible value.

The method for finding absolute extrema points is

1) Derivate the function;

2) Find the values of x that makes f'(x) = 0;

3) Using the interval boundary values and the x found above, determine the function value of each of those points;

4) The highest value is maximum, while the lowest value is minimum;

For the function given, absolute maximum and minimum points are:

$f(x)=xe^{-\frac{x^{2}}{32} }$

Using the product rule, first derivative will be:

$f'(x)=e^{-\frac{x^{2}}{32} }(1-\frac{x^{2}}{16} )$

$f'(x)=e^{-\frac{x^{2}}{32} }(1-\frac{x^{2}}{16} )$ = 0

$1-\frac{x^{2}}{16}=0$

$\frac{x^{2}}{16}=1$

$x^{2}=16$

x = ±4

x can't be -4 because it is not in the interval [-2,8].

$f(-2)=-2e^{-\frac{(-2)^{2}}{32} }=-1.76$

$f(4)=4e^{-\frac{4^{2}}{32} }=2.42$

$f(8)=8e^{-\frac{8^{2}}{32} }=1.08$

Analysing each f(x), we noted when x = -2, f(-2) is minimum and when x = 4, f(4) is maximum.

Therefore, absolute maximum is f(4) = 2.42 and

absolute minimum is f(-2) = -1.76

8 0

3 years ago

– 18q+10q+16q–3= – 19

sergeinik [125]

Step-by-step explanation:

– 18q+10q+16q–3= – 19
8q=-19
q=-19/8
q=-2.375

8 0

3 years ago

Read 2 more answers