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Tanzania [10]
3 years ago
10

How do you write 53% as a fraction in lowest terms

Mathematics
2 answers:
Vaselesa [24]3 years ago
8 0
Well you would take 53% then put it over 100 so you should get 53/100
dedylja [7]3 years ago
8 0
53/100? I guess this is the answer you might be looking for. I don't think you can simplify 53/100. It's already in the simplest form.
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Find the x-coordinates where f ' (x)=0 for f(x)=2x+sin(4x) in the interval [0, pi]
MAVERICK [17]

Answer:

The x-coordinate is \dfrac{\pi}{6}[/tex].

Step-by-step explanation:

We are given a function f(x) as:

f(x)=2x+\sin (4x)

Now on differentiating both side with respect to x we get that:

f'(x)=2+4 \cos (4x)

When f'(x)=0

this means that 2+4\cos (4x)=0\\\\4\cos (4x)=-2\\\\\cos(4x)=\dfrac{-1}{2}

Hence, cosine function takes the negative value in second and third quadrant but we have to only find the value in the interval [0,\pi].

also we know that \cos (\dfrac{2\pi}{3})=\dfrac{-1}{2}----(1) (which lie in the second quadrant)

so on comparing our equation with equation (1) we obtain:

4x=\dfrac{2\pi}{3}\\\\x=\dfrac{\pi}{6}

Hence, the x-coordinates where f'(x)=0 for f(x)=2x+\sin(4x) is \dfrac{\pi}{6}.

5 0
3 years ago
Help me ASAP PLEASE
lina2011 [118]

Answer:

  4. Functions: 1, 2, 4, 5, 7, 8, 11, 12

  5. y = -1/2x+5/2, x ≤ -1; 2x +1, x > -1.

      y = 2, x < 0; x, 0 ≤ x ≤ 3; 3 x > 3.

Step-by-step explanation:

4. A list of ordered pairs is a function of no x-values are re-used. In (3) and (6), the value x=2 is used more than once.

A graph represents a function if it passes the "vertical line test." A vertical line cannot intersect the graph in 2 or more points. (9) and (10) both fail that test.

If the relation is not listed here as being "not a function," then it is one of the answers to question 4.

__

5. The first step in writing the equation of a piecewise function is to identify the pieces. These are generally bounded by points of discontinuity--jumps in the function value, or changes in the slope of lines.

The second step is to identify the section of the function that boundary points belong to. Solid dots are part of the function definition; open circles are not.

<u>Left Graph</u>

The left piece ends at x=-1. There is a solid dot attached to the left piece, so its definition will be for the domain x ≤ -1. That line has slope -1/2, since is drops 1 unit for each 2 to the right. If extended, it would intersect the y-axis at y = 2 1/2 = 5/2. So, the piece on the left is y = -1/2x + 5/2 for x ≤ -1.

The right piece starts at x=-1, but does not include that point. It has a rise of 2 for each 1 to the right, so its slope is 2. It crosses the y-axis at y=1, so the piece on the right is y = 2x + 1 for x > -1.

You can use the method of your textbook author to combine theses pieces into one equation. The method shown above is one way to do it.

__

<u>Right Graph</u>

This graph has a discontinuity at x=0 and a change in slope at x=3. It can be described by 3 pieces. The point at x=0 does not belong to the left piece, but goes with the middle piece.

The left piece of the function is the constant 2, so has the equation y = 2 for x < 0.

The middle piece has a slope of 1 and a y-intercept of 0, so has the equation y = x for 0 ≤ x ≤ 3.

The point at x=3 belongs to both the middle piece and the right piece, so can be part of both function definitions, if you like. Generally, it is better form to include any given x-value in only one of the pieces of the function. So the equation for the right piece can be y = 3 for x > 3.

4 0
3 years ago
The perimeter of a rectangular painting is 262 centimeters. If the width of the painting is 59 centimeters, what is its length?
Liula [17]
The formula of the perimeter of a rectangle is 2l+2w so look at the photo
3 0
3 years ago
The quotient of w and 5 is at least 25
const2013 [10]
I think the answer is 5 because 25 divided by 5=5

8 0
3 years ago
Read 2 more answers
Solve the inequality 2ax-3&lt;4x+5a for all possible values of the parameter a
olga_2 [115]

Answer:

Step-by-step explanation:

2ax−3<4x+5a

Step 1: Add -5a to both sides.

2ax−3+−5a<5a+4x+−5a

2ax−5a−3<4x

Step 2: Add 3 to both sides.

2ax−5a−3+3<4x+3

2ax−5a<4x+3

5 0
3 years ago
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