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2 years ago

2. Which of the following is an example of a proper fraction?

2 answers:

5 0

Answer:

3. 3/4

Step-by-step explanation:

Proper fractions will be fractions that can’t be determined as mixed fractions

Improper fractions can be determined as a mixed fraction

kvasek [131]2 years ago

4 0

Answer:

3/4

Step-by-step explanation:

7/6 and 4/3 are both improper fractions, you can't have the numerator be greater than the denominator (you would have to simplify an improper fraction to a mixed number for it to be acceptable). Lastly, 4/4 would just equal 1. So therefore, 3/4 is the right answer.

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Tara watched a movie at 3:19. The movie she watched was 1:20 long. When was her movie over?

solniwko [45]

About 4:39.
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4 0

3 years ago

Read 2 more answers

Use the triangle below to find cos V.

Semenov [28]

Answer:

$cos\ V = \frac{2\sqrt{29}}{29}$

Step-by-step explanation:

Given

The attached triangle

Required

Find cos V

In trigonometry:

$cos\theta = \frac{Adjacent}{Hypotenuse}$

In this case:

$cos V= \frac{TV}{UV}$

Where

$TV = 6$

and

$UV^2 = TV^2 + TU^2$ -- Pythagoras

$UV^2 = 6^2 + 15^2$

$UV^2 = 261$

Take square roots

$UV = \sqrt{261$

$UV = \sqrt{9*29$

$UV = \sqrt{9} *\sqrt{29$

$UV = 3\sqrt{29$

So:

$cos V= \frac{TV}{UV}$

$cos\ V = \frac{6}{3\sqrt{29}}$

$cos\ V = \frac{2}{\sqrt{29}}$

Rationalize:

$cos\ V = \frac{2}{\sqrt{29}}*\frac{\sqrt{29}}{\sqrt{29}}$

$cos\ V = \frac{2\sqrt{29}}{29}$

8 0

2 years ago

SOME ONE PLEASE HELP !!!!

Andrei [34K]

x = 20 - -5 = 20 + 5 = 25

y = 16 - 1 = 15

7 0

2 years ago

You are a lifeguard and spot a drowning child 60 meters along the shore and 40 meters from the shore to the child. You run along

sukhopar [10]

Answer:

The lifeguard should run across the shore a distance of 48.074 m before jumpng into the water in order to minimize the time to reach the child.

Step-by-step explanation:

This is a problem of optimization.

We have to minimize the time it takes for the lifeguard to reach the child.

The time can be calculated by dividing the distance by the speed for each section.

The distance in the shore and in the water depends on when the lifeguard gets in the water. We use the variable x to model this, as seen in the picture attached.

Then, the distance in the shore is d_b=x and the distance swimming can be calculated using the Pithagorean theorem:

$d_s^2=(60-x)^2+40^2=60^2-120x+x^2+40^2=x^2-120x+5200\\\\d_s=\sqrt{x^2-120x+5200}$

Then, the time (speed divided by distance) is:

$t=d_b/v_b+d_s/v_s\\\\t=x/4+\sqrt{x^2-120x+5200}/1.1$

To optimize this function we have to derive and equal to zero:

$\dfrac{dt}{dx}=\dfrac{1}{4}+\dfrac{1}{1.1}(\dfrac{1}{2})\dfrac{2x-120}{\sqrt{x^2-120x+5200}} \\\\\\\dfrac{dt}{dx}=\dfrac{1}{4} +\dfrac{1}{1.1} \dfrac{x-60}{\sqrt{x^2-120x+5200}} =0\\\\\\ \dfrac{x-60}{\sqrt{x^2-120x+5200}} =\dfrac{1.1}{4}=\dfrac{2}{7}\\\\\\ x-60=\dfrac{2}{7}\sqrt{x^2-120x+5200}\\\\\\(x-60)^2=\dfrac{2^2}{7^2}(x^2-120x+5200)\\\\\\(x-60)^2=\dfrac{4}{49}[(x-60)^2+40^2]\\\\\\(1-4/49)(x-60)^2=4*40^2/49=6400/49\\\\(45/49)(x-60)^2=6400/49\\\\45(x-60)^2=6400\\\\$

$x$

As $d_b=x$ , the lifeguard should run across the shore a distance of 48.074 m before jumpng into the water in order to minimize the time to reach the child.

7 0

3 years ago

Eddie made a video. The video included a 5-minute introduction followed by short segments. • Each segment was 8 minutes long • E

GalinKa [24]

Answer:

B - 18

Step-by-step explanation:

149 total minutes minus 5 for the introductions leaves 144.

144/5=18

5 0

3 years ago