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SVEN [57.7K]
3 years ago
14

Ali and Omar cycle around a cycle track.

Mathematics
1 answer:
kari74 [83]3 years ago
3 0

Answer and step-by-step explanation:

First, we will have to know at what time they will cross the line at the same time. So we can do the equation:

50 * 80 = 400

That means in 400 seconds they will be level again. Now, let's find Ali's amount of laps for 400 seconds, if it takes Ali 50 seconds for one lap. We can do:

400/50 = 8

This means it will take Ali 8 seconds to be leveled with Omar. Now Omar has the equation:

400/80 = 5

Because it takes him 80 seconds to take one lap. Therefore, it will take Ali 8 laps, and Omar 5 laps to be at the same starting line. Hope this helps!

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Answer:

  c.  quadrilateral

Step-by-step explanation:

All of the sides are different lengths, so the quadrilateral cannot be a parallelogram, rhombus, or square.

Its best descriptor is <em>parallelogram</em>.

_____

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5 0
2 years ago
Effie ate 4 peaches. this represents 25% of the bag of peaches. how many peaches were originally in the bag
pav-90 [236]
16 peaches were originally in the bag.
16/25%=4
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3 years ago
Staples pays George Nagovsky an annual salary of $36,000. Today, George's boss informs him that he will received a $4,600 raise.
konstantin123 [22]
George will receive 13% of his old income

3 0
3 years ago
Let $f(x) = x^2$ and $g(x) = \sqrt{x}$. Find the area bounded by $f(x)$ and $g(x).$
Anna [14]

Answer:

\large\boxed{1\dfrac{1}{3}\ u^2}

Step-by-step explanation:

Let's sketch graphs of functions f(x) and g(x) on one coordinate system (attachment).

Let's calculate the common points:

x^2=\sqrt{x}\qquad\text{square of both sides}\\\\(x^2)^2=\left(\sqrt{x}\right)^2\\\\x^4=x\qquad\text{subtract}\ x\ \text{from both sides}\\\\x^4-x=0\qquad\text{distribute}\\\\x(x^3-1)=0\iff x=0\ \vee\ x^3-1=0\\\\x^3-1=0\qquad\text{add 1 to both sides}\\\\x^3=1\to x=\sqrt[3]1\to x=1

The area to be calculated is the area in the interval [0, 1] bounded by the graph g(x) and the axis x minus the area bounded by the graph f(x) and the axis x.

We have integrals:

\int\limits_{0}^1(\sqrt{x})dx-\int\limits_{0}^1(x^2)dx=(*)\\\\\int(\sqrt{x})dx=\int\left(x^\frac{1}{2}\right)dx=\dfrac{2}{3}x^\frac{3}{2}=\dfrac{2x\sqrt{x}}{3}\\\\\int(x^2)dx=\dfrac{1}{3}x^3\\\\(*)=\left(\dfrac{2x\sqrt{x}}{2}\right]^1_0-\left(\dfrac{1}{3}x^3\right]^1_0=\dfrac{2(1)\sqrt{1}}{2}-\dfrac{2(0)\sqrt{0}}{2}-\left(\dfrac{1}{3}(1)^3-\dfrac{1}{3}(0)^3\right)\\\\=\dfrac{2(1)(1)}{2}-\dfrac{2(0)(0)}{2}-\dfrac{1}{3}(1)}+\dfrac{1}{3}(0)=2-0-\dfrac{1}{3}+0=1\dfrac{1}{3}

6 0
3 years ago
eduardo took a survey to find out how many students in his school have a pet, there are 1,000 students in the school. 20% of the
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Answer:

200

Step-by-step explanation: Sorry If wrong!

6 0
2 years ago
Read 2 more answers
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