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3 years ago

The graph shows the results of a race between Damon and Alex. Both boys begin the race at the same time.

Mathematics

2 answers:

Zarrin [17]3 years ago

4 0

Answer:

1st and 3rd statements are true.

Sep by step explanation:

We have been given a graph that shows the results of a race between Damon and Alex. Both boys begin the race at the same time.

Let us see which of the given option are correct.

1. Let us find Alex's speed using slope of the graph.

$\text{Slope}=\frac{80-0}{0-12}$

$\text{Slope}=\frac{80}{-12}$

$\text{Slope}=-6\frac{8}{12}$

$\text{Slope}=-6\frac{2}{3}$

Slope is showing remaining distance per second. The distance is decreasing $6\frac{2}{3}$ yards per second. Therefore, Alex is running at a rate of $6\frac{2}{3}$ yards per second and 1st statement is the true.

2. We can see from our graph that before starting the race Damon's distance from finish line is 100 yards and Alex's distance from finish line is 80 yards, therefore, Alex got a 20 yard head start.

Therefore, 2nd statement is false.

3. Let us find speed of Damon using slope.

$\text{Slope}=\frac{100-0}{0-10}$

$\text{Slope}=\frac{100}{-10}$

$\text{Slope}=-10$

We can see that slope is showing remaining distance per second. The distance is decreasing 10 yards per second therefore, Damon is running at a rate of 10 yards per second and 3rd statement is true.

4. We have already seen that Alex is given a 20 yard head start but he lost the race as he completed the race 2 seconds after Damon, therefore, 4th statement is false.

5. We can see from our graph that Alex completed race in 12 seconds, while Damon completed race in 10 seconds that is 2 minutes before Alex, therefore, 5th statement is false.

6. We have already seen that Damon runs at 10 yards per second and Alex runs at rate of $6\frac{2}{3}$ yards per second, therefore, last option is incorrect.

NeX [460]3 years ago

3 0

The first one, the 3rd one and the 5th one are true.

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leva [86]

Answer:

1. y = 32

2. x = 6

3. -33 = n

4. a = -15.6

Step-by-step explanation:

y-23=55

+23 +23 (23 cancels)

y = 32

x - 14 = -8

+14 +14 (-14 cancels)

x = 6

-15 = n + 18

-18 -18 (18 cancels)

-33 = n

18.3 + a = 2.7

-18.3 -18.3 (18.3 cancels)

a = -15.6

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3 years ago

Emily wants to know the volume of a beach ball. She deduces that the beach ball has diameter of 12 inches, and then uses the fol

Over [174]

Answer:

D. She used the beach's ball's diameter when she should have used the radius

Step-by-step explanation:

To find the volume of the beach ball, using the volume of a sphere is the right formula to use, which is ⁴/3πr³.

The formula she used is correct.

Since the diameter of the ball is assumed to be 12 inches, what is needed to find the volume is the radius.

Radius (r) = ½(diameter) = ½(12) = 6 in.

This is where Emily made a mistake.

She used the diameter of the beach ball instead of its radius (r) which is needed in the equation.

She should have gotten,

V = ⁴/3(3.14)(6)² = 904.32 cubic inches

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2 years ago

9.<br> 8 cm<br> 5.1 cm<br> 5.1 cm<br> 15 cm<br> 5 cm<br> 10 cm<br> Find the area and perimeter

Ganezh [65]

Answer:

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Step-by-step explanation:

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3 years ago

What is a equivalent fraction for 3/9?

kherson [118]

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You can use any, these are just the smallest to the biggest but any are fine but I would use the first one. Hope I helped. Equivalent is what 3/9 equals to.

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3 years ago

Let X be a set of size 20 and A CX be of size 10. (a) How many sets B are there that satisfy A Ç B Ç X? (b) How many sets B are

Svetlanka [38]

Answer:

(a) Number of sets B given that

A⊆B⊆C: 2¹⁰. (That is: A is a subset of B, B is a subset of C. B might be equal to C)
A⊂B⊂C: 2¹⁰ - 2. (That is: A is a proper subset of B, B is a proper subset of C. B≠C)

(b) Number of sets B given that set A and set B are disjoint, and that set B is a subset of set X: 2²⁰ - 2¹⁰.

Step-by-step explanation:

Let $x_1, x_2, \cdots, x_{20}$ denote the 20 elements of set X.

Let $x_1, x_2, \cdots, x_{10}$ denote elements of set X that are also part of set A.

For set A to be a subset of set B, each element in set A must also be present in set B. In other words, set B should also contain $x_1, x_2, \cdots, x_{10}$ .

For set B to be a subset of set C, all elements of set B also need to be in set C. In other words, all the elements of set B should come from $x_1, x_2, \cdots, x_{20}$ .

$\begin{array}{c|cccccccc}\text{Members of X} & x_1 & x_2 & \cdots & x_{10} & x_{11} & \cdots & x_{20}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set A?} & \text{Yes}&\text{Yes}&\cdots &\text{Yes}& \text{No} & \cdots & \text{No}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set B?}& \text{Yes}&\text{Yes}&\cdots &\text{Yes}& \text{Maybe} & \cdots & \text{Maybe}\end{array}$ .

For each element that might be in set B, there are two possibilities: either the element is in set B or it is not in set B. There are ten such elements. There are thus $2^{10} = 1024$ possibilities for set B.

In case the question connected set A and B, and set B and C using the symbol ⊂ (proper subset of) instead of ⊆, A ≠ B and B ≠ C. Two possibilities will need to be eliminated: B contains all ten "maybe" elements or B contains none of the ten "maybe" elements. That leaves $2^{10} -2 = 1024 - 2 = 1022$ possibilities.

Set A and set B are disjoint if none of the elements in set A are also in set B, and none of the elements in set B are in set A.

Start by considering the case when set A and set B are indeed disjoint.

$\begin{array}{c|cccccccc}\text{Members of X} & x_1 & x_2 & \cdots & x_{10} & x_{11} & \cdots & x_{20}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set A?} & \text{Yes}&\text{Yes}&\cdots &\text{Yes}& \text{No} & \cdots & \text{No}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set B?}& \text{No}&\text{No}&\cdots &\text{No}& \text{Maybe} & \cdots & \text{Maybe}\end{array}$ .

Set B might be an empty set. Once again, for each element that might be in set B, there are two possibilities: either the element is in set B or it is not in set B. There are ten such elements. There are thus $2^{10} = 1024$ possibilities for a set B that is disjoint with set A.

There are 20 elements in X so that's $2^{20} = 1048576$ possibilities for B ⊆ X if there's no restriction on B. However, since B cannot be disjoint with set A, there's only $2^{20} - 2^{10}$ possibilities left.

5 0

3 years ago