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Paladinen [302]
3 years ago
9

You are told a professional cyclist eats around 5120 calories everyday for the Tour de Abroad. Estimate how many calories a cycl

ist would eat during a 21 day race
Mathematics
1 answer:
nikitadnepr [17]3 years ago
6 0

Answer:

Total calories eat = 107,520 calories

Step-by-step explanation:

Given:

Number of calories everyday = 5120 calories

Number of days = 21

Find:

Total calories eat

Computation:

Total calories eat = Number of calories everyday × Number of days

Total calories eat = 5120 × 21

Total calories eat = 107,520 calories

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Nathan recently bought a paint sprayer for his painting business. Over the last seven days, he has used 211 gallons of paint to
wolverine [178]

Answer: 16.1 ounces per minute

Step-by-step explanation:

in 7 days, he painted by 4 hours each day, so we have a total of 28 hours.

In a hour we have 60 minutes, so we have a total of:

28*60 = 1680 minutes.

He used 211 gallons, and knowing that 1 gallon is 128 ounces, then 211 gallons is:

211*128 ounces = 27,008 ounces.

Then the ounces dispensed each minute are:

R = 27,008/1680 ounces per minute = 16.1 ounces per minute

3 0
3 years ago
LetT: P2 → P4be the linear transformationT(p) = 4x2p.Find the matrix A for T relative to the basesB = {1, x, x2}andB' = {1, x, x
ExtremeBDS [4]

Answer:  The required matrix A is

A=\left[\begin{array}{ccc}0&0&0\\0&0&0\\4&0&0\\0&4&0\\0&0&4\end{array}\right]_{5\times3} .

Step-by-step explanation:  We are given the following linear transformation :

T:P^2\Rightarrow P^4,~~~~~~~~~~~~~T(p)=4x^2p.

We are to find the matrix A relative to the bases B=\{1,x,x^2\} and B^\prime=\{1,x,x^2,x^3,x^4\}.

We have

T(1)=4x^2\times1=4x^2=0\times1+0\times x+4\times x^2+0\times x^3+0\times x^4,\\\\T(x)=4x^2\times x=4x^3=0\times1+0\times x+0\times x^2+4\times x^3+0\times x^4,\\\\T(x)=4x^2\times x^2=4x^4=0\times1+0\times x+0\times x^2+0\times x^3+4\times x^4.

Therefore, the matrix A is of order 5 × 3, given by

A=\left[\begin{array}{ccc}0&0&0\\0&0&0\\4&0&0\\0&4&0\\0&0&4\end{array}\right]_{5\times3} .

Thus, the required matrix A is

A=\left[\begin{array}{ccc}0&0&0\\0&0&0\\4&0&0\\0&4&0\\0&0&4\end{array}\right]_{5\times3} .

4 0
3 years ago
The graph of g is a vertical translation of the graph of the parent linear function. A right triangle with an area of 8 square u
NNADVOKAT [17]
The graph of g is a vertical translation of the graph of the parent linear function. A right triangle with an area of 8 square units is formed by the x-axis, the y-axis, and the graph of g. Write an equation for g.
4 0
3 years ago
Prove that for all integers a and b. If<br>a^2(b^2 - 2b) is odd, then a and b are odd​
nikklg [1K]

Answer:

  • we have:

a²(b² - 2b) = a²b² - 2a²b

  • a²(b² - 2b) is odd but -2a²b is the even number

=> a²b² must be the odd (because an odd number minus an even number equals an odd number)

=> a and b are odd (proven)

6 0
3 years ago
A point on the terminal side of an angle in standard position is given. Find the exact value of each of the six trigonometric fu
Nataly [62]
Refer to Figure 1 for the diagram of the point and reference triangle needed to answer the problem.

Plot the point (2,-3) which is in quadrant 4

Draw a right triangle such that one leg is resting on the x axis and the other leg is perpendicular to the x axis. The hypotenuse goes from (0,0) to (2,-3). 

The horizontal leg is 2 units long. So a = 2
The vertical leg is 3 units long. So b = 3
The hypotenuse is unknown but we can use the pythagorean theorem to find it. We'll call it c for now

Use Pythagorean theorem to find c
a^2 + b^2 = c^2
2^2 + 3^2 = c^2
4 + 9 = c^2
13 = c^2
c^2 = 13
c = sqrt(13)
The hypotenuse is sqrt(13) which is shown in the attached image (Figure 1)

Using figure 1, now refer to figure 2 to compute the ratios for sine, cosine and tangent
Figure 3 shows how to compute the ratios for cosecant, sectant, and cotangent

Note: With figure 1, the leg furthest from angle theta is the opposite side (in this case -3). The leg closest to the angle theta is 2 units long.

8 0
3 years ago
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