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salantis [7]
3 years ago
8

A plane left on time at noon on Monday. It arrived at its destination at 8:16PM the same day, which was 16 minutes late. On aver

age, how many seconds did the plane fall behind every fifteen minutes?
Mathematics
1 answer:
wel3 years ago
3 0
The plane arrived on 8:16 pm and was 16 minutes late. This means the arrival time of plan was 8:00 pm. The plane left on noon i.e. 12:00 pm.

So total expected time of the journey was = 8 hours

8 hours means 8 x 60 minutes = 480 minutes

480 minutes can also be seen as 32 groups of 15 minutes each as 15 x 32 = 480.

In 32 groups of 15 minutes each the plane was late by a total of 16 minutes.

So, in every 15 minute the plane was late by 16/32 minute = 0.5 minute

Thus, we can conclude that on average the plane fall behind 0.5 minutes in every 15 minutes
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Setler [38]

Answer:

-4

Step-by-step explanation:

4+ (-2)^3

-2 *-2*-2 = -8

4 + -8

-4

4 0
3 years ago
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Find all x-intercepts and y-intercepts of the graph of the function.<br> F(x)=x³-4x² - 4x + 16
Lady bird [3.3K]

Answer: (0;16)  (4;0)  (2;0)  (-2;0).

Step-by-step explanation:

Y-intercepts of the graph of the function F(x)=x³-4x² - 4x + 16:

x=0\\F(x)=0^3-4*0^2-4*0+16\\F(x)=16.\\Hence,\  (0;16).

X-intercepts of the graph of the function F(x)=x³-4x² - 4x + 16:

F(x)=0\\0=x^3-4x^2-4x+16\\0=(x^3-4x^2)-(4x+16)\\0=x^2*(x-4)-4*(x-4)\\0=(x-4)*(x^2-4)\\0=(x-4)*(x^2-2^2)\\0=(x-4)*(x-2)*(x+2)\\x-4=0\\x_1=4.\\Hence,\ (4;0)\\x-2=0\\x_2=2.\\Hence,\ (2;0)\\x+2=0\\x_3=-2.\\Hence,\ (-2;0).

8 0
2 years ago
Convert the following statement using an "if-then" structure.
GaryK [48]

Answer:

  D. If John owns a dog, then he owns a cat

Step-by-step explanation:

The implication p → q (if p, then q) has the same truth table as the logical expression ~p∨q. You have the expression ...

  ~(John owns a dog) ∨ (he owns a cat)

Matching parts of this expression to the components of the expression ~p∨q, we see we can choose ...

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and put those into the structure of the implication: if p, then q.

  If John owns a dog, then he owns a cat. . . . . matches choice D

6 0
3 years ago
Solve and check equation 20 = 2r - 3r + 10
Andreas93 [3]

2r-3r+10=20

Move all terms to the left:

2r-3r+10-(20)=0

add all the numbers together, and all the variables

-1r-10=0

move all terms containing r to the left, all other terms to the right

-r=10

r=10/-1

r=-10

7 0
3 years ago
What is the coefficient of the x^5y^5- term in the biomial expansion of (2x-3y)^10
jasenka [17]

<u>Answer:</u>

 The coefficient of x^{5} \times y^{5} \text { is }=\left(252 \times 2^{5} \times(-3)^{5}\right)=252 \times 32 \times 243=1959552

<u>Solution: </u>

The given expression is (2 x-3 y)^{10}

As per binomial theorem, we know,

(x+y)^{n}=\sum n C_{k} x^{n-k} y^{k}

Now here a = 2x, b = (- 3y) and n = 10 and k = 0,1,2,….10

Now x^{5}\times y^5 will be the 6 term where k =5

Now, \mathrm{T}_{6}=10 \mathrm{C}_{5} \times(2 \mathrm{x})^{(10-5)} \times(-3 \mathrm{y})^{5}=10 \mathrm{C}_{5} 2^{5} \times \mathrm{x}^{5} \times(-3)^{5} \times \mathrm{y}^{5}

So, the coefficient of x^{5} \times y^{5} \text { is }=10\left(5 \times 2^{5} \times(-3)^{5}\right.

10 \mathrm{C}_{5}=\frac{10 !}{5 ! \times(10-5) !}=\frac{10 !}{5 !+5 !}=\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 !}{5 ! \times 5 !}=\frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1}=\left(\frac{30240}{120}\right)=252

The coefficient of x^{5} \times y^{5} \text { is }=\left(252 \times 2^{5} \times(-3)^{5}\right)=252 \times 32 \times 243=1959552

6 0
4 years ago
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