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OleMash [197]
3 years ago
12

Select the correct answer.

Mathematics
1 answer:
Bumek [7]3 years ago
6 0

Answer:

Is the A

Step-by-step explanation:

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On the first day of​ June, there were about 17.71 h of daylight in a city. Five months​ later, there were about 5.30 h of daylig
Andru [333]

Answer:

Percentage increases and decreases, are found by setting the problem up proportionally.

 

You are trying to find what percentage of 17.37 is 8.30 first.

 

so What %/100 = 8.30/17.37

 

Now define a variable:

 

Let x = What%  

 

x/100 = 8.30/17.37

17.37x = 830

x = 830/17.37

x ≈ 47.78

 

8.30 is approximately 47.78% of 17.37, so the percentage decrease is 100% - 47.78% = 52.22%

The percentage decrease is approximately 52.22%

7 0
3 years ago
The gym teacher has $250 to spend on volleyball equipment. She buys 4 volleyball nets for $28 each. Volleyballs cost $7 each. Ho
pav-90 [236]

Answer:

The maximum number of volleyballs that she can buy is 19

Step-by-step explanation:

Let

x ----> the number of volleyballs

we know that

The cost of each volleyball net ($28) by the number of volleyball nets (4) plus the cost of each volleyball ($7) multiplied by the number of volleyballs (x) must be less than or equal to $250

so

The inequality that represent this situation is

28(4)+7x\leq 250

Solve for x

112+7x\leq 250

subtract 112 both sides

7x\leq 250-112

7x\leq 138

Divide by 7 both sides

x\leq 19.7

therefore

The maximum number of volleyballs that she can buy is 19

4 0
3 years ago
Consider the function f(x)=xln(x). Let Tn be the nth degree Taylor approximation of f(2) about x=1. Find: T1, T2, T3. find |R3|
Fynjy0 [20]

Answer:

R3 <= 0.083

Step-by-step explanation:

f(x)=xlnx,

The derivatives are as follows:

f'(x)=1+lnx,

f"(x)=1/x,

f"'(x)=-1/x²

f^(4)(x)=2/x³

Simialrly;

f(1) = 0,

f'(1) = 1,

f"(1) = 1,

f"'(1) = -1,

f^(4)(1) = 2

As such;

T1 = f(1) + f'(1)(x-1)

T1 = 0+1(x-1)

T1 = x - 1

T2 = f(1)+f'(1)(x-1)+f"(1)/2(x-1)^2

T2 = 0+1(x-1)+1(x-1)^2

T2 = x-1+(x²-2x+1)/2

T2 = x²/2 - 1/2

T3 = f(1)+f'(1)(x-1)+f"(1)/2(x-1)^2+f"'(1)/6(x-1)^3

T3 = 0+1(x-1)+1/2(x-1)^2-1/6(x-1)^3

T3 = 1/6 (-x^3 + 6 x^2 - 3 x - 2)

Thus, T1(2) = 2 - 1

T1(2) = 1

T2 (2) = 2²/2 - 1/2

T2 (2) = 3/2

T2 (2) = 1.5

T3(2) = 1/6 (-2^3 + 6 *2^2 - 3 *2 - 2)

T3(2) = 4/3

T3(2) = 1.333

Since;

f(2) = 2 × ln(2)

f(2) = 2×0.693147 =

f(2) = 1.386294

Since;

f(2) >T3; it is significant to posit that T3 is an underestimate of f(2).

Then; we have, R3 <= | f^(4)(c)/(4!)(x-1)^4 |,

Since;

f^(4)(x)=2/x^3, we have, |f^(4)(c)| <= 2

Finally;

R3 <= |2/(4!)(2-1)^4|

R3 <= | 2 / 24× 1 |

R3 <= 1/12

R3 <= 0.083

5 0
2 years ago
PLANTS The table shows the heights of
ser-zykov [4K]

Answer: blue,  to love me, to nice, to i love you, to be a good boy and suck.

Step-by-step explanation:

hi I am girl

6 0
2 years ago
The marine life store would like to set up fish tanks that contain equal numbers of angel fish , sword tales , and guppies. what
Shalnov [3]
How do people......I just don't understand this
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