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3 years ago

Which of the following ordered pairs are solutions to the system of equations below?

Mathematics

1 answer:

spayn [35]3 years ago

7 0

(3.583,-5.833) is your answer to the problem. because that is we’re they meet

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I need help with these 2 problems please, thank you

SpyIntel [72]

Answer:

1. y= (-4/11)x + 1/11

2. y= (3/5)x + 66/5

Step-by-step explanation:

8 0

3 years ago

20-5+15(25)=b+3+65(23)

vlada-n [284]

Answer:

-1108

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Mr. Harris is packaging items to give to his students. He has 48 pencils and 30 notebooks. He wants each package to contain the

guapka [62]

Answer:

The maximum number of packages that can be made with each package have same number of each item is = 6

Step-by-step explanation:

Given:

Mr. Harris has 48 pencils and 30 notebooks.

To find the number of packages he can make with each package have same number of each item.

Solution:

Number of pencils = 48

Number of notebooks = 30

In order to find the number of packages he can make with each package have same number of each item, we will find the greatest common factor of the given numbers.

To find the G.C.F., we will list down the prime factors of each.

$48=2\times 2\times 2\times 2\times 3$

$30=2\times 3\times 5$

We find that the G.C.F. = $2\times 3$ = 6

Thus, the maximum number of packages that can be made with each package have same number of each item is = 6

4 0

3 years ago

What is the value of x? Enter your answer in the box. x =

avanturin [10]

Answer:

x=20

Step-by-step explanation:

do the equation 3x+50=6x-10

7 0

3 years ago

Read 2 more answers

Use an appropriate Taylor series to find the first four nonzero terms of an infinite series that is equal to sin(5/2).

koban [17]

Answer:

2.5, -2.61, 0.81, -0.12

Step-by-step explanation:

The taylor series of the function sin(x) around zero is given by

$sin(x)=\sum_{n=0}^{\infty}\dfrac{(-1)^k}{(2k-1)!}x^{2k+1}=x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}$

Therefore,

$\sin(\frac{5}{2})=\dfrac{5}{2}-\dfrac{[\frac{5}{2}]^3}{3!}+\dfrac{[\frac{5}{2}]^5}{5!}-\dfrac{[\frac{5}{2}]^7}{7!}+...$

hence the first four nonzero terms of the series are

$\dfrac{5}{2}=2.5\\\\-\dfrac{[\frac{5}{2}]^3}{3!} \approx -2.61\\\\\dfrac{[\frac{5}{2}]^5}{5!} \approx 0.81\\\\-\dfrac{[\frac{5}{2}]^7}{7!} \approx -0.12$

6 0

3 years ago