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3 years ago

How do you fine the answer ? Help me pls

Mathematics

2 answers:

statuscvo [17]3 years ago

5 0

Answer:

120.

Step-by-step explanation:

Let the common difference be d. Then these equations are true,

so add all the left sides and all the right sides:

y = 30-d

x = 30-2d

z = 30+d

f = 30+2d

-------------------------

y+x+z+f = 120

Notice that there are lots of different possibilities for x,y,z,f.

For example the arithmetic sequence could be

10 20 30 40 50, which have sum 120 with d=10, or

20 25 30 35 40, which also have sum 120 with d=5, or

54 42 30 18 6, which also have sum 120 with d=-12, or any of

an infinite number of possible arithmetic sequences. But the

sum will always be 120.

Artemon [7]3 years ago

4 0

Answer:

130

Step-by-step explanation:

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Thanks been trying for ages

Drupady [299]

Answer:

x₁ = 3,29296875

x₂ = 3,276659786

x₃ = 3,279420685

Step-by-step explanation:

This is a recurrent relationship between xₙ₊₁ and xₙ

To get x₁, write the formula with n=0:

$x_1 = 3+ \frac{3}{x_0^2}$

Then fill in what you know, x₀=3.2. This gives you x₁.

Repeat for n=1,... and so forth.

Excel can do this effectively, see picture.

3 0

2 years ago

Write the linearization of the function at the points indicated. (Enter your answer as an equation. Let x be the independent var

Natalija [7]

Answer:

$\displaystyle y=5+\frac{x}{10}$

$\displaystyle y=10+\frac{(x-75)}{20}$

Step-by-step explanation:

Linearization

It consists of finding an approximately linear function that behaves as close as possible to the original function near a specific point.

Let y=f(x) a real function and (a,f(a)) the point near which we want to find a linear approximation of f. If f'(x) exists in x=a, then the equation for the linearization of f is

$y=f(x)=f(a)+f'(a)(x-a)$

Let's find the linearization for the function

$y=\sqrt{25+x}$

at (0,5) and (75,10)

Computing f'(x)

$\displaystyle f'(x)=\frac{1}{2\sqrt{25+x}}$

At x=0:

$\displaystyle f'(0)=\frac{1}{2\sqrt{25+0}}=\frac{1}{10}$

We find f(0)

$f(0)=\sqrt{25+0}=5$

Thus the linearization is

$\displaystyle y=f(0)+f'(0)(x-0)=5+\frac{1}{10}x$

$\displaystyle y=5+\frac{x}{10}$

Now at x=75:

$\displaystyle f'(75)=\frac{1}{2\sqrt{25+75}}=\frac{1}{20}$

We find f(75)

$f(75)=\sqrt{25+75}=10$

Thus the linearization is

$\displaystyle y=f(75)+f'(75)(x-75)=10+\frac{1}{20}(x-75)$

$\displaystyle y=10+\frac{(x-75)}{20}$

5 0

3 years ago

Which is true about the degree of the sum and difference of the polynomials 3x^5^y – 2x^3y^4 – 7xy3 and –8x^5y + 2x^3y^4 + xy^3?

Fofino [41]

The degree of a polynomial is the highest exponent in a term
thus it'd be 7 cuz
2x^3y^4
4 + 3 = 7

3 0

3 years ago

Can some one help with domain and range

shusha [124]

Yeah whats the question

6 0

3 years ago

Which of the following numbers is divisible by 11?

Olin [163]

The only number that is divisible by 11 without having a decimal is D. 7337

5 0

3 years ago

Read 2 more answers