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3 years ago

Graph f(x) = -1/2x - 5

Mathematics

2 answers:

vladimir1956 [14]3 years ago

7 0

Write this expression f(x) in geogebra and see for yourself.

I can tell you that this graph is linear (straight) and decreases with 1/2 of x values. It intersex the y-axis in -5

GOOD LUCK ! :)

xeze [42]3 years ago

5 0

Graph -5 on the number line then use your slope (-1/2) and from -5 go one down and 2 right

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What is 36% of 745? Show work

kogti [31]

745/x=100/36
(745/x)*x=(100/36)*x
745=2.7777777777778*x
(2.7777777777778) to get x
745/2.7777777777778=x 
268.2=x 
x=268.2

answer 36% of 745=268.2

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2 years ago

PLS HELP!!!! Pointtssss!! The temperature during a very cold day is

grin007 [14]

The type of polynomial that would best model the data is a cubic polynomial. (Correct choice: D)

<h3>What kind of polynomial does fit best to a set of points?</h3>

In this question we must find a kind of polynomial whose form offers the best approximation to the point set, that is, the least polynomial whose mean square error is reasonable.

In a graphing tool we notice that the least polynomial must be a cubic polynomial, as there is no enough symmetry between (10, 9.37) and (14, 8.79), and the points (6, 3.88), (8, 6.48) and (10, 9.37) exhibits a pseudo-linear behavior.

The type of polynomial that would best model the data is a cubic polynomial. (Correct choice: D)

To learn more on cubic polynomials: brainly.com/question/21691794

#SPJ1

6 0

1 year ago

Please help! Will give brainiest (if i can) if you answer.

german

Answer:

Step-by-step explanation:

5 0

2 years ago

Convert to standard form: -5/2 (x-5) = y+8

nydimaria [60]

-5/2 (x - 5) = y + 8

-5/2 x + 25/2 = y + 8

-5/2x - y = 8 - 25/2

-5/2 x - y = -9/2

Multiply through by 2:-

-5x - 2y = -9 (answer).

7 0

3 years ago

A.Find a formula for

snow_lady [41]

Answer:

a) $\frac{n}{n+1}$

b) Proof in explanation.

Step-by-step explanation:

$\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\cdots+\frac{1}{n(n+1)}$ .

So let's look at the last term for a minute:

$\frac{1}{n(n+1)}$

Let's use partial fractions to see if we can find a way to write this so it is more useful to us.

$\frac{1}{n(n+1)}=\frac{A}{n}+\frac{B}{n+1}$

Multiply both sides by $n(n+1)$ :

$1=A(n+1)+Bn$

Distribute:

$1=An+A+Bn$

Reorder:

$1=An+Bn+A$

Factor:

$1=n(A+B)+A$

This implies $A=1$ and $A+B=0$ which further implies that $B=-1$ .

This means we are saying that:

$\frac{1}{n(n+1)}$ can be written as $\frac{1}{n}+\frac{-1}{n+1}$

We can check by combing the fractions:

$\frac{n+1}{n(n+1)}+\frac{-n}{n(n+1)}$

$\frac{n+1-n}{n(n+1)}$

$\frac{1}{n(n+1)}$

So it does check out.

So let's rewrite our whole expression given to us using this:

$(\frac{1}{1}+\frac{-1}{2})+(\frac{1}{2}+\frac{-1}{3})+(\frac{1}{3}+\frac{-1}{4})+\cdots +(\frac{1}{n}+\frac{-1}{n+1})$

We should see that all the terms in between the first and last are being zeroed out.

That is, this sum is equal to:

$\frac{1}{1}+\frac{-1}{n+1}$

Multiply first fraction by (n+1)/(n+1) so we can combine the fractions:

$\frac{n+1}{n+1}+\frac{-1}{n+1}$

Combine fractions:

$\frac{n}{n+1}$

Proof:

Let's see what happens when n=1.

Original expression gives us $\frac{1}{1 \cdot 2}=\frac{1}{2}$ .

The expression we came up with gives us $\frac{1}{1+1}=\frac{1}{2}$ .

So it is true for the base case.

Let's assume our expression and the expression given is true for some integer k greater than 1.

We want to now show it is true for integer k+1.

So under our assumption we have:

$\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\cdots \frac{1}{k(k+1)}=\frac{k}{k+1}$

So let's add the (k+1)th term of the given series on both sides:

$\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\cdots \frac{1}{k(k+1)}+\frac{1}{(k+1)(k+2)}=\frac{k}{k+1}+\frac{1}{(k+1)(k+2)}$

(Now we are just playing with right hand side to see if we can put it in the form our solution which be if we can $\frac{k+1}{k+2}$ .)

I'm going to find a common denominator which will be (k+1)(k+2):

$\frac{k}{k+1} \cdot \frac{k+2}{k+2}+\frac{1}{(k+1)(k+2)}$

Combine the fractions:

$\frac{k(k+2)+1}{(k+1)(k+2)}$

Distribute:

$\frac{k^2+2k+1}{(k+1)(k+2)}$

Factor the numerator:

$\frac{(k+1)^2}{(k+1)(k+2)}$

Cancel a common factor of $(k+1)$

$\frac{k+1}{k+2}$

We have proven the given expression and our formula for the sum are equal for all natural numbers,n.

6 0

2 years ago