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3 years ago

Help 20 ptssss asap !

Mathematics

1 answer:

ELEN [110]3 years ago

4 0

Answer:

Assuming you want an equation, y = -4/3x - 1

Step-by-step explanation:

Formula is y = mx + b

m = slope= -4/3

b = y-intercept = -1

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Please help me!!!!!!

UkoKoshka [18]

Answer:

Step-by-step explanation:

because u have 5 potatoe bags and it .............

6 0

2 years ago

Can the sides of a triangle have lengths 2, 10, and 11?<br> yes<br> no

madreJ [45]

Answer:

Yes

Step-by-step explanation:

Triangle inequality theorem states that a triangle can exist if

a + b > c; a + c > b; b + c > a

a = 2; b = 10; c = 11

2 + 10 = 12; 12 > 11.

2 + 11 = 13; 13 > 10.

10 + 11 = 21; 21 > 2.

4 0

2 years ago

3-3x6+2 and how is that reached, please!

Dvinal [7]

The order of mathematical operation is MDAS.
M - multiplication
D - division
A - addition
S - subtraction

3 - 3 x 6 + 2

Multiplication : 3 x 6 = 18
Division: none
Addition : + 3 + 2 = 5 (combine positive numbers)
Subtraction : 5 - 18 = -13

3 - 18 + 2 = -13

3 0

3 years ago

Prove that

Dovator [93]

Answer:

see explanation

Step-by-step explanation:

Using the trigonometric identities

secx = $\frac{1}{cosx}$ , cosecx = $\frac{1}{sinx}$

cotx = $\frac{cosx}{sinx}$ , tanx = $\frac{sinx}{cosx}$

Consider the left side

secA cosecA - cotA

= $\frac{1}{cosA}$ × $\frac{1}{sinA}$ - $\frac{cosA}{sinA}$

= $\frac{1}{cosAsinA}$ - $\frac{cosA}{sinA}$

= $\frac{1-cos^2A}{cosAsinA}$

= $\frac{sin^2A}{cosAsinA}$ ( cancel sinA on numerator/ denominator )

= $\frac{sinA}{cosA}$

= tanA = right side ⇒ proven

5 0

2 years ago

Find the Maclaurin series for f(x) using the definition of a Maclaurin series. [Assume that f has a power series expansion. Do n

aliya0001 [1]

Answer:

$f(x)=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}$

Step-by-step explanation:

The Maclaurin series of a function f(x) is the Taylor series of the function of the series around zero which is given by

$f(x)=f(0)+f^{\prime}(0)x+f^{\prime \prime}(0)\dfrac{x^2}{2!}+ ...+f^{(n)}(0)\dfrac{x^n}{n!}+...$

We first compute the n-th derivative of $f(x)=\ln(1+2x)$ , note that

$f^{\prime}(x)= 2 \cdot (1+2x)^{-1}\\f^{\prime \prime}(x)= 2^2\cdot (-1) \cdot (1+2x)^{-2}\\f^{\prime \prime}(x)= 2^3\cdot (-1)^2\cdot 2 \cdot (1+2x)^{-3}\\...\\\\f^{n}(x)= 2^n\cdot (-1)^{(n-1)}\cdot (n-1)! \cdot (1+2x)^{-n}\\$

Now, if we compute the n-th derivative at 0 we get

$f(0)=\ln(1+2\cdot 0)=\ln(1)=0\\\\f^{\prime}(0)=2 \cdot 1 =2\\\\f^{(2)}(0)=2^{2}\cdot(-1)\\\\f^{(3)}(0)=2^{3}\cdot (-1)^2\cdot 2\\\\...\\\\f^{(n)}(0)=2^n\cdot(-1)^{(n-1)}\cdot (n-1)!$

and so the Maclaurin series for f(x)=ln(1+2x) is given by

$f(x)=0+2x-2^2\dfrac{x^2}{2!}+2^3\cdot 2! \dfrac{x^3}{3!}+...+(-1)^{(n-1)}(n-1)!\cdot 2^n\dfrac{x^n}{n!}+...\\\\= 0 + 2x -2^2 \dfrac{x^2}{2!}+2^3\dfrac{x^3}{3!}+...+(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}+...\\\\=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^n\dfrac{x^n}{n}$

3 0

3 years ago