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lara31 [8.8K]
3 years ago
13

Pls help ASAP I rlly need help thank you

Mathematics
1 answer:
yKpoI14uk [10]3 years ago
7 0

Answer:

point B is (0,1)

Step-by-step explanation: A and the midpoint are 4 units away right so count out B and you get (0,1)

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What is the solution to the equation 2x + 4 = 24? <br> x = 1 <br> x = 3 <br> x = 6 <br> x = 10
horsena [70]

Answer:

    x = 14

Step-by-step explanation:

1.1     Pull out like factors :

  2x - 20  =   2 • (x - 10)  

Equation at the end of step  1  :

Step  2  :

Equations which are never true :

2.1      Solve :    2   =  0

This equation has no solution.

A a non-zero constant never equals zero.

Solving a Single Variable Equation :

2.2      Solve  :    x-10 = 0  

Add  10  to both sides of the equation :  

                     x = 10  

One solution was found :

                  x = 10

4 0
3 years ago
Read 2 more answers
Help help please i will mark u brill
Roman55 [17]

Answer:

Step-by-step explanation:

u/4 + 9.9 = 16.9

Putting value of u = 24 in the equation

24/4 + 9.9 = 16.9

6 + 9.9 = 16.9

15.9 = 16.9

Bringing like terms on one

16.9 - 15.9 = 1

7 0
3 years ago
Read 2 more answers
What is angle A equal? <br><br> Thanks :)
Alex Ar [27]
Angle A and angle B sums up to 180 (interior angles)

6x - 35 + 3x + 53 = 180
9x + 18 = 180
9x = 162
x = 18

angle A = 6(18)-35 = 73 degrees
7 0
2 years ago
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Use the definition of a Taylor series to find the first three non zero terms of the Taylor series for the given function centere
Ket [755]

Answer:

e^{4x}=e^4+4e^4(x-1)+8e^4(x-1)^2+...

\displaystyle e^{4x}=\sum^{\infty}_{n=0} \dfrac{4^ne^4}{n!}(x-1)^n

Step-by-step explanation:

<u>Taylor series</u> expansions of f(x) at the point x = a

\text{f}(x)=\text{f}(a)+\text{f}\:'(a)(x-a)+\dfrac{\text{f}\:''(a)}{2!}(x-a)^2+\dfrac{\text{f}\:'''(a)}{3!}(x-a)^3+...+\dfrac{\text{f}\:^{(r)}(a)}{r!}(x-a)^r+...

This expansion is valid only if \text{f}\:^{(n)}(a) exists and is finite for all n \in \mathbb{N}, and for values of x for which the infinite series converges.

\textsf{Let }\text{f}(x)=e^{4x} \textsf{ and }a=1

\text{f}(x)=\text{f}(1)+\text{f}\:'(1)(x-1)+\dfrac{\text{f}\:''(1)}{2!}(x-1)^2+...

\boxed{\begin{minipage}{5.5 cm}\underline{Differentiating $e^{f(x)}$}\\\\If  $y=e^{f(x)}$, then $\dfrac{\text{d}y}{\text{d}x}=f\:'(x)e^{f(x)}$\\\end{minipage}}

\text{f}(x)=e^{4x} \implies \text{f}(1)=e^4

\text{f}\:'(x)=4e^{4x} \implies \text{f}\:'(1)=4e^4

\text{f}\:''(x)=16e^{4x} \implies \text{f}\:''(1)=16e^4

Substituting the values in the series expansion gives:

e^{4x}=e^4+4e^4(x-1)+\dfrac{16e^4}{2}(x-1)^2+...

Factoring out e⁴:

e^{4x}=e^4\left[1+4(x-1)+8}(x-1)^2+...\right]

<u>Taylor Series summation notation</u>:

\displaystyle \text{f}(x)=\sum^{\infty}_{n=0} \dfrac{\text{f}\:^{(n)}(a)}{n!}(x-a)^n

Therefore:

\displaystyle e^{4x}=\sum^{\infty}_{n=0} \dfrac{4^ne^4}{n!}(x-1)^n

7 0
2 years ago
(A) {(-6,2), (3, 4), (4,4), (5, 6)}<br> DOMAIN:<br> RANGE:<br> Is the relation above a function?
vovikov84 [41]

Answer:

Yes

Step-by-step explanation:

Domain: {-6, 3, 4, 5}

Range: {2, 4, 6}

This relation is a function.

6 0
2 years ago
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