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2 years ago

The speed limit near Marjorie's house is 30 miles per hour.

Mathematics

1 answer:

Bess [88]2 years ago

7 0

The answer is 48 km/hr

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Find the range. 4.7 6.3 5.4 3.2 4.9 3.1 –3.1 9.5 –9.5

WINSTONCH [101]

Answer:

The range is 19

Step-by-step explanation:

First, you have to order the numbers from least to greatest:

-9.5, -3.1, 3.1, 3.2, 4.7, 4.9, 5.4, 6.3, 9.5

Then, you have to find the difference between the largest number and and smallest number. You do this by subtracting them:

9.5 - (-9.5)= 19

So, the range is 19

8 0

2 years ago

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

1 year ago

Please help asap 29 pts

Alexeev081 [22]

Answer:

C. 56 is the correct answer, my bad.

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Identify the equivalent expression for each of the expressions below. Root(5, (m + 2) ^ 3). Root(3, (m + 2) ^ 5). Root(5, m ^ 3)

vesna_86 [32]

Answer:

1. $(\sqrt[5]{(m+2)})^{3} = (m+2)^{\frac{3}{5}}$

2. $(\sqrt[3]{(m+2)})^{5} = (m+2)^{\frac{5}{3}}$

3. $\sqrt[5]{(m)}^{3}+2 = m^{\frac{3}{5}}+2$

4. $\sqrt[3]{(m)}^{5}+2 = m^{\frac{5}{3}}+2$

Step-by-step explanation:

Recall that

$(\sqrt[n]{x})^{m} = (x^{\frac{m}{n}})$

Where $x^{m}$ is called radicand and n is called index

1. Root(5, (m + 2) ^ 3)

In this case,

n is 5

m is 3

x = (m + 2)

$(\sqrt[5]{(m+2)})^{3} = (m+2)^{\frac{3}{5}}$

2. Root(3, (m + 2) ^ 5)

In this case,

n is 3

m is 5

x = (m + 2)

$(\sqrt[3]{(m+2)})^{5} = (m+2)^{\frac{5}{3}}$

3. Root(5, m ^ 3) + 2

In this case,

n is 5

m is 3

x = m

$\sqrt[5]{(m)}^{3}+2 = m^{\frac{3}{5}}+2$

4. Root(3, m ^ 5) + 2

In this case,

n is 3

m is 5

x = m

$\sqrt[3]{(m)}^{5}+2 = m^{\frac{5}{3}}+2$

6 0

2 years ago

CAN U GUYS PELASE ANSWER THE SIMULTANEOUS EQUATION QUESTION QUESTIONS ASAP. THIS IS EXTREMELY URGENT. Solve it by using the subs

prisoha [69]

Answer:

a ) y = 1 and x = -1

d) y = 5 and x = -1/2

Step-by-step explanation:

<h2><u>Substitution method</u></h2><h2><u>Question a</u></h2>

y = x+ 2

y = 2x + 3

<em>we can make the first formula in terms of x , so we can place in into the second formula</em>

y = x + 2

x = y - 2

now put y - 2 where x is in the second equation

y = 2x + 3

y = 2(y - 2) + 3

y = 2y - 4 +3

now solve

4 - 3 = 2y -y

y = 1

since y = 1 we can find what x is by putting into the first formula

y = x +2

x = y - 2

x = (1) -2

x = -1

<h3><u>hence y = 1 and x = -1 </u></h3><h3><u /></h3><h2><u>Question d</u></h2>

y = 2x + 6

y = 4 - 2x

<em>we can make the first formula in terms of x , so we can place in into the second formula</em>

y = 2x + 6

y - 6 = 2x

x = (y-6)/ 2

now place (y-6)/2 where x is in the second formula

y = 4 -2x

y = 4 - 2 ( $\frac{y - 6}{2}$ )

now solve

the multiplication by 2 and division by 2 are cancelled out

hence making the simplified equation as:

y = 4 - y + 6

2y = 4 + 6

2y = 10

y = 5

now place this into the first equation

y = 2x + 6

y - 6 = 2x

x = (y-6)/ 2

x = (5-6)/2

x = -1/2

<h3><u>hence y = 5 and x = -1/2</u></h3>

6 0

3 years ago