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3 years ago

12

On a drawing a triangle has sides that equal 4 cm, 5 cm, and 6 cm. If the scale of the drawing is 1 cm = 10 feet, find the sides

of the original large triangle.

2 answers:

kodGreya [7K]3 years ago

4 0

4 * 10 = 40
5 * 10 = 50
6 * 10 = 60

The side length of the original large triangle are 40 feet, 50 feet, and 60 feet.

Oksi-84 [34.3K]3 years ago

3 0

Answer:

A= 40 B=50 c=60

Step-by-step explanation:

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Annie traveled 5 times the sum of the number of hours Brian traveled and 2. together they traveled 20 hours. find the number of

kari74 [83]

Well There isn't enough evidence to tell how many hours each of them worked. Yu stated how long they each traveled separately but did not state the amount of hours worked, or any proof for it.

But Assuming that by worked you mean traveled.
Annie traveled 5 times the sum of the number of hours Brian traveled and 2. together they traveled 20 hours. find the number of hours each person worked.So 20 is 6 times the number of hours Annie & Brian Worked + 10
(From the 10 extra hours Annie worked from +2)
10/6=1 hour and 40 min
So Brain worked/traveled for 1 hour and 40 minuets.
Then we find Annie's time. Add 2 hours to 1 hour and 40 and you get
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So Annie worked/traveled for 18 hours and 20 min
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3 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

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

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2 years ago

What is the least number by which 384 should be multiplied to make it a perfect square?

mel-nik [20]

Answer:

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1 year ago

marin [14]

I hope this helps you

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2 years ago

Graph a line that contains the point (-7,-4)and has a slope of - 2/3

sergeinik [125]

Hi there! :)

Answer:

Given the information, we can write an equation in slope-intercept form

(y = mx + b) to graph the line:

Plug in the slope for 'm', the y-coordinate of the point given for 'y', and the

x-coordinate given for 'x':

-4 = -2/3(-7) + b

-4 = 14/3 + b

Solve for b:

-12/3 = 14/3 + b

-12/3 - 14/3 = b

-26/3 = b

Therefore, the equation of the line is y = -2/3x - 26/3 (Graphed below)

Some points on the line include:

(-7, -4)

(-4, -6)

(0, -26/3)

(2, -10)

(5, -12)

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