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3 years ago

Is 4÷2/7 the same as 4 groups of 2/7 yes or no explain your answer

Mathematics

1 answer:

UkoKoshka [18]3 years ago

8 0

Answer:

Step-by-step explanation:

4/(2/7)=(4/1)(7/2)=28/2=14

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How many liters of gasoline to fill 13.2 gallon tank?

densk [106]

1. Find how many liters of gasoline to fill 13.2 gallon tank.
based on the metric conversion, there are 3.79 liters of liquid in every 1 liquid gallon.
Now, since we have 13.2 gallon tank, we need to multiply it with the given unit of conversion
=> 1 gallon = 3.79 liters
=> 13.2 gallon x 3.79 liters
=> 13.2 x 3.79
=> 50.028, therefore it needs around 50.028 liters of gasoline to be able to fill the 13.2 gallon tank

5 0

3 years ago

-4+2(12) ———— -3(-3)-(-1)

dsp73

Answer:

20/10 is 2

final answer is 2

8 0

3 years ago

If y = 3x+b intersects y= mx +3 at (2, 3), find the value of b and m.

Darina [25.2K]

Answer:

The value of b is -3 and m is 0.

Step-by-step explanation:

You have to substitute the coordinates into both equation in order to find m and b value :

Find m,

$y = mx + 3$

At (2,3),

$3 = m(2) + 3$

$3 - 3 = 2m$

$2m = 0$

$m = 0$

Find b,

$y = 3x + b$

At (2,3),

$3 = 3(2) + b$

$b + 6 = 3$

$b = 3 - 6$

$b = - 3$

3 0

2 years ago

Read 2 more answers

Write an equation for a line that is parallel to 2x+5y=15 and passes through the point (-10, 3)

Lyrx [107]

Answer:

The equation of the line is,

$y = - \frac{2}{5} x - 1$

Step-by-step explanation:

First, you have to write it in a form of y = mx + b :

$2x + 5y = 15$

$5y = 15 - 2x$

$y = 3 - \frac{2}{5} x$

$y = - \frac{2}{3} x + 5$

When both lines are parallel to each other, they will have to same gradient value. So the equation of the line is y = (-2/5)x + b. Next, you have to find the value of b by substutituting (-10,3) into the equation :

$y = - \frac{ 2}{5}x + b$

$let \: x = - 10,y = 3$

$3 = - \frac{2}{5} ( - 10) + b$

$3 = 4 + b$

$3 - 4 = b$

$b = - 1$

3 0

3 years ago

Read 2 more answers

Use this list of Basic Taylor Series and the identity sin2θ= 1 2 (1−cos(2θ)) to find the Taylor Series for f(x) = sin2(3x) based

notsponge [240]

Answer:

The Taylor series for $sin^2(3 x) = - \sum_{n=1}^{\infty} \frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}$ , the first three non-zero terms are $9x^{2} -27x^{4}+\frac{162}{5}x^{6}$ and the interval of convergence is $( -\infty, \infty )$

Step-by-step explanation:

These are the steps to find the Taylor series for the function $sin^2(3 x)$

Use the trigonometric identity:

$sin^{2}(x)=\frac{1}{2}*(1-cos(2x))\\ sin^{2}(3x)=\frac{1}{2}*(1-cos(2(3x)))\\ sin^{2}(3x)=\frac{1}{2}*(1-cos(6x))$

2. The Taylor series of $cos(x)$

$cos(y) = \sum_{n=0}^{\infty}\frac{-1^{n}y^{2n}}{(2n)!}$

Substituting y=6x we have:

$cos(6x) = \sum_{n=0}^{\infty}\frac{-1^{n}6^{2n}x^{2n}}{(2n)!}$

3. Find the Taylor series for $sin^2(3x)$

$sin^{2}(3x)=\frac{1}{2}*(1-cos(6x))$ (1)

$cos(6x) = \sum_{n=0}^{\infty}\frac{-1^{n}6^{2n}x^{2n}}{(2n)!}$ (2)

Substituting (2) in (1) we have:

$\frac{1}{2} (1-\sum_{n=0}^{\infty}\frac{-1^{n}6^{2n}x^{2n}}{(2n)!})\\ \frac{1}{2}-\frac{1}{2} \sum_{n=0}^{\infty}\frac{-1^{n}6^{2n}x^{2n}}{(2n)!}$

Bring the factor $\frac{1}{2}$ inside the sum

$\frac{6^{2n}}{2}=9^{n}2^{2n-1} \\ (-1^{n})(9^{n})=(-9^{n} )$

$\frac{1}{2}-\sum_{n=0}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}$

Extract the term for n=0 from the sum:

$\frac{1}{2}-\sum_{n=0}^{0}\frac{-9^{0}2^{2*0-1}x^{2*0}}{(2*0)!}-\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}\\ \frac{1}{2} -\frac{1}{2} -\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}\\ 0-\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}\\ sin^{2}(3x)=-\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}$

To find the first three non-zero terms you need to replace n=3 into the sum

$sin^{2}(3x)=\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}\\ \sum_{n=1}^{3}\frac{-9^{3}2^{2*3-1}x^{2*3}}{(2*3)!} = 9x^{2} -27x^{4}+\frac{162}{5}x^{6}$

To find the interval on which the series converges you need to use the Ratio Test that says

For the power series centered at x=a

$P(x)=C_{0}+C_{1}(x-a)+C_{2}(x-a)^{2}+...+ C_{n}(x-a)^{n}+...,$

suppose that $\lim_{n \to \infty} |\frac{C_{n}}{C_{n+1}}| = R.$ . Then

If $R=\infty,$ the the series converges for all x
If $0$ then the series converges for all $|x-a|$
If R=0, the the series converges only for x=a

So we need to evaluate this limit:

$\lim_{n \to \infty} |\frac{\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}}{\frac{-9^{n+1}2^{2*(n+1)-1}x^{2*(n+1)}}{(2*(2n+1))!}} |$

Simplifying we have:

$\lim_{n \to \infty} |-\frac{(n+1)(2n+1)}{18x^{2} } |$

Next we need to evaluate the limit

$\lim_{n \to \infty} |-\frac{(n+1)(2n+1)}{18x^{2} } |\\ \frac{1}{18x^{2} } \lim_{n \to \infty} |-(n+1)(2n+1)}|}$

-(n+1)(2n+1) is negative when n -> ∞. Therefore $|-(n+1)(2n+1)}|=2n^{2}+3n+1$

You can use this infinity property $\lim_{x \to \infty} (ax^{n}+...+bx+c) = \infty$ when a>0 and n is even. So

$\lim_{n \to \infty} |-\frac{(n+1)(2n+1)}{18x^{2} } | \\ \frac{1}{18x^{2}} \lim_{n \to \infty} 2n^{2}+3n+1=\infty$

Because this limit is ∞ the radius of converge is ∞ and the interval of converge is $( -\infty, \infty )$ .

6 0

3 years ago