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3 years ago

4 times a number is increased by 15, the answer is the same as when 70 is decreased by the number

Mathematics

2 answers:

pentagon [3]3 years ago

3 0

4n+15=70+n

I'm pretty sure this is the answer

Have a good day!

galina1969 [7]3 years ago

3 0

4x +15 = 70 -x

if you want to solve it

add both sides by x and you will have

5x+15=70
minus both sides 15 and you will have
5x= 55

divide both sides by 5 and you will get

x=11

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Which ordered pairs are solutions to the inequality 2y-x≤-6 ?

Ray Of Light [21]

Answer:

Step-by-step explanation:

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3 years ago

Determine if the following system of equations has no solutions, infinitely many solutions or exactly one solution.

34kurt

Answer:

Infinitely many solutions

Step-by-step explanation:

First, let's organize the equations of the lines into slope-intercept form.

x + 4y = 1
=> 4y = -x + 1
=> y = -x/4 + 1/4

And,

2x + 8y = 2
=> 8y = -2x + 2
=> y = -2x/8 + 2/8
=> y = -x/4 + 1/4

Since both the graphs are the same, the two equations have infinitely many solutions.

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2 years ago

Simplify using distributive property 4(z+2)+2(3-z)

riadik2000 [5.3K]

Answer:

2z+14

Step-by-step explanation:

you multiply

4*z =4z

4*2 =8

and then you multiply

2*3=6

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the answer is 2z+14

5 0

2 years ago

CAN SOMEONE JUST PLEASE ANSWER THIS ASAP FOR BRAINLIEST!!

aksik [14]

Answer:

F= 3x^2

Step-by-step explanation:

1) Multiplying exponents causes them to add, for instance if you were to multiply x^3 * x^4 the final product would be x^7. If you were to divide exponents then they would subtract.

2) Similarly in this problem you would determine the missing factor by dividing the product from -10x^3

3) Using the steps above, begin dividing.

-30x^5 / -10x^3

4) -30 divided by -10 is 3 and as mentioned in step number one, you subtract exponents whenever they are divided by each other. 5-3=2

5) Due to all of the steps mentioned above, The answer is 3x^2

6 0

3 years ago

Read 2 more answers

Consider the function f(x)equalscosine left parenthesis x squared right parenthesis. a. Differentiate the Taylor series about 0

dybincka [34]

I suppose you mean

$f(x)=\cos(x^2)$

Recall that

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

which converges everywhere. Then by substitution,

$\cos(x^2)=\displaystyle\sum_{n=0}^\infty(-1)^n\frac{(x^2)^{2n}}{(2n)!}=\sum_{n=0}^\infty(-1)^n\frac{x^{4n}}{(2n)!}$

which also converges everywhere (and we can confirm this via the ratio test, for instance).

a. Differentiating the Taylor series gives

$f'(x)=\displaystyle4\sum_{n=1}^\infty(-1)^n\frac{nx^{4n-1}}{(2n)!}$

(starting at $n=1$ because the summand is 0 when $n=0$ )

b. Naturally, the differentiated series represents

$f'(x)=-2x\sin(x^2)$

To see this, recalling the series for $\sin x$ , we know

$\sin(x^2)=\displaystyle\sum_{n=0}^\infty(-1)^{n-1}\frac{x^{4n+2}}{(2n+1)!}$

Multiplying by $-2x$ gives

$-x\sin(x^2)=\displaystyle2x\sum_{n=0}^\infty(-1)^n\frac{x^{4n}}{(2n+1)!}$

and from here,

$-2x\sin(x^2)=\displaystyle 2x\sum_{n=0}^\infty(-1)^n\frac{2nx^{4n}}{(2n)(2n+1)!}$

$-2x\sin(x^2)=\displaystyle 4x\sum_{n=0}^\infty(-1)^n\frac{nx^{4n}}{(2n)!}=f'(x)$

c. This series also converges everywhere. By the ratio test, the series converges if

$\displaystyle\lim_{n\to\infty}\left|\frac{(-1)^{n+1}\frac{(n+1)x^{4(n+1)}}{(2(n+1))!}}{(-1)^n\frac{nx^{4n}}{(2n)!}}\right|=|x|\lim_{n\to\infty}\frac{\frac{n+1}{(2n+2)!}}{\frac n{(2n)!}}=|x|\lim_{n\to\infty}\frac{n+1}{n(2n+2)(2n+1)}$

The limit is 0, so any choice of $x$ satisfies the convergence condition.

3 0

3 years ago