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KonstantinChe [14]
3 years ago
13

What is the first step to solve this equation: 12-4x=54

Mathematics
2 answers:
ValentinkaMS [17]3 years ago
4 0

Answer:

B. subtract 12 from both sides

Step-by-step explanation:

12-4x=54

<u>-12 -12</u>

<u>-</u><u>4</u><u>x</u><u>=</u><u>4</u><u>2</u>

arsen [322]3 years ago
4 0

Answer:

B. Subtract 12 from both sides

Step-by-step explanation:

You must be confused on why I say this. But hold tight, I have an explanation.

Because 12 is on the front side of the minus sign, it is actually positive. That means you must subtract it from both sides.

But what about 4x? Well, it is actually negative! Because the 12 is reduced to zero, the equation is technically 0 - 4x = 42. However, we just remove the 0 to make -4x = 42.

By the way, the fully solved equation is: x = -10.5 or x = -10 1/2

Hope this helps!

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natulia [17]
Improper Fraction
12/7 = 1 5/7
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3 years ago
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here is the base of a prism if the height of the prism is 5cm what is it’s surface area? what is it’s volume
denpristay [2]
<h2>Volume Of Prism</h2>

Here is the base of a prism if the height of the prism is 5cm what is it’s surface area? what is it’s volume.

<h3>Solution:</h3>
  • Volume = s²
  • V = 5cm
  • V = 5cm x 5cm
  • V = 25cm²

- So The Surface area of Prism is 25cm².

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1 year ago
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One number is larger than another by 9. If the greater number is increased by 10, and the lesser number is tripled, the sum of t
kkurt [141]

Answer:

The larger number is -6, the smaller number is -15

Step-by-step explanation:

We have two numbers, a and b.

We know that one number is larger than another by 9.

Then we can write:

a = b + 9

then a is larger than b by 9 units.

If the greater number is increased by 10 (a + 10) and the lesser number is tripled (3*b), the sum of the two would be -41:

(a + 10) + 3*b = -41

So we got two equations:

a = b + 9

(a + 10) + 3*b = -41

This is a system of equations.

One way to solve this is first isolate one variable in one of the two equations:

But we can see that the variable "a" is already isolated in the first equation, so we have:

a = b + 9

now we can replace that in the other equation:

(a + 10) + 3*b = -41

(b + 9) + 10 + 3*b = -41

now we can solve this for b.

9 + b + 10 + 3b = -41

(9 + 10) + (3b + b) = -41

19 + 4b = -41

4b = -41 -19 = -60

b = -60/4 = -15

b = -15

then:

a = b + 9

a = -15 + 9 = -6

a = -6

7 0
2 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}

3 0
2 years ago
Please help Let f(x)=2x^2+x-3 and g(x)=x-1. Perform the indicated operation, then find the domain. (F-g)(x)
Vesna [10]
F(x) = 2x² + x - 3
g(x) = x - 1
(f - g)(x) = (2x² + x - 3) - (x - 1)
(f - g)(x) = 2x² + (x - x) + (-3 + 1)
(f - g)(x) = 2x² - 2

The answer is D.
4 0
3 years ago
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