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e-lub [12.9K]
3 years ago
7

4(5x-9)=20x-45 i have no idea how to solve this

Mathematics
1 answer:
Marat540 [252]3 years ago
4 0

Answer: no solution

Step-by-step explanation:

20x - 36 = 20x - 45

-20x.          -20x

-36 = -45 since this is not true, the answer is there is no solution

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Rationalise the denominator of: (√3 + √2)/(√3-√2) = ?<br>​
krok68 [10]

Step-by-step explanation:

<h3><u>Given</u><u>:</u><u>-</u></h3>

(√3+√2)/(√3-√2)

<h3><u>To </u><u>find</u><u>:</u><u>-</u></h3>

<u>Rationalised</u><u> form</u><u> </u><u>=</u><u> </u><u>?</u>

<h3><u>Solution</u><u>:</u><u>-</u></h3>

We have,

(√3+√2)/(√3-√2)

The denominator = √3-√2

The Rationalising factor of √3-√2 is √3+√2

On Rationalising the denominator then

=>[(√3+√2)/(√3-√2)]×[(√3+√2)/(√3+√2)]

=>[(√3+√2)(√3+√2)]×[(√3-√2)(√3+√2)]

=>(√3+√2)²/[(√3-√2)(√3+√2)]

=> (√3+√2)²/[(√3)²-(√2)²]

Since (a+b)(a-b) = a²-b²

Where , a = √3 and b = √2

=> (√3+√2)²/(3-2)

=> (√3-√2)²/1

=> (√3+√2)²

=> (√3)²+2(√3)(√2)+(√2)²

Since , (a+b)² = a²+2ab+b²

Where , a = √3 and b = √2

=> 3+2√6+2

=> 5+2√6

<h3><u>Answer:-</u></h3>

The rationalised form of (√3+√2)/(√3-√2) is 3+2√6+2.

<h3><u>Used formulae:-</u></h3>

→ (a+b)² = a²+2ab+b²

→ (a-b)² = a²-2ab+b²

→ (a+b)(a-b) = a²-b²

→ The Rationalising factor of √a-√b is √a+√b

8 0
2 years ago
Plz help ill give u brainlist
adoni [48]
The problem tells you that a circle with radius 21 inches is painted in two colors. The two colors cover equal areas, so each color is half the area of the circle. You need to find half of the area of the circle. To do that, use the formula for the area of a circle, and then divide by 2.

A = \pi r^2

A = 3.14 \times 21~ inches \times 21~inches

A = 1384.74~square ~inches

Half the circle has area 1384.74 square inches / 2 = 692.37 square inches.

Answer: Each color covers 692.37 square inches.
4 0
3 years ago
How do you divide 3.4÷12.92
Travka [436]
Well 3.4 goes into the house and 12.92 is outside of it. 3.4 is a smaller number so you'd add zeros to get the 12.92 into 3.4. As you add zeros you'd take the decimals out and move them to the right two times. But u actually move the decimal first than adding zeros. Step 1: Take out the decimals. Step 2: Add the zeros. Step 3: Hope this helped.
8 0
3 years ago
Read 2 more answers
a rectangular box of rice is shown below. what is the greatest amount or rice, in cubic inches, that the box can hold?
Gemiola [76]

Answer:

nbbbbbbb

Step-by-step explanation:

5 0
2 years ago
Read 2 more answers
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
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