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3 years ago

Solve the equation: 9x - 5 = 5x + 7

Mathematics

2 answers:

Afina-wow [57]3 years ago

6 0

Answer:

x=3 i think

Step-by-step explanation:

IRISSAK [1]3 years ago

3 0

X=3

9x-5x=4x
5+7=12
12=4x
12/4=3
x=3

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she is increasing the price of her signature honey lemonade from 0.20 cents a cup to 0.25 cents. What is the percent increase?

Ne4ueva [31]

It is a 25% increase.

3 0

3 years ago

The product of some negative number and 5 less than three times that number is 12. Find the number.

miv72 [106K]

(x)(3x - 5) = 12
3x^2 - 5x = 12
3x^2 - 5x - 12 = 0
(x - 3)(3x + 4) = 0
x = 3, -4/3
The answer must be negative so the number is -4/3.

Hope this helps!

4 0

3 years ago

<img src="https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Csf%5Clim_%7Bx%20%5Cto%200%20%7D%20%5Cfrac%7B1%20-%20%5Cprod%20%5Climits_%

xxTIMURxx [149]

To demonstrate a method for computing the limit itself, let's pick a small value of n. If n = 3, then our limit is

$\displaystyle \lim_{x \to 0 } \frac{1 - \prod \limits_{k = 2}^{3} \sqrt[k]{\cos(kx)} }{ {x}^{2} }$

Let a = 1 and b the cosine product, and write them as

$\dfrac{a - b}{x^2}$

with

$b = \sqrt{\cos(2x)} \sqrt[3]{\cos(3x)} = \sqrt[6]{\cos^3(2x)} \sqrt[6]{\cos^2(3x)} = \left(\cos^3(2x) \cos^2(3x)\right)^{\frac16}$

Now we use the identity

$a^n-b^n = (a-b)\left(a^{n-1}+a^{n-2}b+a^{n-3}b^2+\cdots a^2b^{n-3}+ab^{n-2}+b^{n-1}\right)$

to rationalize the numerator. This gives

$\displaystyle \frac{a^6-b^6}{x^2 \left(a^5+a^4b+a^3b^2+a^2b^3+ab^4+b^5\right)}$

As x approaches 0, both a and b approach 1, so the polynomial in a and b in the denominator approaches 6, and our original limit reduces to

$\displaystyle \frac16 \lim_{x\to0} \frac{1-\cos^3(2x)\cos^2(3x)}{x^2}$

For the remaining limit, use the Taylor expansion for cos(x) :

$\cos(x) = 1 - \dfrac{x^2}2 + \mathcal{O}(x^4)$

where $\mathcal{O}(x^4)$ essentially means that all the other terms in the expansion grow as quickly as or faster than x⁴; in other words, the expansion behaves asymptotically like x⁴. As x approaches 0, all these terms go to 0 as well.

Then

$\displaystyle \cos^3(2x) \cos^2(3x) = \left(1 - 2x^2\right)^3 \left(1 - \frac{9x^2}2\right)^2$

$\displaystyle \cos^3(2x) \cos^2(3x) = \left(1 - 6x^2 + 12x^4 - 8x^6\right) \left(1 - 9x^2 + \frac{81x^4}4\right)$

$\displaystyle \cos^3(2x) \cos^2(3x) = 1 - 15x^2 + \mathcal{O}(x^4)$

so in our limit, the constant terms cancel, and the asymptotic terms go to 0, and we end up with

$\displaystyle \frac16 \lim_{x\to0} \frac{15x^2}{x^2} = \frac{15}6 = \frac52$

Unfortunately, this doesn't agree with the limit we want, so n ≠ 3. But you can try applying this method for larger n, or computing a more general result.

Edit: some scratch work suggests the limit is 10 for n = 6.

6 0

2 years ago

What is this in simplest form?

katrin [286]

40/50

80 divided by 2
100 divided by 2

5 0

2 years ago

SRRY IK LAST QUESTION FOR REAL THIS TIME

Effectus [21]

Answer:

Option (A)

Step-by-step explanation:

Volume of a cylinder is given by the formula,

V = πr²h

Here r = radius of the cylinder

h = height of the cylinder

Therefore, volume of a cylinder given in the question,

V = π( $\frac{18}{2}$ )²(8)

= 648π

= 2035.8 km²

Volume of the given cylinder is 2035.8 km².

Option (A) will be the answer.

5 0

2 years ago