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Assoli18 [71]
3 years ago
13

Choose the correct simplification of (7x3 − 8x − 5) + (3x3 + 7x + 1).

Mathematics
2 answers:
Travka [436]3 years ago
8 0

the correct simplification of (7x^3 − 8x − 5) + (3x^3 + 7x + 1),

Is 10x^3 − x − 4

Ganezh [65]3 years ago
4 0
<span>10x3 − x − 4 is the correct answer to your question</span>
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Solve the equation
Klio2033 [76]

Answer:

x = 25

Step-by-step explanation:

.20x + .68(120 - x) = .54(120) (Distribute .68 to 120 and -x)

.20x + 81.6 - .68x = .54(120) (multiply .54 and 120)

.20x + 81.6 - .68x = 69.6 ( add .20x and -.68x)

-.48x + 81.6 = 69.6 ( subtract 81.6 on each side)

-.48x = -12 (divide -12 by -.48)

x = 25

4 0
3 years ago
In the right triangle above, a = 99 and c= 124. Find the value of b, rounded to the nearest tenth.
Murljashka [212]

Answer:

B= -40

Step-by-step explanation:

So a right triangle has to = 180

So we write the given, which 124+99

we are trying to get 180 so lets subtract that by the equation, 180-124+99

180-(124+99)

12+99=223

180-223

=-43

But in your case it has to be rounded to the nearest tenth

So the answer is -40

8 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
Researches notice a grey whale cruising at a depth of 233 feet below sea level. It rises 73 feet and then comes up another 40 fe
musickatia [10]

Answer:

The whale is at a depth 120 feet below the sea level.

Step-by-step explanation:

A grey whale is cruising at a depth of 233 feet below sea level. The depth below the sea level has the negative position, thus the initial position of the whale is -233.

It rises 73 feet, the position of the whale becomes 73 feet higher. This means we should add 73.

Then it comes up another 40 feet, so we need to add 40.

Mathematically,

-233+73+40=-160+40=-120

The whale is at a depth 120 feet below the sea level.

3 0
3 years ago
HELLPPPPPPPP A Blue Ray Player costs $150 and each disc costs $15.50.
Sliva [168]

Answer:

150 + 15.50 x 0.075

Step-by-step explanation:

4 0
2 years ago
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