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nata0808 [166]
3 years ago
6

Free brainleist What is the product of five square root five times six square root four ? Simplify your answer. (5 points)

Mathematics
2 answers:
SSSSS [86.1K]3 years ago
6 0
Sixty square root five (simplified) There ya go hope it helps
snow_lady [41]3 years ago
4 0

Answer:

The correct option is B. Sixty square root five

Step-by-step explanation:

Five square root five : 5√5

Six square root four : 6√4

Now, √4 = 2

⇒ 6√4 = 6 × 2 = 12

We need to find the product of these two numbers

⇒ 5√5 × 12 = 60√5

Therefore, The correct option is B. Sixty square root five

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Lindsey plans on visiting her friend two states over. The distance to her friend is 1188 miles. If she leaves on Friday and driv
podryga [215]

Answer:

Do you mean 7 hours a day? If so:

7*60=420

1188/420=2.82ish which rounds to 3 days.

Step-by-step explanation:

4 0
3 years ago
Math question up here
Gemiola [76]
The answer is 11 because you do the smallest parenthesis. and work your way out.
7 0
3 years ago
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I need help help help quickly
harkovskaia [24]

Answer:  24

Step-by-step explanation: Height × width × length

4 0
2 years ago
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I need help doing everything.
ohaa [14]

Using set notations, we have that:

1. The range of the function is:

  • Inequality: x ≥ -15.
  • Set-builder: {x | x ≥ -15}
  • Interval: [-15, ∞).

2. The function is increasing on:

  • Inequality: x > 3.
  • Set-builder: {x | x > 3}
  • Interval: (3, ∞).

3. The function is decreasing on:

  • Inequality: x < 3
  • Set-builder: {x | x < 3}
  • Interval: (-∞, 3).

4. The function is negative on:

  • Inequality: 0.918 < x < 5.082.
  • Set-builder: {x | 0.918 < x < 5.082}.
  • Interval: (0.918, 5.082).

<h3>What is the standard interval notation?</h3>

The standard interval notation of an interval of lower bound a and upper bound b is given by:

[a,b].

The inequality notation is:

a ≤ x ≤ b

The set builder notation is:

{x | a ≤ x ≤ b}.

<h3>What is the range of the function?</h3>

The range of a function is the set that contains all possible output values for the function. In a graph, it is given by the values of y of the function.

Hence, for this problem, the range is given by these following notations:

  • Inequality: x ≥ -15.
  • Set-builder: {x | x ≥ -15}
  • Interval: [-15, ∞).

<h3>When is a quadratic function increasing and when it is decreasing?</h3>

Considering a concave up quadratic function, as given in this problem, we have that:

  • It decays for the values of x before the vertex.
  • It increases for the values of x after the vertex.

In this problem, the vertex is at x = 3, hence the intervals are given as follows:

2. The function is increasing on:

  • Inequality: x > 3.
  • Set-builder: {x | x > 3}
  • Interval: (3, ∞).

3. The function is decreasing on:

  • Inequality: x < 3
  • Set-builder: {x | x < 3}
  • Interval: (-∞, 3).

<h3>When a function is negative?</h3>

A function is negative when it's graph is below the x-axis. Hence, for this problem, we have that:

4. The function is negative on:

  • Inequality: 0.918 < x < 5.082.
  • Set-builder: {x | 0.918 < x < 5.082}.
  • Interval: (0.918, 5.082).

More can be learned about set notations at brainly.com/question/24462379

#SPJ1

6 0
1 year ago
If x = a cosθ and y = b sinθ , find second derivative
Olin [163]

I'm guessing the second derivative is for <em>y</em> with respect to <em>x</em>, i.e.

\dfrac{\mathrm d^2y}{\mathrm dx^2}

Compute the first derivative. By the chain rule,

\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dy}{\mathrm d\theta}\dfrac{\mathrm d\theta}{\mathrm dx}=\dfrac{\frac{\mathrm dy}{\mathrm d\theta}}{\frac{\mathrm dx}{\mathrm d\theta}}

We have

y=b\sin\theta\implies\dfrac{\mathrm dy}{\mathrm d\theta}=b\cos\theta

x=a\cos\theta\implies\dfrac{\mathrm dx}{\mathrm d\theta}=-a\sin\theta

and so

\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{b\cos\theta}{-a\sin\theta}=-\dfrac ba\cot\theta

Now compute the second derivative. Notice that \frac{\mathrm dy}{\mathrm dx} is a function of \theta; so denote it by f(\theta). Then

\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm df}{\mathrm dx}

By the chain rule,

\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm df}{\mathrm d\theta}\dfrac{\mathrm d\theta}{\mathrm dx}=\dfrac{\frac{\mathrm df}{\mathrm d\theta}}{\frac{\mathrm dx}{\mathrm d\theta}}

We have

f=-\dfrac ba\cot\theta\implies\dfrac{\mathrm df}{\mathrm d\theta}=\dfrac ba\csc^2\theta

and so the second derivative is

\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\frac ba\csc^2\theta}{-a\sin\theta}=-\dfrac b{a^2}\csc^3\theta

4 0
3 years ago
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