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Fudgin [204]
2 years ago
8

Use the definition of continuity and the properties of limits to show that the function

Mathematics
1 answer:
Zepler [3.9K]2 years ago
7 0

Answer:

Applied the definition and the limit.

They had the same result, so the function is continuous.

Step-by-step explanation:

At function f(x) is continuous at x = a if:

\lim_{x \to a} f(x) = f(a)

In this question:

f(x) = x^{2} + 5(x-2)^{7}

At x = 3.

\lim_{x \to 3} x^{2} + 5(x-2)^{7} = 3^{2} + 5(3-2)^{7} = 14

f(3) = 3^{2} + 5(3-2)^{7} = 14

Since \lim_{x \to 3} f(x) = f(3), f(x) is continuous at x = 3.

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Help on this please.
Komok [63]

Answer:

80 or 81

Step-by-step explanation:

6 0
2 years ago
Talia has a square bedroom. Her family is moving and her bedroom in her new apartment will be 3 feet shorter in one direction an
djverab [1.8K]

Answer (x represents the first direction and the y represents the second direction):

  1. (x - 3) × (y + 5)
  2. New bedroom: 7 ft × 15 ft. Her new bedroom is 5 sq. ft larger than her old bedroom

I hope this helps!

3 0
3 years ago
What is the equation of this graphed line?
Veseljchak [2.6K]

Answer:

Slope Intercept form of the equation is y  = \frac{1}{3} x - 9

Step-by-step explanation:

Here,  the two point line are given as  is A(-6,-3) and B(6,-7)

The slope of the line AB =  m = \frac{y_2 - y_1}{x_2 - x_1}

m = \frac{-7-(-3)}{6-(-6)}   = \frac{-7 + 3}{6 + 6}   = \frac{4}{12} \\\implies m = \frac{1}{3}

⇒ the slope of AB  is m =  (4/3)

By SLOPE  INTERCEPT FORMULA:

The equation of a line with slope m and a point (x0, y0) is given as

(y-y0)=  m (x-x0)

⇒ The equation of line with point (6,-7) is:  

y + 7 = \frac{1}{3} (x-6)  \implies  3y + 21  - x + 6 - 0\\or, -x + 3y + 27 = 0

Now, the given equation is -x + 3y  = -27

Convert it in the  SLOPE INTERCEPT FORM  y = mx + c

We get, 3y = x - 27

or, y = \frac{1}{3} x - \frac{27}{3} \\\implies y = \frac{1}{3} x - 9

Hence, the Slope Intercept form of the equation is y  = \frac{1}{3} x - 9

3 0
2 years ago
Use lagrange multipliers to find the maximum volume of a rectangular box that is inscribed in a sphere of radius r
vesna_86 [32]

For the derivative tests method, assume that the sphere is centered at the origin, and consider the circular projection of the sphere onto the xy-plane. An inscribed rectangular box is uniquely determined

1

by the xy-coordinate of its corner in the first octant, so we can compute the z coordinate of this corner by

x2+y2+z2=r2 =⇒z= r2−(x2+y2). Then the volume of a box with this coordinate for the corner is given by

V = (2x)(2y)(2z) = 8xy r2 − (x2 + y2),

and we need only maximize this on the domain x2 + y2 ≤ r2. Notice that the volume is zero on the boundary of this domain, so we need only consider critical points contained inside the domain in order to carry this optimization out.

For the method of Lagrange multipliers, we optimize V(x,y,z) = 8xyz subject to the constraint x2 + y2 + z2 = r2<span>. </span>

7 0
3 years ago
Which expression is equivalent to 3d + 1?
weqwewe [10]

Answer:

3d + 1

I'm pretty sure it's the same thing.

4 0
2 years ago
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