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3 years ago

Every possible cross section of a three-dimensional figure is a circle. What is the figure?

Mathematics

2 answers:

Veseljchak [2.6K]3 years ago

5 0

Answer:

its cylinder

Step-by-step explanation:

dsp733 years ago

3 0

A cross section is the shape we get when cutting straight through an object. In geometry it is the shape made when a solid is cut through by a plane. cylinder cross section.The cross section of this circular cylinder is a circle.

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Find the value of x, 110°, 4x+2, and if you can help with steps that would be great :,)

Helen [10]

Answer:

x = 17

Step-by-step explanation:

The two angles with measures form a linear pair and are supplementary.

The measures of supplementary angles has a sum of 180°.

If you add the two given measures, the sum must equal 180°.

4x + 2 + 110 = 180

4x + 112 = 180

4x = 68

x = 68/4

x = 17

7 0

3 years ago

What is the solution to the equation squareroot 4t+5=3

zubka84 [21]

Answer:

Step-by-step explanation:

√(4t+5) = 3

4t + 5 = 9

4t = 4

t = 1

8 0

2 years ago

Read 2 more answers

Find the number c that satisfies the conclusion of the Mean Value Theorem on the given interval. (Enter your answers as a comma-

ivanzaharov [21]

Answer:

$c = \displaystyle\frac{9}{4}$

Step-by-step explanation:

The following information is missing in the given question:

$f(x) = \sqrt{x}$

Using this we may solve the question as:

We are given the following in the question:

$f(x) = \sqrt{x}, x \in [0,9]$

We have to find the number c such that f(x) satisfies the Mean value theorem.

Mean Value theorem:

It states that if the function is differentiable in the closed interval [a,b], differentiable in the interval (a,b), then there exist c in (a,b) such that:

$f'(c) = \displaystyle\frac{f(b)-f(a)}{b-a}$

Now,

$f(x) = \sqrt{x}\\f'(x) = \frac{1}{2\sqrt{x}}$

Continuity in [0,9]

Since a polynomial function is continuous everywhere, f(x) is continuous in [0,9]

Differentiability in (0,9)

Since a polynomial function is differentiable everywhere the given function is differentiable in interval (0,9)

Then, by mean value theorem:

$f'(c) = \displaystyle\frac{f(b)-f(a)}{b-a}\\\\\frac{1}{2\sqrt{c}} = \frac{f(9) - f(0)}{9-0} = \frac{3}{9}\\\\\frac{1}{2\sqrt{c}} = \frac{1}{3}\\\\c= \frac{9}{4}$