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3 years ago

8

If a right circular cone is intersected by a plane only at its vertex, as in the

1 answer:

Anuta_ua [19.1K]3 years ago

4 0

Answer:

D. A point

Step-by-step explanation:

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QUESTION 33

The length of the legs of the right triangle are given as,

6 centimeters and 8 centimeters.

The length of the hypotenuse can be found using the Pythagoras Theorem.

${h}^{2} = {6}^{2} + {8}^{2}$

${h}^{2} = 36+ 64$

${h}^{2} = 100$

$h = \sqrt{100}$

$h = 10cm$

Answer: C

QUESTION 34

The triangle has a hypotenuse of length, 55 inches and a leg of 33 inches.

The length of the other leg can be found using the Pythagoras Theorem,

${l}^{2} + {33}^{2} = {55}^{2}$

${l}^{2} = {55}^{2} - {33}^{2}$

${l}^{2} = 1936$

$l = \sqrt{1936}$

$l = 44cm$

Answer:B

QUESTION 35.

We want to find the distance between,

(2,-1) and (-1,3).

Recall the distance formula,

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Substitute the values to get,

$d=\sqrt{( - 1-2)^2+(3- - 1)^2}$

$d=\sqrt{( - 3)^2+(4)^2}$

$d=\sqrt{9+16}$

$d=\sqrt{25}$

$d = 5$

Answer: 5 units.

QUESTION 36

We want to find the distance between,

(2,2) and (-3,-3).

We use the distance formula again,

$d=\sqrt{( - 3-2)^2+( - 3- 2)^2}$

$d=\sqrt{( - 5)^2+( - 5)^2}$

$d=\sqrt{25+25}$

$d=\sqrt{50}$

$d=5\sqrt{2}$

Answer: D

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3 years ago

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