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3 years ago

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Lines l and m are parallel. Parallel lines l and m are intersected by lines s and t. At the intersection of lines l, s, and t, c

lockwise from the top left, the angles are blank, 50 degrees, (x + 25) degrees, (2 x) degrees, 1, blank. Lines t, s, and m create a triangle with angles 1, 2, 3. Using the diagram, determine which statements are true. Select all that apply. m∠1 = 50° m∠3 = (2x + x + 25)° m∠2 = (x + 25)° m∠1 + m∠2 + m∠3 = 180° 50 + 2x+ x + 25 = 180

2 answers:

larisa [96]3 years ago

4 0

Answer:

A, c, d, e

Step-by-step explanation:

on edg

Dovator [93]3 years ago

3 0

Answer:

All of them are true except for the second one.

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In triangle ABC, the measure of ∠B is 90°, BC=16, and AC=20. Triangle DEF is similar to triangle ABC, where vertices D, E, and F

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The third side of triangle ABC is AB. Using the Pythagorean Theorem, its length is 12.
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Read 2 more answers

Arrange the cones in order from lease volume to greatest volume

bearhunter [10]

Answer:

Volume of the cone in ascending order.

$V_{2}=270\pi\ units^{3}$

cone with DIAMETER of 18 & height of 10

cone with RADIUS of 10 & height of 9

cone with RADIUS of 11 & height of 9

cone with DIAMETER of 20 & height of 12

Step-by-step explanation:

Let $V_{2}. V_{3}. and\ V_{4}.$ be the volume of the cone.

Let d, r and h be the diameter, radius and height of the cone.

Given:

$d_{1} = 20\ and\ h_{1}=12$

$d_{2} = 18\ and\ h_{2}=10$

$r_{3} = 10\ and\ h_{3}=9$

$r_{4} = 11\ and\ h_{14}=9$

Arrange the cones in order from lease volume to greatest volume.

Solution:

The volume of the cone is given below.

$V=\pi r^{2} \frac{h}{3}$ ----------------(1)

where: r is radius of the base of cone.

and h is height of the cone.

The volume of the cone for $d_{1} = 20\ and\ h_{1}=12$

$r_{1} = \frac{d_{1}}{2}$

$r_{1} = \frac{20}{2}=10\ units$

$V_{1}=\pi (r_{1})^{2} \frac{h_{1}}{3}$

$V_{1}=\pi (10)^{2} \frac{12}{3}$

$V_{1}=\pi\times 100\times 4$

$V_{1}=400\pi\ units^{3}$

Similarly, for volume of the cone for $d_{2} = 18\ and\ h_{2}=10$

$r_{2} = \frac{d_{2}}{2}$

$r_{2} = \frac{18}{2}=9\ units$

$V_{2}=\pi (r_{2})^{2} \frac{h_{2}}{3}$

$V_{2}=\pi (9)^{2} \frac{10}{3}$

$V_{2}=\pi\times 81\times \frac{10}{3}$

$V_{2}=\pi\times 27\times 10$

$V_{2}=270\pi\ units^{3}$

Similarly, for volume of the cone for $r_{3} = 10\ and\ h_{3}=9$

$V_{3}=\pi (r_{3})^{2} \frac{h_{3}}{3}$

$V_{3}=\pi (10)^{2} \frac{9}{3}$

$V_{3}=\pi\times 100\times 3$

$V_{3}=\pi\times 300$

$V_{3}=300\pi\ units^{3}$

Similarly, for volume of the cone for $r_{4} = 11\ and\ h_{4}=9$

$V_{4}=\pi (r_{4})^{2} \frac{h_{4}}{3}$

$V_{4}=\pi (11)^{2} \frac{9}{3}$

$V_{4}=\pi\times 121\times 3$

$V_{4}=\pi\times 363$

$V_{4}=363\pi\ units^{3}$

So, the volume of the cone in ascending order.

$V_{2}=270\pi\ units^{3}$

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3 years ago

Use the information to answer the following questions.

Katena32 [7]

70/3

6x > 220 - 80
220 - 80 = 140

6x > 140

Divide 6 on all terms & simplify.

6x/6 > 140/6

Divide x by 1 and cancel common factor 140/6

x > 140/6

Factor 2 out of 140

x > 2(70) / 6

Cancel common factors
Factor 2 out of 6

x > 2 • 70 / 2 • 3

Cross out both of the 2’s and rewrite expression

x > 70/3 or (70/3 , ♾ )

5 0

3 years ago