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IRISSAK [1]
2 years ago
8

In the triangle XYZ, the measure of angle X is 68 , XY= 2.5

Mathematics
1 answer:
Yuliya22 [10]2 years ago
6 0

Answer: Answer:Correct option is C.)  X, V, and Y are collinear points.Step-by-step explanation:Given ΔXYZ with sides YZ and YX intersected by line VW. we have to tell Which fact is not used to prove that XYZ is similar to VYW.As we know the similarity rules, to prove the two triangles similar we need to prove angles or sides congruent or in proportion.Option 1: The reflexive property showing an angle is congruent to itself helps to prove triangle congruentOption 2: The Corresponding Angles Postulate helps to prove angles are congruent.Option 3: A transversal can help prove angle congruence.From all the given options, all the three options A,B and D are related to angles congruence, only the option 3 that is  X, V, and Y are collinear points is the option which does not give any result with the help of that we can prove the angles or sides congruent or we can say triangles congruent.Hence, the correct option is C.

Step-by-step explanation:

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charle [14.2K]
C is right I thinkkkk
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Evaluate the expressions and match them to the integers.
alexira [117]

Answer:

See below.

Step-by-step explanation:

-7(6) = -42

-1(-10) = 10

2(-5) = -10

-3(-14) = 42

-32/-4 = 8

-2(-2)(-2) = -8

4 0
3 years ago
What is the slope of the line containing the midpoint of the segment with endpoints at $(2, 4)$ and $(0, -2)$ and the midpoint o
oksian1 [2.3K]

The Midpoint 1 = (1,1), Mid point 2 = (3,3 ) and slope is 1 of the given points of the line.

<h3>What is the slope of line?</h3>

The slope of a line segment is a measure of the steepness of the line segment. It is the ratio of rise (the change in vertical height between the endpoints) over run (the change in horizontal length between the endpoints.

The slope of a line is a measure of its steepness. Mathematically, slope is calculated as "rise over run" (change in y divided by change in x).

Whenever the equation of a line is written in the form y = mx + b, it is called the slope-intercept form of the equation. The m is the slope of the line.

Midpoints of 1 line segment are -

= (2+0 / 2 , 4-2 / 2)

= (1,1)

Mid points of line 2 =

= (5+1 / 2, 1+5 / 2)

= (3,3)

slope  -

The required slope = rise / run

= (y1-y2)/(x1-x2)

= (1-3)/(1-3)

= 1

Therefore, the Midpoint 1 = (1,1), Mid point 2 = (3,3 ) and slope is 1 of the given points of the line.

To learn more about slope and mind-points from the given link

brainly.com/question/5792883

#SPJ4

6 0
1 year ago
Consider the equation below. (If an answer does not exist, enter DNE.) f(x) = x3 − 6x2 − 15x + 4 (a) Find the interval on which
kozerog [31]

Answer:

a) The function, f(x) is increasing at the intervals (x < -1.45) and (x > 3.45)

Written in interval form

(-∞, -1.45) and (3.45, ∞)

- The function, f(x) is decreasing at the interval (-1.45 < x < 3.45)

(-1.45, 3.45)

b) Local minimum value of f(x) = -78.1, occurring at x = 3.45

Local maximum value of f(x) = 10.1, occurring at x = -1.45

c) Inflection point = (x, y) = (1, -16)

Interval where the function is concave up

= (x > 1), written in interval form, (1, ∞)

Interval where the function is concave down

= (x < 1), written in interval form, (-∞, 1)

Step-by-step explanation:

f(x) = x³ - 6x² - 15x + 4

a) Find the interval on which f is increasing.

A function is said to be increasing in any interval where f'(x) > 0

f(x) = x³ - 6x² - 15x + 4

f'(x) = 3x² - 6x - 15

the function is increasing at the points where

f'(x) = 3x² - 6x - 15 > 0

x² - 2x - 5 > 0

(x - 3.45)(x + 1.45) > 0

we then do the inequality check to see which intervals where f'(x) is greater than 0

Function | x < -1.45 | -1.45 < x < 3.45 | x > 3.45

(x - 3.45) | negative | negative | positive

(x + 1.45) | negative | positive | positive

(x - 3.45)(x + 1.45) | +ve | -ve | +ve

So, the function (x - 3.45)(x + 1.45) is positive (+ve) at the intervals (x < -1.45) and (x > 3.45).

Hence, the function, f(x) is increasing at the intervals (x < -1.45) and (x > 3.45)

Find the interval on which f is decreasing.

At the interval where f(x) is decreasing, f'(x) < 0

from above,

f'(x) = 3x² - 6x - 15

the function is decreasing at the points where

f'(x) = 3x² - 6x - 15 < 0

x² - 2x - 5 < 0

(x - 3.45)(x + 1.45) < 0

With the similar inequality check for where f'(x) is less than 0

Function | x < -1.45 | -1.45 < x < 3.45 | x > 3.45

(x - 3.45) | negative | negative | positive

(x + 1.45) | negative | positive | positive

(x - 3.45)(x + 1.45) | +ve | -ve | +ve

Hence, the function, f(x) is decreasing at the intervals (-1.45 < x < 3.45)

b) Find the local minimum and maximum values of f.

For the local maximum and minimum points,

f'(x) = 0

but f"(x) < 0 for a local maximum

And f"(x) > 0 for a local minimum

From (a) above

f'(x) = 3x² - 6x - 15

f'(x) = 3x² - 6x - 15 = 0

(x - 3.45)(x + 1.45) = 0

x = 3.45 or x = -1.45

To now investigate the points that corresponds to a minimum and a maximum point, we need f"(x)

f"(x) = 6x - 6

At x = -1.45,

f"(x) = (6×-1.45) - 6 = -14.7 < 0

Hence, x = -1.45 corresponds to a maximum point

At x = 3.45

f"(x) = (6×3.45) - 6 = 14.7 > 0

Hence, x = 3.45 corresponds to a minimum point.

So, at minimum point, x = 3.45

f(x) = x³ - 6x² - 15x + 4

f(3.45) = 3.45³ - 6(3.45²) - 15(3.45) + 4

= -78.101375 = -78.1

At maximum point, x = -1.45

f(x) = x³ - 6x² - 15x + 4

f(-1.45) = (-1.45)³ - 6(-1.45)² - 15(-1.45) + 4

= 10.086375 = 10.1

c) Find the inflection point.

The inflection point is the point where the curve changes from concave up to concave down and vice versa.

This occurs at the point f"(x) = 0

f(x) = x³ - 6x² - 15x + 4

f'(x) = 3x² - 6x - 15

f"(x) = 6x - 6

At inflection point, f"(x) = 0

f"(x) = 6x - 6 = 0

6x = 6

x = 1

At this point where x = 1, f(x) will be

f(x) = x³ - 6x² - 15x + 4

f(1) = 1³ - 6(1²) - 15(1) + 4 = -16

Hence, the inflection point is at (x, y) = (1, -16)

- Find the interval on which f is concave up.

The curve is said to be concave up when on a given interval, the graph of the function always lies above its tangent lines on that interval. In other words, if you draw a tangent line at any given point, then the graph seems to curve upwards, away from the line.

At the interval where the curve is concave up, f"(x) > 0

f"(x) = 6x - 6 > 0

6x > 6

x > 1

- Find the interval on which f is concave down.

A curve/function is said to be concave down on an interval if, on that interval, the graph of the function always lies below its tangent lines on that interval. That is the graph seems to curve downwards, away from its tangent line at any given point.

At the interval where the curve is concave down, f"(x) < 0

f"(x) = 6x - 6 < 0

6x < 6

x < 1

Hope this Helps!!!

5 0
3 years ago
Select the property of equality used to arrive at the conclusion. If x = 3, then x 2 = 3x
VARVARA [1.3K]
I think it would be the Multiplication property of equality
4 0
3 years ago
Read 2 more answers
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