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3 years ago

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In AJKL.m._J = 40and m K = 55°. In AMNP, m M = 40° and m P = 85 A student concludes that the triangles are not similar. Do you a

gree or disagree? Why? (1 point)
O Disagree, by the Interior Angles Theorem, m N = 55°, so the triangles are similar by the AA similarity criterion

O Disagree, by the Interior Angles Theorem, mZN = 45°, so the triangles are similar by the AA similarity criterion

O Agree by the interior Angles Theorem, m_N = 45 Togo the triangles are not similar by the AA

O Agree, since m_K #m_P, the triangles are not similar by the AA similarity criterion

1 answer:

MaRussiya [10]3 years ago

6 0

Answer: A

Step-by-step explanation:

Source: trust me bro

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Been a while since I’ve needed to not use a calculator. Why is -2X10^-18 (3/16) = 4.09X10^-19?

Anestetic [448]

Answer:

see the explanation

Step-by-step explanation:

we have

$2.18*10^{-18}(\frac{3}{16})$

we have that

$\frac{3}{16}=0.1875$

Convert to scientific notation

$0.1875=1.875*10^{-1}$

substitute

$2.18*10^{-18}(1.875*10^{-1}})$

Remember that

To multiply two numbers in scientific notation, multiply their coefficients and add their exponents

$2.18*10^{-18}1.875*10^{-1}}=(2.18*1.875)*10^{(-18-1)}=4.0875*10^{-19}$

Round the number 4.0875 two decimal places

so

$4.0875*10^{-19}=4.09*10^{-19}$

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3 years ago

Graph x^4-3x^2+x. Identify the x-intercepts and the points where the local maximums and local minimums occur. Determine the inte

Brums [2.3K]

Functions, equations and tables can be represented using graphs

The x-intercepts are: 0, 0.35 and 1.53
The local maximum is at point (0.17, 0.08), and the local minimum is at point (1.13, -1.07)

The expression is given as:

$x^4-3x^2+x$

Rewrite the expression as a function

$f(x) = x^4-3x^2+x$

See attachment for the graph of the function.

<h3>The x-intercepts</h3>

This is the point where the graph cross the x-axis.

From the attached graph, the x-intercepts are: 0, 0.35 and 1.53

<h3>The local maximum</h3>

From the attached graph, the local maximum is at point (0.17, 0.08)

<h3>The local minimum</h3>

From the attached graph, the local minimum is at point (1.13, -1.07)

<h3>Intervals</h3>

The graph increases at interval (-1.30, 0) and (1,13, infinity)
The graph decreases at interval (- infinity, 1.30) and (0.35, 1.13)

Read more about graphs and functions at:

brainly.com/question/13136492

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2 years ago

Decide whether or not it is possible for a triangle to have three angle measurements or three side lengths given. If it is possi

shutvik [7]

Answer:

First one is not possible

Second one is possible, and would be an isocèles triangle

Third one is possible, would be scalene

Last one is not possible

Comment any further questions :)

Step-by-step explanation:

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2 years ago

Prove that in general (x + a)<br> 2 ≠ x2 + a2

qwelly [4]

Answer:

yes they are not equal because (x + a) = 2x + 2a

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3 years ago

Find the distance between a point (– 2, 3 – 4) and its image on the plane x+y+z=3 measured across a line (x + 2)/3 = (2y + 3)/4

dimaraw [331]

Answer:

Distance of the point from its image = 8.56 units

Step-by-step explanation:

Given,

Co-ordinates of point is (-2, 3,-4)

Let's say

$x_1\ =\ -2$

$y_1\ =\ 3$

$z_1\ =\ -4$

Distance is measure across the line

$\dfrac{x+2}{3}\ =\ \dfrac{2y+3}{4}\ =\ \dfrac{3z+4}{5}$

So, we can write

$\dfrac{x-x_1+2}{3}\ =\ \dfrac{2(y-y_1)+3}{4}\ =\ \dfrac{3(z-z_1)+4}{5}\ =\ k$

$=>\ \dfrac{x-(-2)+2}{3}\ =\ \dfrac{2(y-3)+3}{4}\ =\ \dfrac{3(z-(-4))+4}{5}\ =\ k$

$=>\ \dfrac{x+4}{3}\ =\ \dfrac{2y-3}{4}\ =\ \dfrac{3z+16}{5}\ =\ k$

$=>\ x\ =\ 3k-4,\ y\ =\ \dfrac{4k+3}{2},\ z\ =\ \dfrac{5k-16}{3}$

Since, the equation of plane is given by

x+y+z=3

The point which intersect the point will satisfy the equation of plane.

So, we can write

$3k-4+\dfrac{4k+3}{2}+\dfrac{5k-16}{3}\ =\ 3$

$=>6(3k-4)+3(4k+3)+2(5k-16)\ =\ 18$

$=>18k-24+12k+9+10k-32\ =\ 18$

$=>\ k\ =\dfrac{13}{8}$

So,

$x\ =\ 3k-4$

$=\ 3\times \dfrac{13}{8}-4$

$=\ \dfrac{7}{4}$

$y\ =\ \dfrac{4k+3}{2}$

$=\ \dfrac{4\times \dfrac{13}{8}+3}{2}$

$=\ \dfrac{19}{4}$

$z\ =\ \dfrac{5k-16}{3}$

$=\ \dfrac{5\times \dfrac{13}{8}-16}{3}$

$=\ \dfrac{-21}{8}$

Now, the distance of point from the plane is given by,

$d\ =\ \sqrt{(x-x_1)^2+(y-y_1)^2+(z-z_1)^2}$

$=\ \sqrt{(-2-\dfrac{7}{4})^2+(3-\dfrac{19}{4})^2+(-4+\dfrac{21}{8})^2}$

$=\ \sqrt{(\dfrac{-15}{4})^2+(\dfrac{-7}{4})^2+(\dfrac{9}{8})^2}$

$=\ \sqrt{\dfrac{225}{16}+\dfrac{49}{16}+\dfrac{81}{64}}$

$=\ \sqrt{\dfrac{1177}{64}}$

$=\ 4.28$

So, the distance of the point from its image can be given by,

D = 2d = 2 x 4.28

= 8.56 unit

So, the distance of a point from it's image is 8.56 units.

4 0

3 years ago