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3 years ago

14

Select the postulate of equality or inequality that is illustrated.

2 answers:

slamgirl [31]3 years ago

8 0

Answer:

Option 3 is correct.

By symmetric postulates of equality

Step-by-step explanation:

Symmetric postulates of equality states that if one value is equal to another value, then the second value is also equal to the first value i.e,

if x =y , then y =x

Given:

If 5 =x +2

then by symmetric property of equality;

x+2 = 5

Zanzabum3 years ago

4 0

The symmetric property of equality.

This property states that: if a = b, b = a.

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Write the equation of the line that has a slope of -4 and goes through (3,-3)

Sindrei [870]

The functional equation of the line is f (x)=a*x+b
you have already a: -4 -->f (x)=-4 x+b
Now you can use the point (3;-3)
3 is the x-coordinate and -3 the f (x)-coordinate
--> -3 = -4*(3) + b Now you can solve it
-3 = -12+b (+12)
b = 9
ANSWER: f (x) = -4x + 9

3 0

3 years ago

What is the answer of:<br> z^2 - 4z^2

Novosadov [1.4K]

$z^2 - 4 z^2$

$z - 4z$

$= - 3 z^2$

hope this helps!.

4 0

2 years ago

The circumstances of Z is 84cm. What is the length of XY (the mirror arc)

Serga [27]

Answer:

B

Step-by-step explanation:

arc length = circumference × fraction of circle

= 84 × $\frac{90}{360}$ = 21 cm

8 0

3 years ago

Find the angle between the given vectors. Round your answer, in degrees, to two decimal places. u=⟨2,−6⟩u=⟨2,−6⟩, v=⟨4,−7⟩

NISA [10]

Answer:

$\theta = 108.29$

Step-by-step explanation:

Given

$u =$

$v =$

Required:

Calculate the angle between u and v

The angle $\theta$ is calculated as thus:

$cos\theta = \frac{u.v}{|u|.|v|}$

For a vector

$A =$

$A = a * b$

$cos\theta = \frac{u.v}{|u|.|v|}$ becomes

$cos\theta = \frac{.}{|u|.|v|}$

$cos\theta = \frac{2*6+4*-7}{|u|.|v|}$

$cos\theta = \frac{12-28}{|u|.|v|}$

$cos\theta = \frac{-16}{|u|.|v|}$

For a vector

$A =$

$|A| = \sqrt{a^2 + b^2}$

So;

$|u| = \sqrt{2^2 + 6^2}$

$|u| = \sqrt{4 + 36}$

$|u| = \sqrt{40}$

$|v| = \sqrt{4^2+(-7)^2}$

$|v| = \sqrt{16+49}$

$|v| = \sqrt{65}$

So:

$cos\theta = \frac{-16}{|u|.|v|}$

$cos\theta = \frac{-16}{\sqrt{40}*\sqrt{65}}$

$cos\theta = \frac{-16}{\sqrt{2600}}$

$cos\theta = \frac{-16}{\sqrt{100*26}}$

$cos\theta = \frac{-16}{10\sqrt{26}}$

$cos\theta = \frac{-8}{5\sqrt{26}}$

Take arccos of both sides

$\theta = cos^{-1}(\frac{-8}{5\sqrt{26}})$

$\theta = cos^{-1}(\frac{-8}{5 * 5.0990})$

$\theta = cos^{-1}(\frac{-8}{25.495})$

$\theta = cos^{-1}(-0.31378701706)$

$\theta = 108.288386087$

<em></em> $\theta = 108.29$ <em> (approximated)</em>

4 0

2 years ago

Please help!! asap!!!

Lubov Fominskaja [6]

Solution:

Rewrite 3 ≥ y - 2 for easier calculation:

y - 2 ≤ 3

Move the term:

y ≤ 3 + 2

y ≤ 5

When graphing the solution, filled dot (black arrow) is used for ≤ and ≥ while empty dot (white arrow) is used for < >

6 0

2 years ago