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3 years ago

Complete the statement using < >, or =. |-18| __ 19 A: < B: > C: =

Mathematics

1 answer:

eduard3 years ago

4 0

Answer:

Step-by-step explanation:

the inverse of -18 is +18. +18 is less than 19

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What is the number of faces (planes) for the following cubes?

Llana [10]

Cube has 6 faces/planes :)

4 0

3 years ago

2 tan 30°<br>II<br>1 + tan- 300

shusha [124]

Question:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$

Answer:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= sin(60^{\circ})$

Step-by-step explanation:

Given

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$

Required

Simplify

In trigonometry:

$tan(30^{\circ}) = \frac{1}{\sqrt{3}}$

So, the expression becomes:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2 * \frac{1}{\sqrt{3}}}{1 + (\frac{1}{\sqrt{3}})^2}$

Simplify the denominator

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2 * \frac{1}{\sqrt{3}}}{1 + \frac{1}{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\frac{2}{\sqrt{3}}}{1 + \frac{1}{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\frac{2}{\sqrt{3}}}{ \frac{3+1}{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\frac{2}{\sqrt{3}}}{ \frac{4}{3}}$

Express the fraction as:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2}{\sqrt 3} / \frac{4}{3}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2}{\sqrt 3} * \frac{3}{4}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{1}{\sqrt 3} * \frac{3}{2}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{3}{2\sqrt 3}$

Rationalize

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{3}{2\sqrt 3} * \frac{\sqrt{3}}{\sqrt{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{3\sqrt{3}}{2* 3}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\sqrt{3}}{2}$

In trigonometry:

$sin(60^{\circ}) = \frac{\sqrt{3}}{2}$

Hence:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= sin(60^{\circ})$

3 0

3 years ago

0-6

dybincka [34]

<h3>Given :-</h3>

Pranav ran 20.3 km more than Branda

Pravin ran 38.6 km

$\\ \\$

To find :

No. of kilometers did Brenda run.

$\\ \\$

No. of kilometers ran by Brenda be x.

$\\ \\$

<h3>Solution:</h3>

$\\ \\$

Equation formed:-

$\\ \\$

Total distance covered by Pravan = More distance covered by Pravan + distance covered by Brenda.

Therefore:-

$\\ \\$

$\leadsto \sf38.6 = 20.3 + x$

Write the equation

$\\ \\$

$\leadsto \sf38.6 - 20.3 =x$

When we transfer 20.3 to left side the positive sign (+) will change into negative sign (–)

$\\ \\$

$\leadsto \sf x = 38.6 - 20.3$

Arrange the equation because x is always represented at left side.

$\\ \\$

$\leadsto \boxed {\pmb{\sf x = 18.3}}\star$

After subtracting 38.6 with 20.3 we will get result as 18.3 .

$\\ \\$

$\therefore \red{ \underline{ \pmb{\frak{Distance ~ covered ~by ~Brenda~ is ~equal ~to~18.3~kilometers}}}}$

7 0

3 years ago

Understanding proportional relationships lesson 3 session 1 mason says you can find an equivalent ratio by adding the same numbe

Naily [24]

Answer:

I don't get this at all

4 0

3 years ago

If the measure of angle AXC =8x-7 and the measure of angle AXB=3x+10 what is the measure of angle AXC?

PSYCHO15rus [73]

Answer:

<AXC = 101 degrees.

Step-by-step explanation:

The question says that XB is a bisector which means that <AXB = 1/2 of AXC

You can't do the question if that assumption isn't true.

<AXC = 8x - 7

<AXB = 3x + 10

3x + 10 = 1/2 ( 8x - 7) Multiply both sides by 2

2(3x + 10) = 8x - 7 Remove the brackets

6x + 20 = 8x - 7 Add 7 to both sides

6x + 20 + 7 = 8x Combine

6x + 27 = 8x Subtract 6x from both sides

27 = 8x - 6x Combine

27 = 2x Switch

2x = 27 Divide by 2

x = 27/2

x = 13.5

AXC = 8x - 7

AXC = 8*13.5 - 7

AXC = 108 -7

AXC = 101

3 0

2 years ago