Write an inequality that represents the graph below:

This question is much too hard would anyone please help me

NeX [460]

Answer:

B and C are the same angles so if B is 60 so is C

4 0

3 years ago

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*HELP ASAP*

k0ka [10]

Using unit concepts, it is found that:

a) Grams.
b) Grams.
c) Grams squared.
d) Grams squared.

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For a data-set, the standard deviation has the same unit as the data-set, both for the sample and the population.
The variance has the unit squared, both for the sample and the population.
For example, if the data-set is in metres, the standard deviation will be in metres while the variance will be in squared metres.

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In this question, the data-set is in grams.

The standard deviation, both for the sample and the population, in items a and b, will be in grams.
The variance, both for the sample and the population, in items c and d, will be in grams squared.

A similar problem is given at brainly.com/question/14524219

5 0

3 years ago

I Need help <br><br><br><br> Find the length of ad

Bingel [31]

Answer:

AD = 104

Step-by-step explanation:

Since is the perpendicular bisector of AD, then

DB = AB, that is

9x - 2 = 7x + 10 ( subtract 7x from both sides )

2x - 2 = 10 ( add 2 to both sides )

2x = 12 ( divide both sides by 2 )

x = 6

Thus

AD = 7x + 10 + 9x - 2 = 16x + 8 ← substitute x = 6

AD = 16(6) + 8 = 96 + 8 = 104

7 0

3 years ago

Craig won first prize in Mrs. Short's class

Bogdan [553]

Answer:

1/11

Step-by-step explanation:

There are 7 gift card options so that makes it 7/77 which simplified is 1/11

6 0

3 years ago

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Verify that each equation is an identity (1 - sin^(2)((x)/(2)))/(1+sin^(2)((x)/(2)))= (1+cosx)/(3-cosX)

Allisa [31]

Answer:

Given that we have;

$sin \left (\dfrac{x}{2} \right ) = \sqrt{\dfrac{1 - cos (x)}{2} }$

By the application of the law of indices and algebraic process of adding a and subtracting a fraction from a whole number, we have;

$\therefore \dfrac{\left ( 1 - sin^2 \left (\dfrac{x}{2} \right ) \right )}{\left ( 1 + sin^2 \left (\dfrac{x}{2} \right ) \right )} =\dfrac{\left ( \dfrac{1 + cos (x)}{2} \right)}{\left (\dfrac{3 - cos (x)}{2} \right ) } =\dfrac{\left ( 1 + cos (x))}{(3 - cos (x))}$

Step-by-step explanation:

An identity is a valid or true equation for all variable values

The given equation is presented as follows;

$\dfrac{\left ( 1 - sin^2 \left (\dfrac{x}{2} \right ) \right )}{\left ( 1 + sin^2 \left (\dfrac{x}{2} \right ) \right )} =\dfrac{\left ( 1 + cos (x))}{(3 - cos (x))}$

From trigonometric identities, we have;

$sin \left (\dfrac{x}{2} \right ) = \sqrt{\dfrac{1 - cos (x)}{2} }$

$\therefore sin^2 \left (\dfrac{x}{2} \right ) = \dfrac{1 - cos (x)}{2}$

$1 - sin^2 \left (\dfrac{x}{2} \right ) = 1 - \dfrac{1 - cos (x)}{2} = \dfrac{2 - (1 - cos (x))}{2} = \dfrac{1 + cos (x))}{2}$

$1 + sin^2 \left (\dfrac{x}{2} \right ) = 1 + \dfrac{1 - cos (x)}{2} = \dfrac{2 + 1 - cos (x))}{2} = \dfrac{3 - cos (x))}{2}$

$\therefore \dfrac{\left ( 1 - sin^2 \left (\dfrac{x}{2} \right ) \right )}{\left ( 1 + sin^2 \left (\dfrac{x}{2} \right ) \right )} =\dfrac{\left ( \dfrac{1 + cos (x)}{2} \right)}{\left (\dfrac{3 - cos (x)}{2} \right ) } =\dfrac{\left ( 1 + cos (x))}{(3 - cos (x))}$

$\therefore \dfrac{\left ( 1 - sin^2 \left (\dfrac{x}{2} \right ) \right )}{\left ( 1 + sin^2 \left (\dfrac{x}{2} \right ) \right )} =\dfrac{\left ( 1 + cos (x))}{(3 - cos (x))}$

3 0

3 years ago