The sum of 4 consecutive odd numbers is 40. What is the greatest of the 4 numbers

PLEASE HELP RIGHT ANSWER GETS BRAINLIST

IgorLugansk [536]

Answer:

what is the question

Step-by-step explanation:

5 0

3 years ago

If f(x)=2x^4-3x^3+2x^2+x-3, explain how you know that the equation f(x)=0 has at least one real root between x=1 and x=2

hodyreva [135]

Answer:

Step-by-step explanation:

by theorem :

f continu in [1 , 2]

f(1)×f(2) < 0 because : f(1) = -1 and f(2) = 15

so the equation f(x)=0 has at least one real root between x=1 and x=2

so :

7 0

3 years ago

Khalid wants to buy a long sandwich for a party. Store A sells a 5 foot sandwich for $42.50. Store B sells a 6 foot sandwich for

Alex17521 [72]

Store A: 1 foot= 42.50÷5 = $8.50

Store B= 1 foot= 49.50÷6 = $8.25

Answer:

Store B has a better buy because the price for 1 foot sandwich is cheaper than Store A.

3 0

3 years ago

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

Graphs of the following equations are straight lines expect?

natka813 [3]

I think you would do that

7 0

3 years ago