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3 years ago

How do you simplify x^2-4x-21 divided by x+3

1 answer:

6 0

We can factor x^2-4x-21 to get (x+3)(x-7)

Notice how
3 plus -7 = -4
3 times -7 = -21
which helps us factor the trinomial

After you've factored it, we can cancel out a pair of (x+3) terms leaving us with the final answer of x-7

Note: the cancellation happens because (x+3)/(x+3) = 1. Divide anything (except 0) over itself and you'll get 1. A simpler example of this is 5/5 = 1.

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Lilit [14]

Answer:

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Step-by-step explanation:

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4 0

3 years ago

Help a brother out please

lara [203]

Answer:

Bethlehem has temperatures ranging from 33.8 Fahrenheit to 55.4 Fahrenheit.

Step-by-step explanation:

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4 0

3 years ago

assume that 0<x<pi/2 and 0<y<pi/2. find the exact value of sin(x+y) if cosx=8/17 and siny= 12/37

max2010maxim [7]

Answer:

621 / 629

Step-by-step explanation:

We are given the following information -

cos x = $\frac{8}{17}$ , and sin y = $\frac{12}{37}$

Respectively we can use the following information -

sin( x + y ) = sin x ( cos y ) + sin y ( cos x ),

$cos^2x + sin^2x = 1,\\cos^2y + sin^2y = 1$

Knowing that cos^2x + sin^2x = 1, cos^2y + sin^2y = 1, we can calculate the value of sin x and cos y, plugging it into the first bit " sin( x + y ) = sin x ( cos y ) + sin y ( cos x ) "

$sin^2x = 1 - cos^2x,\\sin^2x = 1 - ( 8 / 17 )^2,\\sin^2x = 15^2 / 17^2\\----------------\\sin ( x ) = 15 / 17$

Respectively cos y should be 35 / 37 -

$cos^2y = 1 - sin^2y,\\cos^2y = 1 - ( 12 / 37 )^2,\\cos^2y = 35^2 / 37^2\\----------------\\cos ( y ) = 35 / 37$

Thus,

sin( x + y ) = ( 15 / 17 ) * ( 35 / 37 ) + ( 12 / 37 ) * ( 8 / 17 ),

sin( x + y ) = 621 / 629

Hope that helps!

5 0

3 years ago

What is the known and unknown in 6 multiple by 2 are are 12 devided by 6

77julia77 [94]

6 ÷12 = 0.5
12 ÷ 6 =. 2

6 × 2 = 12
totally answer

7 0

3 years ago

Identify the range of the function shown in the graph

sladkih [1.3K]

The correct answer is: [A]: " -5 ≤ y ≤ 5 " .
_________________________________________________
The "range" refers to the the values of a function that encompass the "y-axis" of the [graphed] function.

By looking at the graph provided in the image attached, we see that for "y" values (when "y" is "positive", or "greater than "0"; the highest value for "y" is "5"; yet "y" never "exceeds "5". We see that for "y" values (when y is "negative", or "less than "0", the lowest values for "y" is "-5"; yet "y" never goes "lower than "-5".

However, "y" values are aways within, and encompassing, all "y" values between and including: "y = -5" and "y = 5" .

As such, Answer choice: [A]: " -5 ≤ y ≤ 5 " ; is correct.
________________________________________________________

4 0

3 years ago