Ask question

Ask question

All categories

3 years ago

11

What are the zero of the functions?

1 answer:

EleoNora [17]3 years ago

8 0

Answer:

x= -4 x=-2

Step-by-step explanation:

factor the function and get

(x+4)(x+2)=0

therefore

x=-4 or x=-2

You might be interested in

Which equals the product of (x – 3)(2x + 1)?

hram777 [196]

Answer:

2x^2 – 5x – 3

Step-by-step explanation:

<u>Apply FOIL method: </u>

=x*2x+x*1+(-3)*2x+(-3)*1

<u>Apply minus-plus rules:</u>

=2xx+1*x-3*2x-3*1

Simplify:

=2x^2 – 5x – 3

Your Answer Is 2x^2 – 5x – 3

plz mark me as brainliest :)

6 0

3 years ago

Read 2 more answers

Which of the following words mean subtraction? Check all that apply.

anastassius [24]

Answer is C please give brainliest

5 0

2 years ago

Read 2 more answers

There are 25 computers in a math lab. There are 6 activity books for every 2 computers in the math lab. How many activity bookle

kondaur [170]

Answer:

b

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Im taking a test & i don’t understand can someone help

Shtirlitz [24]

62 units hope this helps

4 0

3 years ago

Solve the oblique triangle where side a has length 10 cm, side c has length 12 cm, and angle beta has measure thirty degrees. Ro

Ugo [173]

Answer:

$Side\ B = 6.0$

$\alpha = 56.3$

$\theta = 93.7$

Step-by-step explanation:

Given

Let the three sides be represented with A, B, C

Let the angles be represented with $\alpha, \beta, \theta$

[See Attachment for Triangle]

$A = 10cm$

$C = 12cm$

$\beta = 30$

What the question is to calculate the third length (Side B) and the other 2 angles ( $\alpha\ and\ \theta$ )

Solving for Side B;

When two angles of a triangle are known, the third side is calculated as thus;

$B^2 = A^2 + C^2 - 2ABCos\beta$

Substitute: $A = 10$ , $C =12$ ; $\beta = 30$

$B^2 = 10^2 + 12^2 - 2 * 10 * 12 *Cos30$

$B^2 = 100 + 144 - 240*0.86602540378$

$B^2 = 100 + 144 - 207.846096907$

$B^2 = 36.153903093$

Take Square root of both sides

$\sqrt{B^2} = \sqrt{36.153903093}$

$B = \sqrt{36.153903093}$

$B = 6.0128115797$

$B = 6.0$ <em>(Approximated)</em>

Calculating Angle $\alpha$

$A^2 = B^2 + C^2 - 2BCCos\alpha$

Substitute: $A = 10$ , $C =12$ ; $B = 6$

$10^2 = 6^2 + 12^2 - 2 * 6 * 12 *Cos\alpha$

$100 = 36 + 144 - 144 *Cos\alpha$

$100 = 36 + 144 - 144 *Cos\alpha$

$100 = 180 - 144 *Cos\alpha$

Subtract 180 from both sides

$100 - 180 = 180 - 180 - 144 *Cos\alpha$

$-80 = - 144 *Cos\alpha$

Divide both sides by -144

$\frac{-80}{-144} = \frac{- 144 *Cos\alpha}{-144}$

$\frac{-80}{-144} = Cos\alpha$

$0.5555556 = Cos\alpha$

Take arccos of both sides

$Cos^{-1}(0.5555556) = Cos^{-1}(Cos\alpha)$

$Cos^{-1}(0.5555556) = \alpha$

$56.25098078 = \alpha$

$\alpha = 56.3$ <em>(Approximated)</em>

Calculating $\theta$

Sum of angles in a triangle = 180

Hence;

$\alpha + \beta + \theta = 180$

$30 + 56.3 + \theta = 180$

$86.3 + \theta = 180$

Make $\theta$ the subject of formula

$\theta = 180 - 86.3$

$\theta = 93.7$

5 0

3 years ago