Divide.

Solve for x HELP ASAP

Andre45 [30]

Answer:

Step-by-step explanation:well I don’t quit know but try looking for people who already wrote the answrr

7 0

2 years ago

Read 2 more answers

-(-2) and -/-2/ simplify to different numbers

kow [346]

I suppose the / are absolute value lines.... so ill answer it that way, -(-2) simplifies to 2. -/-2/ simplifies to -2

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

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

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

$100 = 180 - 144 *Cos\alpha$

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

Part 1

svetlana [45]

Answer:

A: This polynomial has a degree of 2 , so the equation 12x2+5x−2=0 has two or fewer roots.

B: The quadratic equation 12x2+5x−2=0 has two real solutions, x=−2/3 or x=1/4 , and therefore has two real roots.

Step-by-step explanation:

f(x) = 12x^2 + 5x - 2.

Since this is a quadratic equation, or a polynomial of second degree, one can easily conclude that this equation will have at most 2 roots. At most 2 roots mean that the function can have either 2 roots at maximum or less than 2 roots. Therefore, in the A category, 2nd option is the correct answer (This polynomial has a degree of 2 , so the equation 12x^2 + 5x − 2 = 0 has two or fewer roots).

To find the roots of f(x), set f(x) = 0. Therefore:

12x^2 + 5x - 2 = 0. Solving the question using the mid term breaking method shows that 12*2=24. The factors of 24 whose difference is 5 are 8 and 3. Therefore:

12x^2 + 8x - 3x - 2 = 0.

4x(3x + 2) -1(3x+2) = 0.

(4x-1)(3x+2) = 0.

4x-1 = 0 or 3x+2 = 0.

x = 1/4 or x = -2/3.

It can be seen that f(x) has two distinct real roots. Therefore, in the B category, 1st Option is the correct answer (The quadratic equation 12x2+5x−2=0 has two real solutions, x=−2/3 or x=1/4 , and therefore has two real roots)!!!

8 0

3 years ago

Pls help me!!!! I'm stuck on how to solve this!!??

Luda [366]

Let A represent the value of the car after each year.
A= initial value (P)×(1+percent increase(r)) ^time
A=P×(1+r)^t
A=18710×(1+(-12%))^8
A=18710×(1-12%)^8
A=18710×(1-0.12)^8
A=18710×(0.88)^8
A= 6728.7619591115
The best approximation is 6729
Therefore the value of the car will be about $6729 after 8 years
Your answer is B.

6 0

3 years ago