Ask question

Ask question

All categories

3 years ago

10

Plzz help me with this question

1 answer:

Olegator [25]3 years ago

8 0

Answer:

4

Step-by-step explanation:

The degree is equal to the largest exponent on a variable. In this instance, it is 4.

You might be interested in

Olhe a imagem e responda

matrenka [14]

Answer:

i cant see the attachment

8 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

The answers are either: 12, 2, 8, or 10 please helppp

blsea [12.9K]

Answer: 10 I believe

Step-by-step explanation:

3 0

3 years ago

How many extraneous solutions exist for the logarithmic equation below if it is solved in the most efficient way possible?log_(2

Nadusha1986 [10]

Answer:

No extraneous solution

Step-by-step explanation:

We have the logarithmic equation given by,

$\log_{2}[\log_{2}(\sqrt{4x})]=1$

i.e. $\log_{2}(\sqrt{4x})=2^{1}$

i.e. $\sqrt{4x}=2^{2}$

i.e. $\sqrt{4x}=4$

i.e. $4x=4^{2}$

i.e. $4x=16$

i.e. $x=4$

So, the solution of the given equation is x=4.

Now, as we domain of square root function is x > 0 and also, the domain of logarithmic function is $( 0,\infty )$ .

Therefore, the domain of the given function is x > 0.

We know that the extraneous solution is the solution which does not belong to the domain.

But as x=4 belongs to the domain x > 0.

Thus, x = 4 is not an extraneous solution.

Hence, this equation does not have any extraneous solution.

7 0

3 years ago

Read 2 more answers

Write an equation perpendicular to y=1/5 x+9 that passes through the point (-2,-2)

Mamont248 [21]

Answer:

-2= 1/5x-2 +9

-2= -0.4 +9

Step-by-step explanation:

4 0

3 years ago