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3 years ago

Help! Select all of the potential solution(s) of the equation 2log5x = log54.

Mathematics

1 answer:

Nastasia [14]3 years ago

7 0

Answer:

x = 2

Step-by-step explanation:

2 log₅ x = log₅ 4
2 log₅ x = log₅ 2²
2 log₅ x = 2 log₅ 2
log₅ x = log₅ 2
x = 2

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Simplify (6x² - x - 4) + (2x² + 5x – 5)

harina [27]

Answer:

NaN -28

Step-by-step explanation:

7 0

3 years ago

6x+9/x-5 x is not equal to 5 find inverse

algol [13]

You have f(x)=(<span>6x+9)/(x-5) and want to find the inverse function? Please note that the parentheses are obligatory here.

1. Replace "f(x)" with y: y = </span>(6x+9)/(x-5)
<span>
2. Interchange the variables x and y: x</span> = (6y+9)/(y-5)

3. Solve this result for y: x(y-5) = (6y+9), or xy-5x = 6y+9. Then:

xy-6y = 9 + 5x, or y(x-6) = 9+5x
9+5x
Then y = --------------
x-6

Here x may not = 6, because division by zero would occur otherwise.

3 0

3 years ago

Please answer this correctly

atroni [7]

The first box is 3 since you are distributing from 3 or whatever number is outside of the paranthesis.

The second box is 24 because 3•8 is 24.

5 0

4 years ago

Geometry!!!!!!!!!!!!!!!!!!!!

svlad2 [7]

Answer:

The two triangles are related by Side-Side-Side (SSS), so the triangles can be proven congruent.

Step-by-step explanation:

There are no angles that can be shown to be congruent to one another, so this eliminates all answer choices with angles (SSA, SAS, ASA, AAA, AAS).

This leaves you with either the HL (Hypotenuse-Leg) Theorem or SSS (Side-Side-Side) Theorem. We could claim that the triangles can be proven congruent by HL, however, we aren't exactly sure as to whether or not the triangles have a right angle. There is no indicator, and in this case, we cannot assume so.

This leaves you with the SSS Theorem.

8 0

3 years ago

let sin(θ) =3/5 and tan(y) =12/5 both angels comes from 2 different right trianglesa)find the third side of the two tringles b)

statuscvo [17]

In a right triangle, we haev some trigonometric relationships between the sides and angles. Given an angle, the ratio between the opposite side to the angle by the hypotenuse is the sine of this angle, therefore, the following statement

$\sin (\theta)=\frac{3}{5}$

Describes the following triangle

To find the missing length x, we could use the Pythagorean Theorem. The sum of the squares of the legs is equal to the square of the hypotenuse. From this, we have the following equation

$x^2+3^2=5^2$

Solving for x, we have

$\begin{gathered} x^2+3^2=5^2 \\ x^2+9=25 \\ x^2=25-9 \\ x^2=16 \\ x=\sqrt[]{16} \\ x=4 \end{gathered}$

The missing length of the first triangle is equal to 4.

For the other triangle, instead of a sine we have a tangent relation. Given an angle in a right triangle, its tanget is equal to the ratio between the opposite side and adjacent side.The following expression

$\tan (y)=\frac{12}{5}$

Describes the following triangle

Using the Pythagorean Theorem again, we have

$5^2+12^2=h^2$

Solving for h, we have

$\begin{gathered} 5^2+12^2=h^2 \\ 25+144=h^2 \\ 169=h^2 \\ h=\sqrt[]{169} \\ h=13 \end{gathered}$

The missing side measure is equal to 13.

Now that we have all sides of both triangles, we can construct any trigonometric relation for those angles.

The sine is the ratio between the opposite side and the hypotenuse, and the cosine is the ratio between the adjacent side and the hypotenuse, therefore, we have the following relations for our angles

$\begin{gathered} \sin (\theta)=\frac{3}{5} \\ \cos (\theta)=\frac{4}{5} \\ \sin (y)=\frac{12}{13} \\ \cos (y)=\frac{5}{13} \end{gathered}$

To calculate the sine and cosine of the sum

$\begin{gathered} \sin (\theta+y) \\ \cos (\theta+y) \end{gathered}$

We can use the following identities

$\begin{gathered} \sin (A+B)=\sin A\cos B+\cos A\sin B \\ \cos (A+B)=\cos A\cos B-\sin A\sin B \end{gathered}$

Using those identities in our problem, we're going to have

$\begin{gathered} \sin (\theta+y)=\sin \theta\cos y+\cos \theta\sin y=\frac{3}{5}\cdot\frac{5}{13}+\frac{4}{5}\cdot\frac{12}{13}=\frac{63}{65} \\ \cos (\theta+y)=\cos \theta\cos y-\sin \theta\sin y=\frac{4}{5}\cdot\frac{5}{13}-\frac{3}{5}\cdot\frac{12}{13}=-\frac{16}{65} \end{gathered}$

4 0

1 year ago