All categories

3 years ago

What is the true solution In 20+In 5=2 In x?

Mathematics

1 answer:

S_A_V [24]3 years ago

7 0

Answer:

x = 10

Step-by-step explanation:

Using the rules of logarithms

• log x + log y ⇔ log(xy)

• log $x^{n}$ ⇔ n log x

• log x = log y ⇒ x = y

Given

ln20 + ln5 = 2 lnx , then

ln(20 × 5) = ln $x^{2}$

ln 100 = ln $x^{2}$ , hence

x² = 100 ( take the square root of both sides )

x = 10

You might be interested in

Help 1 A / 7 (x + 2) -5 = 30

DaniilM [7]

Answer:

x = 3

Step-by-step explanation:

Given

7(x + 2) - 5 = 30 ( add 5 to both sides )

7(x + 2) = 35 ( divide both sides by 7 )

x + 2 = 5 ( subtract 2 from both sides )

x = 3

3 0

3 years ago

Algebraic Reasoning please help!

stiv31 [10]

Answer:

$\left[\begin{array}{ccc}7\\4\\2\end{array}\right]$

The answer is a single-column matrix (7,4,2)

Step-by-step explanation:

In such multiplication of matrices, you have to proceed by multiplying each ROW of the first matrix by the COLUMN of the second matrix. So,

$\left[\begin{array}{ccc}3&6&1\end{array}\right] * \left[\begin{array}{ccc}2\\0\\1\end{array}\right] = (3 * 2) + (6 * 0) + (1 * 1) = 6 + 0 + 1 = 7$

then...

$\left[\begin{array}{ccc}2&4&0\end{array}\right] * \left[\begin{array}{ccc}2\\0\\1\end{array}\right] = (2 * 2) + (4 * 0) + (0 * 1) = 4 + 0 + 0 = 4$

and

$\left[\begin{array}{ccc}0&6&2\end{array}\right] * \left[\begin{array}{ccc}2\\0\\1\end{array}\right] = (0 * 2) + (6 * 0) + (2 * 1) = 0 + 0 + 2= 2$

I hope it helps.

5 0

3 years ago

Is the triangle with side lengths of 20 mm, 24 mm, and 36 mm a right triangle

adoni [48]

Answer:

no it's an acute triangle

Step-by-step explanation:

why? because any angle below 90 degrees is considered an acute triangle which in your case 20,24, and 36 are all very far from being a right triangle

7 0

4 years ago

What is the least number of colors you need to correct color in the sections of these pictures so that no two touching sections

Kobotan [32]

You would assume that in this figure, the number of colored sections with which are not colored with respect to a " touching " colored section, would be half of the total colored sections. However that is not the case, the sections are not alternating as they still meet at a common point. After all, it notes no two touching sections, not adjacent sections. Their is no equation to calculate this requirement with respect to the total number of sections.

Let's say that we take one triangle as the starting. This triangle will be the start of a chain of other triangles that have no two touching sections, specifically 7 triangles. If a square were to be this starting shape, there are 5 shapes that have no touching sections, 3 being a square, the other two triangles. This is presumably a lower value as a square occupies two times as much space, but it also depends on the positioning. Therefore, the least number of colored sections you can color in the sections meeting the given requirement, is 5 sections for this first figure.

Respectively the solution for this second figure is 5 sections as well.

7 0

4 years ago

P.7 Is an even number or Q: 3 is square root of 10 ( P v Q ) answer asap please

Arte-miy333 [17]

The correct answer is the Last Choice ft equals truth

7 0

4 years ago