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3 years ago

What is the value of x if 5^x+2=5^9?

Mathematics

1 answer:

Mkey [24]3 years ago

7 0

In this equation, x = 7.

In order to solve this you are going to need to use logarithmic functions. By taking a logarithm you can eliminate exponents.

$5^{x + 2} = 5^{9}$

First we're going to take the logarithm of both sides, which allows us to put the exponents in front of the terms.

(x + 2)Log5 = 9Log5

Now that we've done this, we can divide both sides by Log5 and eliminate that term.

x + 2 = 9

From here, we simply need to subtract in order to solve.

x = 7

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An article is bought for Rs. 125 what is the profit percentage?

Aneli [31]

Step-by-step explanation:

Solution given;

cost price=Rs125

profit%=?

we have

profit%=[Selling price-cost price]/cost price×100%

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3 years ago

100 points for correct answer

Tems11 [23]

Answer:

$\dfrac{x}{3}+\dfrac{3}{y}$

Step-by-step explanation:

Given fraction:

$\dfrac{xy+9}{3y}$

$\textsf{Apply the fraction rule} \quad \dfrac{a+b}{c}=\dfrac{a}{c}+\dfrac{b}{c}:$

$\implies \dfrac{xy}{3y}+\dfrac{9}{3y}$

Rewrite 9 as 3 · 3:

$\implies \dfrac{xy}{3y}+\dfrac{3 \cdot 3}{3y}$

Cancel the common factor y in the first fraction and the common factor 3 in the second fraction:

$\implies \dfrac{x\diagup\!\!\!\!y}{3\diagup\!\!\!\!y}+\dfrac{\diagup\!\!\!\!3 \cdot 3}{\diagup\!\!\!\!3y}$

$\implies \dfrac{x}{3}+\dfrac{3}{y}$

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1 year ago

A train travels 95 kilometers in 1 hours, and then 97 kilometers in 1 hours. What is its average speed?

aleksandrvk [35]

Average speed is given by [Total Distance] ÷ [Total Time]

Average speed = $\frac{95+97}{1+1} = \frac{192}{2}=96$ km/hour

8 0

3 years ago

I don't know to find the answer to 8. Can someone explain to me?

lapo4ka [179]

Perhaps the easiest way to find the midpoint between two given points is to average their coordinates: add them up and divide by 2.

A) The midpoint C' of AB is
.. (A +B)/2 = ((0, 0) +(m, n))/2 = ((0 +m)/2, (0 +n)/2) = (m/2, n/2) = C'
The midpoint B' is
.. (A +C)/2 = ((0, 0) +(p, 0))/2 = (p/2, 0) = B'
The midpoint A' is
.. (B +C)/2 = ((m, n) +(p, 0))/2 = ((m+p)/2, n/2) = A'

B) The slope of the line between (x1, y1) and (x2, y2) is given by
.. slope = (y2 -y1)/(x2 -x1)
Using the values for A and A', we have
.. slope = (n/2 -0)/((m+p)/2 -0) = n/(m+p)

C) We know the line goes through A = (0, 0), so we can write the point-slope form of the equation for AA' as
.. y -0 = (n/(m+p))*(x -0)
.. y = n*x/(m+p)

D) To show the point lies on the line, we can substitute its coordinates for x and y and see if we get something that looks true.
.. (x, y) = ((m+p)/3, n/3)
Putting these into our equation, we have
.. n/3 = n*((m+p)/3)/(m+p)
The expression on the right has factors of (m+p) that cancel*, so we end up with
.. n/3 = n/3 . . . . . . . true for any n

_____
* The only constraint is that (m+p) ≠ 0. Since m and p are both in the first quadrant, their sum must be non-zero and this constraint is satisfied.

The purpose of the exercise is to show that all three medians of a triangle intersect in a single point.

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3 years ago

Prove that the triangle EDF is isosceles. Give reasons for your answer.

Gekata [30.6K]

Answer:

$\triangle EDF$ is isosceles.

Step-by-step explanation:

Please have a look at the attached figure.

We are given the following things:

$\angle EDF = y$

$\text{External }\angle DFG = 90 +\dfrac{y}{2}$

Let us try to find out $\angle E$ and $\angle DFE$ . After that we will compare them.

Finding  $\angle DFE$ :

Side EG is a straight line so $\angle GFE = 180$

$\angle GFE$ is sum of internal $\angle DFE$ and external $\angle DFG$

$\angle GFE = 180 = \angle DFE + \angle DFG\\\Rightarrow 180 = \angle DFE + (90+\dfrac{y}{2})\\\Rightarrow \angle DFE = 180 - 90 - \dfrac{y}{2}\\\Rightarrow \angle DFE = 90 - \dfrac{y}{2} ....... (1)$