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3 years ago

What is an equivalent exponential equation? log5(1/25)=−2

Mathematics

1 answer:

jenyasd209 [6]3 years ago

7 0

$\log_ab=c\iff a^c=b\\\\\log_5\dfrac{1}{25}=-2\iff5^{-2}=\dfrac{1}{25}\\\\5^{-2}=\dfrac{1}{5^2}=\dfrac{1}{25}$

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2.248 to the whole number

Viktor [21]

2.248 rounded to the whole number would be 2, because if we look after the number 2, the 2nd 2 it is less than 5. Therefore we round backwards, and the answer is 2.

So, 2.248 rounded to the nearest whole number would be 2.

Hope I helped ya!! xD

4 0

4 years ago

In the xy-plane, The line P is perpendicular to the graph of the equation 4x+3y=12. Which of the following is the slope of line

gladu [14]

A slope is also known as the gradient of a line. The slope of line P is 3/4.

<h3>What is Slope?</h3>

A slope also known as the gradient of a line is a number that helps to know both the direction and the steepness of the line.

As it is known that for two perpendicular lines the slope of the lines will be negative reciprocal of each other. The slope of line 4x+3y=12 can be written as,

y = 4 - (4/3)x

m₁ = -1/m₂

-(4/3) = -1/m₂

m₂ = 3/4

Hence, the slope of line P is 3/4.

Learn more about Slope of Line:

brainly.com/question/14511992

#SPJ1

6 0

2 years ago

I need help with what is x 4x=20

Elodia [21]

Answer:

Step-by-step explanation:

You are looking for x so what you are really looking for is what times 4 =20 and that is 5.

5x4=20

3 0

2 years ago

Read 2 more answers

Find the circumference of the circle and round to the nearest tenth

VLD [36.1K]

Answer:

The circumference is 45.2yd

Step-by-step explanation:

Given

$d = 14.4$ --- diameter

Required

Determine the circumference

The circumference is calculated as:

$C =\pi d$

So, we have:

$C =\pi * 14.4$

Take pi a 3.14

$C =3.14 * 14.4$

$C =45.216$

Approximate

$C =45.2yd$

4 0

3 years ago

Help me to answer now ineed this <br> Please...

Vera_Pavlovna [14]

ANSWER TO QUESTION 1

$\frac{\frac{y^2-4}{x^2-9}} {\frac{y-2}{x+3}}$

Let us change middle bar to division sign.

$\frac{y^2-4}{x^2-9}\div \frac{y-2}{x+3}$

We now multiply with the reciprocal of the second fraction

$\frac{y^2-4}{x^2-9}\times \frac{x+3}{y-2}$

We factor the first fraction using difference of two squares.

$\frac{(y-2)(y+2)}{(x-3)(x+3)}\times \frac{x+3}{y-2}$

We cancel common factors.

$\frac{(y+2)}{(x-3)}\times \frac{1}{1}$

This simplifies to

$\frac{(y+2)}{(x-3)}$

ANSWER TO QUESTION 2

$\frac{1+\frac{1}{x}} {\frac{2}{x+3}-\frac{1}{x+2}}$

We change the middle bar to the division sign

$(1+\frac{1}{x}) \div (\frac{2}{x+3}-\frac{1}{x+2})$

We collect LCM to obtain

$(\frac{x+1}{x})\div \frac{2(x+2)-1(x+3)}{(x+3)(x+2)}$

We expand and simplify to obtain,

$(\frac{x+1}{x})\div \frac{2x+4-x-3}{(x+3)(x+2)}$

$(\frac{x+1}{x})\div \frac{x+1}{(x+3)(x+2)}$

We now multiply with the reciprocal,

$(\frac{(x+1)}{x})\times \frac{(x+2)(x+3)}{(x+1)}$

We cancel out common factors to obtain;

$(\frac{1}{x})\times \frac{(x+2)(x+3)}{1}$

This simplifies to;

$\frac{(x+2)(x+3)}{x}$

ANSWER TO QUESTION 3

$\frac{\frac{a-b}{a+b}} {\frac{a+b}{a-b}}$

We rewrite the above expression to obtain;

$\frac{a-b}{a+b}\div {\frac{a+b}{a-b}}$

We now multiply by the reciprocal,

$\frac{a-b}{a+b}\times {\frac{a-b}{a+b}}$

We multiply out to get,

$\frac{(a-b)^2}{(a+b)^2}$

ANSWER T0 QUESTION 4

To solve the equation,

$\frac{m}{m+1} +\frac{5}{m-1} =1$

We multiply through by the LCM of $(m+1)(m-1)$

$(m+1)(m-1) \times \frac{m}{m+1} + (m+1)(m-1) \times \frac{5}{m-1} =(m+1)(m-1) \times 1$

This gives us,

$(m-1) \times m + (m+1) \times 5}=(m+1)(m-1)$

$m^2-m+ 5m+5=m^2-1$

This simplifies to;

$4m-5=-1$

$4m=-1-5$

$4m=-6$

$\Rightarrow m=-\frac{6}{4}$

$\Rightarrow m=-\frac{3}{2}$

ANSWER TO QUESTION 5

$\frac{3}{5x}+ \frac{7}{2x}=1$

We multiply through with the LCM of $10x$

$10x \times \frac{3}{5x}+10x \times \frac{7}{2x}=10x \times1$

We simplify to get,

$2 \times 3+5 \times 7=10x$

$6+35=10x$

$41=10x$

$x=\frac{41}{10}$

$x=4\frac{1}{10}$

Method 1: Simplifying the expression as it is.

$\frac{\frac{3}{4}+\frac{1}{5}}{\frac{5}{8}+\frac{3}{10}}$

We find the LCM of the fractions in the numerator and those in the denominator separately.

$\frac{\frac{5\times 3+ 4\times 1}{20}}{\frac{(5\times 5+3\times 4)}{40}}$

We simplify further to get,

$\frac{\frac{15+ 4}{20}}{\frac{25+12}{40}}$

$\frac{\frac{19}{20}}{\frac{37}{40}}$

With this method numerator divides(cancels) numerator and denominator divides (cancels) denominator

$\frac{\frac{19}{1}}{\frac{37}{2}}$

Also, a denominator in the denominator multiplies a numerator in the numerator of the original fraction while a numerator in the denominator multiplies a denominator in the numerator of the original fraction.

That is;

$\frac{19\times 2}{1\times 37}$

This simplifies to

$\frac{38}{37}$

Method 2: Changing the middle bar to a normal division sign.

$(\frac{3}{4}+\frac{1}{5})\div (\frac{5}{8}+\frac{3}{10})$

We find the LCM of the fractions in the numerator and those in the denominator separately.

$(\frac{5\times 3+ 4\times 1}{20})\div (\frac{(5\times 5+3\times 4)}{40})$

We simplify further to get,

$(\frac{15+ 4}{20})\div (\frac{(25+12)}{40})$

$\frac{19}{20}\div \frac{(37)}{40}$

We now multiply by the reciprocal,

$\frac{19}{20}\times \frac{40}{37}$

$\frac{19}{1}\times \frac{2}{37}$

$\frac{38}{37}$

5 0

3 years ago