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3 years ago

The number 31 is increased to 44. What is the percentage by which the number was increased, to the nearest tenth of a percent?

Mathematics

1 answer:

Sophie [7]3 years ago

3 0

Answer:

41.9 %

Step-by-step explanation:

when you look at the to numbers you will see that the two has a 41.9 % difference.

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Solve for x write the exact answer using either base -10 or base-e logarithms

jeka57 [31]

<u>Given</u>:

The given expression is $2^{x+5}=13^{2 x}$

We need to determine the value of x using either base - 10 or base - e logarithms.

<u>Value of x:</u>

Let us determine the value of x using the base - e logarithms.

Applying the log rule that if $f(x)=g(x)$ then $\ln (f(x))=\ln (g(x))$

Thus, we get;

$\ln \left(2^{x+5}\right)=\ln \left(13^{2 x}\right)$

Applying the log rule, $\log _{a}\left(x^{b}\right)=b \cdot \log _{a}(x)$ , we get;

$(x+5) \ln (2)=2 x \ln (13)$

Expanding, we get;

$x \ln (2)+5 \ln (2)=2 x \ln (13)$

Subtracting both sides by $5 \ln (2)$ , we get;

$x \ln (2)=2 x \ln (13)-5 \ln (2)$

Subtracting both sides by $2 x \ln (13)$ , we get;

$x \ln (2)-2 x \ln (13)=-5 \ln (2)$

Taking out the common term x, we have;

$x( \ln (2)-2 \ln (13))=-5 \ln (2)$

$x=\frac{-5 \ln (2)}{\ln (2)-2 \ln (13)}$

$x=\frac{5 \ln (2)}{2 \ln (13)-\ln (2)}$

Thus, the value of x is $x=\frac{5 \ln (2)}{2 \ln (13)-\ln (2)}$

6 0

3 years ago

Help me guys, Due now please

Otrada [13]

Answer:

about 8.3 miles an hour.

Step-by-step explanation:

34/4 is 8.5

90/11 is 8.2

average: 8.3

6 0

2 years ago

I need to find all real values of x from this inequality x^2>2x-2 and also this question x^2>2x+2. I need the answer fast

sergeinik [125]

$x^2\gt2x-2$

$x^2-2x+2\gt0$

You are essentially asking at this point where is the quadratic function greater than 0. To get the interval find the roots. Between the roots the function is guaranteed to be LESS THAN zero. Everywhere else the function is greater than 0.

$x^2-2x+2=0$

Use Vieta's rule to get the roots,

$(x-1)(x-1)=0$

Both roots are the same so no value is less than 0 which means all values except at the point of root are strictly greater, the point of root is at $x_{1,2}=1$ which means the set for which the quadratic function is greater than 0 is $\mathbb{R}-\{1\}$ "all real numbers except 1".