All categories

3 years ago

Select the correct answer. Find the value of x. Please answer!!!

Mathematics

1 answer:

olga nikolaevna [1]3 years ago

6 0

C. 64

Rewrite the logarithm:

8=x^.5

Solve by raising both sides of the equation to the power of the inverse (in this case, the inverse of .5, which is 2):

8^2=x^(.5*2)

64=x

You might be interested in

Selected-Response

maks197457 [2]

Answer: 6 pints

Step-by-step explanation:

32 times 3 in 96

so you must multiply 2 times 3

5 0

3 years ago

Absolute value /-3/-/-2/=

Anni [7]

Answer:

I think it is 1

Step-by-step explanation:

8 0

3 years ago

In the angle of elevation from the tip of the shadow to the top of the tree is 32 degrees, what is the height of the tree to the

LenaWriter [7]

Answer:

it might be -40.0

Step-by-step explanation:

i did the worl

3 0

3 years ago

(4,7,12,19) How do you right this in recursive and explicit equations??????? pls help

Soloha48 [4]

You can see how the difference between two consecutive terms is constantly increasing:

$a_1 = 4$

$a_2 = 7 = 4+3$

$a_3 = 12 = 7+5$

$a_4 = 19 = 12+7$

So, for the next terms we'll have to add +9, +11, +13 and so on.

Also, note that $a_2$ is obtained by adding the 2nd odd number to $a_1$ , $a_3$ is obtained by adding the 3rd odd number to $a_2$ , and so on.

So, the recursive formula is

$a_n = a_{n-1}+(2n-1),\quad a_1=1$

For the explicit formula, recall that the sum of the first n odd numbers is n squared. Taking into account the fact that we're not starting from 1, we have

$a_n = n^2+3$

7 0

3 years ago

Let <img src="https://tex.z-dn.net/?f=i" id="TexFormula1" title="i" alt="i" align="absmiddle" class="latex-formula"> be the imag

VLD [36.1K]

$Hey~freckledspots!\\----------------------$

$We~will~solve~for~i^{425}!$

$Rule~of~exponent: a^{b + c} = a^ba^c\\Apply:~i^{425}~=~i^{424}i\\ \\Rule~of~exponent: a^{bc} = (a^{b})^c\\Apply: i^{424} = i(i^2)^{212} \\\\Rule~of~imaginary~number: i^2 = -1\\Apply: i(i^2)^{212} = -1^{212}i\\\\Rule~of~exponent~if~n~is~even: -a^n = a^n\\Apply: -1^{212}i = 1^{212}i\\\\Simplify: 1^{212}i = 1i\\Multiply: 1i * 1 = i\\----------------------\\$

$Now~let's~solve~1^{14}!\\\\Rule~of~exponent: a^{b + c} = a^ba^c\\Apply: i^{14} = (i^2)^7\\\\Rule~of~imaginary~number: i^2 = -1\\Apply: (i^2)^7 = -1^7\\\\Rule~of~exponent~if~n~is~odd: (-a)^n = -a^n\\Apply: -1^7 = -1^7\\\\Simplify: -1^7 = -1\\----------------------\\Now,~we~have: i-1+i^{-14}+i^{44}\\----------------------$

$Now~lets~solve~i^{-14}\\\\Rule~of~exponent: a^{-b} = \frac{1}{a^b} \\Apply: i^{-14} = \frac{1}{i^{14}} \\\\Rule~of~exponent: a^{bc} = (a^b)^c\\Apply: \frac{1}{i^{14}} = \frac{1}{(i^2)^7}\\ \\Rule~of~imagianry~number: i^2 = -1\\Apply: \frac{1}{(i^2)^7} = \frac{1}{-1^7} \\\\Simplify: \frac{1}{-1^7} = \frac{1}{-1} \\\\Rule~of~fractions: \frac{a}{-b} = -\frac{a}{b} \\Apply: \frac{1}{-1} = -\frac{1}{1} = -1\\----------------------\\Now,~we~have: i-1-1+i^44\\----------------------$

$Now~let's~solve~i^{44}!\\\\Rule~of~exponent: a^{bc} = (a^b)^c\\Apply: i^{44} = (i^2)^{22}\\\\Rule~of~imaginary~numbers: i^2 = -1\\Apply: (i^2)^{22} = -1^{22}\\\\Rule~of~exponent~if~n~is~even: (-a)^n = a^n\\Apply: -1^{22} = 1^{22}\\\\Simplify: 1^{22} = 1\\----------------------\\Now,~we~have~i-1-1+1\\----------------------$

$Now~let's~simplify~the~expression!\\\\= i-1-1+1 \\= 1 + i -2\\= -1+i\\----------------------$

$Answer:\\\large\boxed{-1+i}\\----------------------$

$Hope~This~Helped!~Good~Luck!$

8 0

3 years ago

Read 2 more answers