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3 years ago

Simplify Will mark brainliest, please show work if possible

Mathematics

2 answers:

Reil [10]3 years ago

5 0

The answer for this question is 8/1

kondaur [170]3 years ago

4 0

Answer:

Step-by-step explanation:

Add the top numbers together and get 64

add the bottom numbers together and get 8

8 divided by 64 equals 8.

Brainliest?

You might be interested in

Find the distance between 2 3/5 and -1 on a number line

True [87]

Answer:

The distance between these two is 3 3/5

Step-by-step explanation:

In order to find the distance between two points, we subtract one from the other. Then if the result is negative, we take the absolute value.

2 3/5 - -1

2 3/5 + 1

3 3/5

7 0

3 years ago

Help me with this pls

Dvinal [7]

Answer:

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Given

6x + 10y = 8 ( subtract 6x from both sides )

10y = - 6x + 8 ( divide each term by 10 )

y = - $\frac{6}{10}$ x + $\frac{8}{10}$ = - $\frac{3}{5}$ x 6 + $\frac{4}{5}$ ← in slope- intercept form

with slope m = - $\frac{3}{5}$ → C

7 0

3 years ago

What exponential function is the best fit for the data in the table?

gogolik [260]

Answer: fourth option

$\frac{1}{4} 4^{x-1}-4$

Explanation:

1) the pair x = 3 f(x) = 0, leads you to probe this:

f(3) = 0 = A [4 ^ (3 - 1) ] + C = 0

=> A [4^2] = - C

A[16] = - C

if A = 1/4

16 / 4 = 4 => C = - 4

That leads you to the function f(x) = [1/4] 4 ^( x - 1) - 4

2) Now you verify the images for that function for all the x-values of the table:

x = 2 => f(2) + [1/4] 4 ^ (2 - 1) - 4 = [1/4] 4 - 4 = 4 / 4 - 4 = 1 - 4 = - 3 => check

x = 3 => f(3) = [1/4] 4^ (3 - 1) - 4 = [1/4] 4^2 - 4 = 16 / 4 - 4 = 4 - 4 = 0 => check

x = 4 -> f(4) = [1/4] 4^ (4-1) - 4 = [1/4] 4^(3) - 4 = (4^3) / 4 - 4 = 4^2 - 4 = 16 - 4 = 12 => check.

Therefore, you have proved that the answer is the fourth option.

6 0

3 years ago

Read 2 more answers

How to solve this trig

n200080 [17]

Hi there!

To find the Trigonometric Equation, we have to isolate sin, cos, tan, etc. We are also given the interval [0,2π).

First Question

What we have to do is to isolate cos first.

$\displaystyle \large{ cos \theta = - \frac{1}{2} }$

Then find the reference angle. As we know cos(π/3) equals 1/2. Therefore π/3 is our reference angle.

Since we know that cos is negative in Q2 and Q3. We will be using π + (ref. angle) for Q3. and π - (ref. angle) for Q2.

Find Q2

$\displaystyle \large{ \pi - \frac{ \pi}{3} = \frac{3 \pi}{3} - \frac{ \pi}{3} } \\ \displaystyle \large \boxed{ \frac{2 \pi}{3} }$

Find Q3

$\displaystyle \large{ \pi + \frac{ \pi}{3} = \frac{3 \pi}{3} + \frac{ \pi}{3} } \\ \displaystyle \large \boxed{ \frac{4 \pi}{3} }$

Both values are apart of the interval. Hence,

$\displaystyle \large \boxed{ \theta = \frac{2 \pi}{3} , \frac{4 \pi}{3} }$

Second Question

Isolate sin(4 theta).

$\displaystyle \large{sin 4 \theta = - \frac{1}{ \sqrt{2} } }$

Rationalize the denominator.

$\displaystyle \large{sin4 \theta = - \frac{ \sqrt{2} }{2} }$

The problem here is 4 beside theta. What we are going to do is to expand the interval.

$\displaystyle \large{0 \leqslant \theta < 2 \pi}$

Multiply whole by 4.

$\displaystyle \large{0 \times 4 \leqslant \theta \times 4 < 2 \pi \times 4} \\ \displaystyle \large \boxed{0 \leqslant 4 \theta < 8 \pi}$

Then find the reference angle.

We know that sin(π/4) = √2/2. Hence π/4 is our reference angle.

sin is negative in Q3 and Q4. We use π + (ref. angle) for Q3 and 2π - (ref. angle for Q4.)

Find Q3

$\displaystyle \large{ \pi + \frac{ \pi}{4} = \frac{ 4 \pi}{4} + \frac{ \pi}{4} } \\ \displaystyle \large \boxed{ \frac{5 \pi}{4} }$

Find Q4

$\displaystyle \large{2 \pi - \frac{ \pi}{4} = \frac{8 \pi}{4} - \frac{ \pi}{4} } \\ \displaystyle \large \boxed{ \frac{7 \pi}{4} }$

Both values are in [0,2π). However, we exceed our interval to < 8π.

We will be using these following:-

$\displaystyle \large{ \theta + 2 \pi k = \theta \: \: \: \: \: \sf{(k \: \: is \: \: integer)}}$

Hence:-

For Q3

$\displaystyle \large{ \frac{5 \pi}{4} + 2 \pi = \frac{13 \pi}{4} } \\ \displaystyle \large{ \frac{5 \pi}{4} + 4\pi = \frac{21 \pi}{4} } \\ \displaystyle \large{ \frac{5 \pi}{4} + 6\pi = \frac{29 \pi}{4} }$

We cannot use any further k-values (or k cannot be 4 or higher) because it'd be +8π and not in the interval.

For Q4

$\displaystyle \large{ \frac{ 7 \pi}{4} + 2 \pi = \frac{15 \pi}{4} } \\ \displaystyle \large{ \frac{ 7 \pi}{4} + 4 \pi = \frac{23\pi}{4} } \\ \displaystyle \large{ \frac{ 7 \pi}{4} + 6 \pi = \frac{31 \pi}{4} }$

Therefore:-

$\displaystyle \large{4 \theta = \frac{5 \pi}{4} , \frac{7 \pi}{4} , \frac{13\pi}{4} , \frac{21\pi}{4} , \frac{29\pi}{4}, \frac{15 \pi}{4} , \frac{23\pi}{4} , \frac{31\pi}{4} }$

Then we divide all these values by 4.

$\displaystyle \large \boxed{\theta = \frac{5 \pi}{16} , \frac{7 \pi}{16} , \frac{13\pi}{16} , \frac{21\pi}{16} , \frac{29\pi}{16}, \frac{15 \pi}{16} , \frac{23\pi}{16} , \frac{31\pi}{16} }$

Let me know if you have any questions!

3 0

3 years ago

If there are 160 calories in 8 ounces of milk then, how many calories are there in a 15 ounce cup of milk.

jenyasd209 [6]

300, because 8x20=160 so 1 ounces= 20 calories
15x20=300

8 0

3 years ago

Read 2 more answers