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3 years ago

What is the product in simplest form?

Mathematics

2 answers:

Alinara [238K]3 years ago

5 0

Answer:

1. -10/39

Step-by-step explanation:

-5/12 * 8/13 = -40/156

as it is in the simplest form divide by 4

-10/39

mixas84 [53]3 years ago

3 0

Yes -10/39 is correcto

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A new novel is selling for $21.95. If it is on sale at 20% off, find the amount of the discount and the sale price.

Finger [1]

21.95×.2=4.39
21.95-4.39=17.56

the discount is $4.39
and the total sale price is $17.56

7 0

4 years ago

Write the equation of a line through (-1,-5) parallel to 3x-y=5, and in slope intercept form

MAXImum [283]

Answer:

y =3 x

Step-by-step explanation:

The standard equation of a line in point slope form is expressed as;

y -y0 = m(x-x0)

m is the slope

(x0, y0) is the point on the line

Given the line 3x - y = 5

Get the slope

- y = -3x + 5

y = 3x - 5

Slope m = 3

Substitute m = 3 and (-1, -5) into the given formula

y - (-5) = 3(x-(-1))

y+5 = 3(x+1)

y + 3 = 3x+3

y = 3x + 3 - 3

y = 3x

Hence the equation in slope intercept form is y =3 x

4 0

3 years ago

Raju's father is 26 years younger than raju's grandfather and 29 years older than raju. The sum of the ages of all the three is

Musya8 [376]

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c 135

d jsjbsbdhejeujsvevevdvvdhdhdhdvevvevdvdvevvevegehhehehebevevvehdhrhevd

8 0

3 years ago

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

One and half times a number equals 6

Zinaida [17]

Answer:

1/2 times 1/3

Step-by-step explanation:

7 0

3 years ago