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2 years ago

Using the side lengths of △PQR and △STU, which angle has a sine ratio of 4/5?

Mathematics

2 answers:

marishachu [46]2 years ago

8 0

Given the two triangles, and the sine ratio formula is equal to:

Sin(x) = O/H

Where x is one angle of the triangle

O is the opposite side of the angle x

H is the hypotenuse of the triangle

To solve this we have to fine O/H = 4/5

So the answer is

Sin (P) = 16/20 = 4/5

<span>So angle P has the sine ratio of 4/5</span>

Olegator [25]2 years ago

8 0

Answer:

It’s <P

Step-by-step explanation:

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2/5 divides by -4/7=?<br> Can I please get a specified answer?

iVinArrow [24]

Answer:

7/10

Step-by-step explanation:

(2/5)%(-4/7)

We will change the second fraction to make multiplication

So it will be 2/5×-7/4

=. -14/20

= -7/10

6 0

2 years ago

The first term of a gemotetric sequence is -15 and the common ratio is 6 What is the twelfth term of the sequence

nikklg [1K]

Answer:

Step-by-step explanation:

a₁₂ = a₁r¹²⁻¹ = -15×6¹¹

8 0

2 years ago

What is a solution to the equation 3 / m + 3 - M / 3 - M equals m^2 + 9 / m^2-9?

Mnenie [13.5K]

Answer: Last option.

Step-by-step explanation:

Given the equation:

$\frac{3}{m+3}-\frac{m}{3-m}=\frac{m^2+9}{m^2-9}$

Follow these steps to solve it:

- Subtract the fractions on the left side of the equation:

$\frac{3(3-m)-m(m+3)}{(m+3)(3-m)}=\frac{m^2+9}{m^2-9}\\\\\frac{9-3m-m^2-3m}{(m+3)(3-m)}=\frac{m^2+9}{m^2-9}\\\\\frac{-m^2-6m+9}{(m+3)(3-m)}=\frac{m^2+9}{m^2-9}$

- Using the Difference of squares formula ( $a^2-b^2=(a+b)(a-b)$ ) we can simplify the denominator of the right side of the equation:

$\frac{-m^2-6m+9}{(m+3)(3-m)}=\frac{m^2+9}{(m+3)(m-3)}$

- Multiply both sides of the equation by $(m+3)(3-m)$ and simplify:

$\frac{(-m^2-6m+9)(m+3)(3-m)}{(m+3)(3-m)}=\frac{(m^2+9)(m+3)(3-m)}{(m+3)(m-3)}\\\\-m^2-6m+9=\frac{(m^2+9)(3-m)}{(m-3)}$

- Multiply both sides by $m-3$ :

$(-m^2-6m+9)(m-3)=\frac{(m^2+9)(3-m)(m-3)}{(m-3)}\\\\(-m^2-6m+9)(m-3)=(m^2+9)(3-m)$

- Apply Distributive property and simplify:

$(-m^2-6m+9)(m-3)=(m^2+9)(3-m)\\\\-m^3-6m^2+9m+3m^2+18m-27=3m^2+27-m^3-9m\\\\-m^3-3m^2+27m-27+m^3-3m^2+9m-27=0\\\\-6m^2+36m-54=0$

- Divide both sides of the equation by -6:

$\frac{-6m^2+36m-54}{-6}=\frac{0}{-6}\\\\m^2-6m+9=0$

- Factor the equation and solve for "m":

$(m-3)^2=0\\\\m=3$

In order to verify it, you must substitute $m=3$ into the equation and solve it:

$\frac{3}{3+3}-\frac{3}{3-3}=\frac{3^2+9}{3^2-9}\\\\\frac{3}{6}-\frac{3}{0}=\frac{18}{0}$

<em>NO SOLUTION</em>

7 0

2 years ago

HELP PLEASE PLEASE!!!

12345 [234]

Answer:

-1

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

An equation is shown below. 12 - 9 + c = 12 What value of c makes the equation true?

kvasek [131]

<h3>♫ - - - - - - - - - - - - - - - ~Hello There!~ - - - - - - - - - - - - - - - ♫</h3>

➷ 12 - 9 = 3

3 + c = 12

c = 12 - 3

c = 9

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

↬ ʜᴀɴɴᴀʜ ♡

7 0

2 years ago

Read 2 more answers