What is the solution to the linear equation?

In the accompanying diagram of right triangle ABC, the hypotenuse if AB, AC = 3, BC = 4, and AB = 5. Sin B is equal to

Trava [24]

Answer: $sin(B)=\frac{3}{5}$

Step-by-step explanation:

We are given that a triangle ABC is a Right Angled Triangle. The side AB is hypotenuse, so the angle opposite to side AB which will be angle C is a Right Angle (measures 90 degrees)

We have the side length of all 3 sides. Based on this information, we can construct a triangle with given measures. The triangle is shown in the attached image.

We have to find the value of Sin(B). Sin of an angle is defined as:

$sin(\theta)=\frac{\text{Opposite Side}}{\text{Hypotenuse}}$

The side opposite to angle B is AC with a length of 3 and hypotenuse is side AB with length 5. So Sin of angle B would be:

$sin(B)=\frac{3}{5}$

8 0

3 years ago

I need help , I don’t understand this

marta [7]

#2. First, we factor each polynomial. Then, if any terms on both the top and the bottom of the fraction match, they cancel out. So... we do just that. You end up with:

$\frac{x(x-4)}{(x+9)(x-4)}$

Notice there's an (x-4) on both top and bottom. So they cancel out. That leaves us with your answer of $\frac{x}{(x+9)}$

#3. We do the same thing as above then multiply and simplify. In the interest of space, I'll cut straight to some simplification.

$\frac{2(x+2)^{3} }{6x(x+2)} ( \frac{5}{(x-2)^{2} } )$

Now we start cancelling. For the first fraction, there are 3 (x+2)'s on top and 1 on the bottom so we will cancel out the one on the bottom and leave 2 (x+2)'s on top. There are no more polynomials to cancel out so now we multiply across:

$\frac{10(x+2)^{2} }{6x(x-2)^{2} }$

10 and 6 share a GCF of 2 so we divide both of those by 2. This leaves us with the final answer of:

$\frac{5(x+2)^{2} }{3x(x-2)^{2} }$

#4. This equation introduces division and because of it, we must flip the second fraction to make the division sign into a multiplication symbol. Again for space, I'll flip the fraction and simplify in one step.

$\frac{3(x+2)(x-2)}{(x+4)(x-2)} ( \frac{x+4}{6(x+3)})$

Now we do our cancelling. First fraction has (x - 2) in the top and bottom. They're gone. The first fraction has a (x + 4) on the bottom and the second fraction has one on the top. Those will also cancel. This leaves you with:

$\frac{3(x+2)}{6(x+3)}$

3 and 6 share a GCF of 3 so we divide both numbers by this. This leaves you with your final answer:

$\frac{x+2}{2(x+3)}$

#5. We are adding so we first factor both fractions and see what we need to multiply by to make the denominators the same. I'll do the former first. (10 - x) and (x - 10) are not the same so we multiply the first equation (top and bottom) by (x - 10) and the second equation by (10 - x). Because they will now have the same denominator we can combine them already. This gives us:

$\frac{(3+2x)(x-10)+(13+x)(10-x)}{(10-x)(x-10)}$

Now we FOIL each to expand and then simplify by combining like terms. Again for space, I'm just showing the result of this; you end up with:

$\frac{x^{2}-20x+100}{(10-x)(x-10)}$

Now we factor the top. This gives you 2 (x - 10)'s on top and one on bottom. So we just leave one on the top and cancel the bottom one out. This leaves you with your answer:

$\frac{x+10}{10-x}$

#6. Same process for this one so I won't repeat. I'll just show the work.

$\frac{3}{(x-3)(x+2)} + \frac{2}{(x-3)(x-2)}$ becomes

$\frac{3(x-2) + 2(x+2)}{(x-3)(x+2)(x-2)}$ which equals

$\frac{3x - 6 + 2x + 4}{(x-3)(x+2)(x-2)}$ giving you the final answer

$\frac{5x - 2}{(x-3)(x+2)(x-2)}$

#7. For this question we find the least common denominator to make the denominators match. For 5, x, and 2x, the LCD is 10x. So we multiply top and bottom of each fraction by what would make the bottom equal 10x. This rewrites the fraction as:

$\frac{3x}{5} ( \frac{2x}{2x}) * ( \frac{5}{x}( \frac{10}{10}) - \frac{5}{2x} ( \frac{5}{5}))$

Simplify to get:

$\frac{3x}{5} * ( \frac{25}{10x})$

After simplifying again, you end up with your final answer:

$\frac{3}{2}$

8 0

3 years ago

Priya noticed in the last activity that between the 2 triangles, you only need to know 4 angles to show that they are similar. S

ValentinkaMS [17]

Answer:

Please find attached a drawing of the triangles ΔRST and EFG showing the angles

The angle on ΔEFG that would prove the triangles are similar is ∠F = 25°

Step-by-step explanation:

In order to prove that two triangles are similar, two known angles of each the triangles need to be shown to be equal

Given that triangle ∠R and ∠S of triangle ΔRST are 95° and 25°, respectively, and that ∠E of ΔEFG is given as 90°, then the corresponding angle on ΔEFG to angle ∠S = 25° which is ∠F should also be 25°

Therefore, the angle on ΔEFG that would prove the triangles are similar is ∠F = 25°.

3 0

3 years ago

A road crew must repave a road that is 3/4 miles long. They can repave 1/48 miles each hour. How long will it take the crew to r

chubhunter [2.5K]

Answer:

36 hours

Step-by-step explanation:

3/4 divided by 1/48

KCF

3/4*48/1= 144/4= 36

3 0

3 years ago

$350, 6.2%, 3 years plzz help

KIM [24]

Answer:

not much info

Step-by-step explanation:

8 0

3 years ago