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Leya [2.2K]
3 years ago
13

HELP PLEASE! SORRY FOR THE RUSH

Mathematics
2 answers:
Setler79 [48]3 years ago
8 0
3 and 4 are both c as for number five just Google it if you can
Artemon [7]3 years ago
5 0

Answer:  the answer for number 3 might be  Cor D but i think it is D              

Step-by-step explanation:

2 QUATERS = 50

14 DIMES= 140

5PENNIES=5                    

5 dollor bill= 5                      

13NICLES= 65

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let sin(θ) =3/5 and tan(y) =12/5 both angels comes from 2 different right trianglesa)find the third side of the two tringles b)
statuscvo [17]

In a right triangle, we haev some trigonometric relationships between the sides and angles. Given an angle, the ratio between the opposite side to the angle by the hypotenuse is the sine of this angle, therefore, the following statement

\sin (\theta)=\frac{3}{5}

Describes the following triangle

To find the missing length x, we could use the Pythagorean Theorem. The sum of the squares of the legs is equal to the square of the hypotenuse. From this, we have the following equation

x^2+3^2=5^2

Solving for x, we have

\begin{gathered} x^2+3^2=5^2 \\ x^2+9=25 \\ x^2=25-9 \\ x^2=16 \\ x=\sqrt[]{16} \\ x=4 \end{gathered}

The missing length of the first triangle is equal to 4.

For the other triangle, instead of a sine we have a tangent relation. Given an angle in a right triangle, its tanget is equal to the ratio between the opposite side and adjacent side.The following expression

\tan (y)=\frac{12}{5}

Describes the following triangle

Using the Pythagorean Theorem again, we have

5^2+12^2=h^2

Solving for h, we have

\begin{gathered} 5^2+12^2=h^2 \\ 25+144=h^2 \\ 169=h^2 \\ h=\sqrt[]{169} \\ h=13 \end{gathered}

The missing side measure is equal to 13.

Now that we have all sides of both triangles, we can construct any trigonometric relation for those angles.

The sine is the ratio between the opposite side and the hypotenuse, and the cosine is the ratio between the adjacent side and the hypotenuse, therefore, we have the following relations for our angles

\begin{gathered} \sin (\theta)=\frac{3}{5} \\ \cos (\theta)=\frac{4}{5} \\ \sin (y)=\frac{12}{13} \\ \cos (y)=\frac{5}{13} \end{gathered}

To calculate the sine and cosine of the sum

\begin{gathered} \sin (\theta+y) \\ \cos (\theta+y) \end{gathered}

We can use the following identities

\begin{gathered} \sin (A+B)=\sin A\cos B+\cos A\sin B \\ \cos (A+B)=\cos A\cos B-\sin A\sin B \end{gathered}

Using those identities in our problem, we're going to have

\begin{gathered} \sin (\theta+y)=\sin \theta\cos y+\cos \theta\sin y=\frac{3}{5}\cdot\frac{5}{13}+\frac{4}{5}\cdot\frac{12}{13}=\frac{63}{65} \\ \cos (\theta+y)=\cos \theta\cos y-\sin \theta\sin y=\frac{4}{5}\cdot\frac{5}{13}-\frac{3}{5}\cdot\frac{12}{13}=-\frac{16}{65} \end{gathered}

4 0
1 year ago
Solve this linear equation for p.
DedPeter [7]

Answer:

2.6(5.5p - 12.4) = 127.92 \\ 5.5p - 12.4 = 49.2 \\ 5.5p = 61.6 \\ p = 11.2

8 0
3 years ago
Read 2 more answers
Harvey is 3 times as old as Jane. The sum of their ages is 48 years. Find the age of each.
Marta_Voda [28]
Alright, let's answer this -

Half of 48 is 24... 

Half of 24 is 12...

12 three times is 36...

36 + 12 is equal to 48. =)
6 0
3 years ago
Read 2 more answers
Jan receive -22 points on her exam she got 11 questions wrong out of 50 how much was Jan penalized for each wrong answer
AveGali [126]
Jan was penalized -2 points fo every question she got wrong.
8 0
3 years ago
Read 2 more answers
Use the divergence theorem to find the outward flux of the vector field F(x,y,z)=2x2i+5y2j+3z2k across the boundary of the recta
aksik [14]

Answer:

The answer is "120".

Step-by-step explanation:

Given values:

F(x,y,z)=2x^2i+5y^2j+3z^2k \\

differentiate the above value:

div F =2 \frac{x^2i}{\partial x}+5 \frac{y^2j}{\partial y}+3 \frac{z^2k }{\partial z}  \\

        = 4x+10y+6z

\ flu \ of \ x = \int  \int div F dx

              = \int\limits^1_0 \int\limits^3_0 \int\limits^1_0 {(4x+10y+6z)} \, dx \, dy \, dz \\\\ = \int\limits^1_0 \int\limits^3_0 {(4xz+10yz+6z^2)}^{1}_{0} \, dx \, dy  \\\\ = \int\limits^1_0 \int\limits^3_0 {(4x+10y+6)} \, dx \, dy  \\\\ = \int\limits^1_0  {(4xy+10y^2+6y)}^3_{0} \, dx   \\\\ = \int\limits^1_0  {(12x+90+18)}\, dx   \\\\= {(12x^2+90x+18x)}^{1}_{0}   \\\\= {(12+90+18)}   \\\\=30+90\\\\= 120

6 0
3 years ago
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