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3 years ago

Which trigonometric ratio is defined as the length of the adjacent leg divided by the length of the hypotenuse?

Mathematics

2 answers:

AfilCa [17]3 years ago

3 0

SOH CAH TOA

Cosine is adjacent over hypotenuse so you're answer is C.

:)))

weqwewe [10]3 years ago

3 0

Answer:

Step-by-step explanation:

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There is 12/15 of a pizza left.How many 3/10 pieces can be made from the leftover pizza?

Katena32 [7]

So 12/15 simplifies to 4/5, and if you multiply both sides by two you have 8/10 so i guess 2 ( and some extra )

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3 years ago

Can somebody pls help me with this question I dont get it

alekssr [168]

Answer:

1. 195=6.50·c

2. 30

Step-by-step explanation:

195=6.5·c

Divide both sides by 6.5

30=c

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3 years ago

A cd usually sells for $12.00. If the cd is 10% off, and sales tax is 5%, what is the total price of the cd, including tax?

Kazeer [188]

Answer:

$11.34 Discounted Price Including Tax

Step-by-step explanation:

CD is $12.00

10 percent off of 12 dollars is:

12 x .10 = $1.2

12 - 1.2 = $10.8 WITHOUT TAX!!

5/100 = 0.05

0.05 x 10.8 = .54 (Tax itself)

10.8 (Discounted price) + .54 (Tax) = $11.34 Discounted Price Including Tax

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3 years ago

Colin ties 5 groups of balloons to the fence. There are 3 orange balloons

maxonik [38]

Answer:

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3 years ago

Read 2 more answers

Find the solution of the problem (1 3. (2 cos x - y sin x)dx + (cos x + sin y)dy=0.

lakkis [162]

Answer:

$2*sin(x)+y*cos(x)-cos(y)=C_1$

Step-by-step explanation:

Let:

$P(x,y)=2*cos(x)-y*sin(x)$

$Q(x,y)=cos(x)+sin(y)$

This is an exact differential equation because:

$\frac{\partial P(x,y)}{\partial y} =-sin(x)$

$\frac{\partial Q(x,y)}{\partial x}=-sin(x)$

With this in mind let's define f(x,y) such that:

$\frac{\partial f(x,y)}{\partial x}=P(x,y)$

and

$\frac{\partial f(x,y)}{\partial y}=Q(x,y)$

So, the solution will be given by f(x,y)=C1, C1=arbitrary constant

Now, integrate $\frac{\partial f(x,y)}{\partial x}$ with respect to x in order to find f(x,y)

$f(x,y)=\int\ 2*cos(x)-y*sin(x)\, dx =2*sin(x)+y*cos(x)+g(y)$

where g(y) is an arbitrary function of y

Let's differentiate f(x,y) with respect to y in order to find g(y):

$\frac{\partial f(x,y)}{\partial y}=\frac{\partial }{\partial y} (2*sin(x)+y*cos(x)+g(y))=cos(x)+\frac{dg(y)}{dy}$

Now, let's replace the previous result into $\frac{\partial f(x,y)}{\partial y}=Q(x,y)$ :

$cos(x)+\frac{dg(y)}{dy}=cos(x)+sin(y)$

Solving for $\frac{dg(y)}{dy}$

$\frac{dg(y)}{dy}=sin(y)$

Integrating both sides with respect to y:

$g(y)=\int\ sin(y) \, dy =-cos(y)$

Replacing this result into f(x,y)

$f(x,y)=2*sin(x)+y*cos(x)-cos(y)$

Finally the solution is f(x,y)=C1 :

$2*sin(x)+y*cos(x)-cos(y)=C_1$

7 0

3 years ago