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3 years ago

What is tanA? A. 15/17 B. 8/17 C. 8/15 D. 15/8

Mathematics

2 answers:

musickatia [10]3 years ago

8 0

Answer:

D. 15/8

Step-by-step explanation:

In the diagram above, using trig ratio;

SOH CAH TOA

sinФ = opposite / hypotenuse

cosФ = adjacent / hypotenuse

tanФ = opposite / adjacent

But you need to know how to identify all the side;

In this case, since our major concern is angle A, that is; since we are to find tan of angle A, then this makes AC to become our adjacent, BC to be our opposite and AC will always be the hypotenuse regardless of whether we are to solve for angle A or angle B. But of course, if we are to solve for angle A the adjacent will not be AC anymore, it will definitely change.

But since we are solving for angle A;

adjacent = AC =8

opposite = BC = 15

Applying our trig. formula;

tanФ = opposite / adjacent

tan A = 15/8

Hoochie [10]3 years ago

4 0

Tangent(Angle) = Opposite / adjacent

Tangent(A) = 15 / 8

The answer is D. 15/8

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Consider the implicit differential equation <img src="https://tex.z-dn.net/?f=%2849%20y%5E%7B3%7D%20%2B%2045%20xy%29%20dx%20%2B%

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We're looking for an integrating factor $\mu(x,y)=x^py^q$ such that

$\mu\underbrace{(49y^3+45xy)}_M\,\mathrm dx+\mu\underbrace{(98xy^2+50x^2)}_N\,\mathrm dy=0$

is exact, which would require that

$(\mu M)_y=(\mu N)_x$
$(49x^py^{q+3}+45x^{p+1}y^{q+1})_y=(98x^{p+1}y^{q+2}+50x^{p+2}y^q)_x$
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You can verify that $(\mu M)_y=(\mu N)_x$ if you'd like. With the ODE now exact, we have a solution $F(x,y)=C$ such that

$F_x=\mu M$
$F=\displaystyle\int(49y^3+45xy)x^{5/2}y^4\,\mathrm dx$
$F=10x^{9/2}y^5+14x^{7/2}y^7+f(y)$

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$f'(y)=0$
$\implies f(y)=C$

and so the general solution is

$F(x,y)=10x^{9/2}y^5+14x^{7/2}y^7=C$

8 0

3 years ago

The vertices of the image of GHJK after a translation are given. Choose the vector that describes the translation.

Lady_Fox [76]

Answer:

Option B

Step-by-step explanation:

Let's take a vertex G of quadrilateral GHJK to understand the rule for the translation,

From the graph attached,

Coordinates of G → (-5, 5)

Coordinates of the image point G' after the translation → G'(-5, 5)

Let the point G is translated 'a' units to the right and 'b' units down,

Rule for this translation,

G(x, y) → G'(x + a, y - a)

By this rule,

G(-5, 5) → G'(-5 + a, 5 - a)

-5 + a = -5 ⇒ a = 0

5 - a = 5 ⇒ a = 0

Therefore, vector that defines the translation will be,

<0, 0>

Option B is the the correct option.

8 0

3 years ago