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3 years ago

In ΔABC, the measure of ∠C=90°, BA = 41, CB = 40, and AC = 9. What ratio represents the cosine of ∠B?

Mathematics

1 answer:

blsea [12.9K]3 years ago

3 0

Answer:

40/41

Step-by-step explanation:

From the given diagram;

BA is the hypotenuse = 41

AC is the opposite = 9

CB is the adjacent = 40

Cos<B = adjaccent/hyp

Cos<B = CB/BA

Cos<B = 40/41

Hence the required ratio is 40/41

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If 6000 dollars is invested in a bank account at an interest rate of 9 per cent per

Volgvan

Answer:

(1) $11955.38

(2) $12228.62

(3) $12293.527

(4) $12326.6

Step-by-step explanation:

Compound interest is given by

A= P( 1 + r/n)^n*t

A= Final amount

P = initial amount = 6000

r = interest rate = 9% = 0.09

n = number of times interest applied per time period

t = number of time periods elapsed

(1) Compounded annually

n = 1 , t = 8

A = 6000( 1+ 0.09/1) ^1*8

A = 6000( 1.09) ^8 = $11955.38

(2) compounded quarterly

n = 4 , t = 8

A = 6000( 1+ 0.09/4) ^4*8

= 6000( 1.0225)^32 = $12228.62

(3) compounded monthly

n = 12 , t = 8

A = 6000( 1+0.09/12)^12*8 = $12293.527

(4) compounded continuously

A = P* e^rt

r = 0.09 , t = 8

= 6000* e^0.09*8

= 6000* e^0.72 = $12326.6

5 0

3 years ago

Read 2 more answers

What is the equation for the locus of points 10 units from the origin?

Lesechka [4]

Answer:

d or b no a tho

Step-by-step explanation:

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3 years ago

Suppose h(x)=3x-2 and j(x) = ax +b. Find a relationship between a and b such that h(j(x)) = j(h(x))

sergejj [24]

Answer:

$\displaystyle a = \frac{1}{3} \text{ and } b = \frac{2}{3}$

Step-by-step explanation:

We can use the definition of inverse functions. Recall that if two functions, f and g are inverses, then:

$\displaystyle f(g(x)) = g(f(x)) = x$

So, we can let j be the inverse function of h.

Function h is given by:

$\displaystyle h(x) = y = 3x-2$

Find its inverse. Flip variables:

$x = 3y - 2$

Solve for y. Add:

$\displaystyle x + 2 = 3y$

Hence:

$\displaystyle h^{-1}(x) = j(x) = \frac{x+2}{3} = \frac{1}{3} x + \frac{2}{3}$

Therefore, a = 1/3 and b = 2/3.

We can verify our solution:

$\displaystyle \begin{aligned} h(j(x)) &= h\left( \frac{1}{3} x + \frac{2}{3}\right) \\ \\ &= 3\left(\frac{1}{3}x + \frac{2}{3}\right) -2 \\ \\ &= (x + 2) -2 \\ \\ &= x \end{aligned}$

And:

$\displaystyle \begin{aligned} j(h(x)) &= j\left(3x-2\right) \\ \\ &= \frac{1}{3}\left( 3x-2\right)+\frac{2}{3} \\ \\ &=\left( x- \frac{2}{3}\right) + \frac{2}{3} \\ \\ &= x \stackrel{\checkmark}{=} x\end{aligned}$