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3 years ago

10

Sin A Cot A = Cos A, Cos A CosecA= Cot A, Sec A Cot A=CosecA, Cosec A Tan A =Sec A their r total 4 question. I will give brainli

est.

1 answer:

EastWind [94]3 years ago

7 0

Hey mate hope its help you.......

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For ΔABC, ∠A = 2x - 2, ∠B = 2x + 2, and ∠C = 5x. If ΔABC undergoes a dilation by a scale factor of 1 2 to create ΔA'B'C' with ∠A

Zolol [24]

<h3>Answer:</h3>

A) ∠A = ∠A' = 38° and ∠B = ∠B' = 42°

<h3>Explanation:</h3>

The sum of angles in ∆ABC is 180°, so ...

... (2x -2) + (2x +2) + (5x) = 180

... 9x = 180

... x = 20

and the angles of ∆ABC are ∠A = 38°, ∠B = 42°, ∠C = 100°.

___

The sum of angles of ∆A'B'C' is 180°, so ...

... (58 -x) +(3x -18) +(120 -x) = 180

... x +160 = 180

... x = 20

and ∠A' = 38°, ∠B' = 42°, ∠C' = 100°.

_____

The values of angle measures of ∆ABC match those of ∆A'B'C', so we can conclude ...

... A) ∠A = ∠A' = 38° and ∠B = ∠B' = 42°

8 0

3 years ago

Compound interest

aev [14]

9514 1404 393

Answer:

4254.31

Step-by-step explanation:

The compound interest multiplier is ...

m = (1 +r/n)^(nt) . . . . annual rate r compounded n times per year, t years

For 11% compounded quarterly for 18 years, the multiplier is ...

m = (1 +0.11/4)^(4·18) = 1.0275^72 ≈ 7.0516671

If 30,000 is the future value, then the present value is ...

PV = FV/m = 30,000/7.0516671

PV ≈ 4254.31

3 0

2 years ago

A water well is to be drilled in the desert where the soil is either rock, clay or sand. The probability of rock P(R)equals=0.

GaryK [48]

The tree diagram for the probability is shown below

P(Clay|Positive) is read 'Probability of Clay given the result is Positive'.

This is a case of conditional probability.

The formula for conditional probability is given as
P(Clay|Positive) = P(Clay∩Positive) ÷ P(Positive)

P(Clay∩Positive) = 0.21×0.48 = 0.1008

P(Positive) = P(Rock∩Positive) + P(Clay∩Positive) + P(Sand∩Positive)
P(Positive) = (0.53×0.53) + (0.21×0.48) + (0.26×0.75)
P(Positive) = 0.2809 + 0.1008 + 0.195
P(Positive) = 0.5767

Hence,
P(Clay|Positive) = 0.1008÷0.5767 = 0.175 (rounded to 3 decimal place)

5 0

3 years ago

This is a question on my partial fractions homework, but no matter what I try I can't figure it out..

Ierofanga [76]

$\dfrac{x^2+x+1}{(x+1)^2(x+2)}=\dfrac{a_1x+a_0}{(x+1)^2}+\dfrac b{x+2}$
$\implies\dfrac{x^2+x+1}{(x+1)^2(x+2)}=\dfrac{(a_1x+a_0)(x+2)+b(x+1)^2}{(x+1)^2(x+2)}$
$\implies x^2+x+1=(a_1+b)x^2+(2a_1+a_0+2b)x+(2a_0+b)$
$\implies\begin{cases}a_1+b=1\\2a_1+a_0+2b=1\\2a_0+b=1\end{cases}\implies a_1=-2,a_0=-1,b=3$

So you have

$\displaystyle\int_0^2\frac{x^2+x+1}{(x+1)^2(x+2)}\,\mathrm dx=-2\int_0^2\frac x{(x+1)^2}\,\mathrm dx-\int_0^2\frac{\mathrm dx}{(x+1)^2}+3\int_0^2\frac{\mathrm dx}{x+2}$
$=\displaystyle-2\int_1^3\dfrac{x-1}{x^2}\,\mathrm dx-\int_0^2\frac{\mathrm dx}{(x+1)^2}+3\int_0^2\frac{\mathrm dx}{x+2}$

where in the first integral we substitute $x\mapsto x+1$ .

$=\displaystyle-2\int_1^3\left(\frac1x-\frac1{x^2}\right)\,\mathrm dx-\frac1{1+x}\bigg|_{x=0}^{x=2}+3\ln|x+2|\bigg|_{x=0}^{x=2}$
$=-2\left(\ln|x|+\dfrac1x\right)\bigg|_{x=1}^{x=3}-\dfrac23+3(\ln4-\ln2)$
$=-2\left(\ln3+\dfrac13-1\right)-\dfrac23+3\ln2$
$=\dfrac23+\ln\dfrac89$

4 0

3 years ago

Find the equations of the tangents to the curve x=9t2+3, y=6t3+3 that pass through the point (12,9).

Xelga [282]

Answer:

The equation will be " $y=x-3$ ".

Step-by-step explanation:

Given:

Points (12, 9) = (x, y)

⇒ $x=9t^2+3$

then,

$\frac{dy}{dt}=18t$

or,

⇒ $y=6t^3+3$

then,

$\frac{dy}{dt}=18t^2$

⇒ $\frac{dy}{dx}=\frac{18t^2}{18t}$

$=t$

By using the point slope form.

The equation of tangent will be:

⇒ $y-9=1(x-12)$

$y-9=x-12$

$y=x-12+9$

$y=x-3$

4 0

3 years ago