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2 years ago

Which statements are true about exponential functions? Check all that apply.

Mathematics

1 answer:

Alika [10]2 years ago

3 0

Domain is all real numbers
The input to an exponential function is the exponent
The base represents the multiplicative rate of change

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Suppose that θ is an acute angle of a right triangle and that sec(θ)=52. Find cos(θ) and csc(θ).

insens350 [35]

Answer:

$\cos{\theta} = \dfrac{1}{52}$

$\csc{\theta} = \dfrac{52}{\sqrt{2703}}$

Step-by-step explanation:

To solve this question we're going to use trigonometric identities and good ol' Pythagoras theorem.

a) Firstly, sec(θ)=52. we're gonna convert this to cos(θ) using:

$\sec{\theta} = \dfrac{1}{\cos{\theta}}$

we can substitute the value of sec(θ) in this equation:

$52 = \dfrac{1}{\cos{\theta}}$

and solve for for cos(θ)

$\cos{\theta} = \dfrac{1}{52}$

side note: just to confirm we can find the value of θ and verify that is indeed an acute angle by $\theta = \arccos{\left(\dfrac{1}{52}\right)} = 88.8^\circ$

b) since right triangle is mentioned in the question. We can use:

$\cos{\theta} = \dfrac{\text{adj}}{\text{hyp}}$

we know the value of cos(θ)=1\52. and by comparing the two. we can say that:

length of the adjacent side = 1
length of the hypotenuse = 52

we can find the third side using the Pythagoras theorem.

$(\text{hyp})^2=(\text{adj})^2+(\text{opp})^2$

$(52)^2=(1)^2+(\text{opp})^2$

$\text{opp}=\sqrt{(52)^2-1}$

$\text{opp}=\sqrt{2703}$

length of the opposite side = √(2703) ≈ 51.9904

we can find the sin(θ) using this side:

$\sin{\theta} = \dfrac{\text{opp}}{\text{hyp}}$

$\sin{\theta} = \dfrac{\sqrt{2703}}{52}}$

and since $\csc{\theta} = \dfrac{1}{\sin{\theta}}$

$\csc{\theta} = \dfrac{52}{\sqrt{2703}}$

4 0

3 years ago

5m^{2}\sqrt{25m^{8}+15m^{6}+5m^{2}}

xxTIMURxx [149]

You can square the whole problem to cancel out the square root. Might make things easier.

5 0

3 years ago

A fund earns a nominal rate of interest of 6\% compounded every two years. Calculate the amount that must be contributed now to

Nesterboy [21]

Answer:

$712.

Step-by-step explanation:

We have been given that a fund earns a nominal rate of interest of 6% compounded every two years. We are asked to find the amount that must be contributed now to have 1000 at the end of six years.

We will use compound interest formula to solve our given problem.

$A=P(1+\frac{r}{n})^{nt}$ , where,

A = Final amount,

P = Principal amount,

r = Annual interest rate in decimal form,

n = Number of times interest is compounded per year,

t = Time in years.

$6\%=\frac{6}{100}=0.06$

Since interest is compounded each two years, so number of compounding per year would be 1/2 or 0.5.

$1000=P(1+\frac{0.06}{0.5})^{0.5*6}$

$1000=P(1+0.12)^{3}$

$1000=P(1.12)^{3}$

$1000=P*1.404928$

$\frac{1000}{1.404928}=\frac{P*1.404928}{1.404928}$

$P=711.7802478$

$P\approx 712$

Therefore, an amount of $712 must be contributed now to have 1000 at the end of six years.

7 0

3 years ago

Could someone help me out?

Brums [2.3K]

Answer:

y = 120°

Step-by-step explanation:

y = 120° because opposite angles of two intersecting lines are equal

7 0

3 years ago

What is the sum of all the integers between 19 and 77

victus00 [196]

Think of it this way: Lets add numbers in pairs, starting at the very outer 2 numbers (19 and 77) then go in by one and add the second number and the second to last (20 and 76), then (21 and 75) and so on. The sum of all of these pairs are all the same: 96. How many 96s will we have? Well since we're coming from each end toward the middle adding pairs we will have half the distance between 19 and 77, that is (77-19)/2 = 29. So we can actually just take 96*29 = 2784. This is the sum of all numbers between 19 and 77

7 0

3 years ago

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