Is anyone good at algebra 2? if so do u have insta or snapcl?

Which of the g-values satisfy the following inequality? 7≥12−g

sergejj [24]

Answer:

g=6

Step-by-step explanation:

g=4 and g=5 will be more than 7 but g=6 is less.

4 0

2 years ago

10/2x5 using order of operations

Jlenok [28]

Answer:

25

Step-by-step explanation:

10 / 2 = 5 x 5 = 25

7 0

2 years ago

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Please help with any of this Im stuck and having trouble with pre calc is it basic triogmetric identities using quotient and rec

german

How I was taught all of these problems is in terms of r, x, and y. Where sin = y/r, cos = x/r, tan = y/x, csc = r/y, sec = r/x, cot = x/y. That is how I will designate all of the specific pieces in each problem.

#3

Let's start with sin here. $\frac{2\sqrt{5}}{5} = \frac{2}{\sqrt{5}}$ Therefore, because sin is y/r, r = $\sqrt{5}$ and y = +2. Moving over to cot, which is x/y, x = -1, and y = 2. We know y has to be positive because it is positive in our given value of sin. Now, to find cos, we have to do x/r.

cos = $\frac{-1}{\sqrt{5}} = \frac{-\sqrt{5}}{5}$

#4

Let's start with secant here. Secant is r/x, where r (the length value/hypotenuse) cannot be negative. So, r = 9 and x = -7. Moving over to tan, x must still equal -7, and y = $4\sqrt{2}$ . Now, to find csc, we have to do r/y.

csc = $\frac{9}{4\sqrt{2}} = \frac{9\sqrt{2}}{8}$

The pythagorean identities are

sin^2 + cos^2 = 1,

1 + cot^2 = csc^2,

tan^2 + 1 = sec^2.

#5

Let's take a look at the information given here. We know that cos = -3/4, and sin (the y value), must be greater than 0. To find sin, we can use the first pythagorean identity.

sin^2 + (-3/4)^2 = 1

sin^2 + 9/16 = 1

sin^2 = 7/16

sin = $\sqrt{7/16}$ = $\frac{\sqrt{7}}{4}$

Now to find tan using a pythagorean identity, we'll first need to find sec. sec is the inverse/reciprocal of cos, so therefore sec = -4/3. Now, we can use the third trigonometric identity to find tan, just as we did for sin. And, since we know that our y value is positive, and our x value is negative, tan will be negative.

tan^2 + 1 = (-4/3)^2

tan^2 + 1 = 16/9

tan^2 = 7/9

tan = - $\sqrt{7/9} = \frac{-\sqrt{7}}{3}$

#6

Let's take a look at the information given here. If we know that csc is negative, then our y value must also be negative (r will never be negative). So, if cot must be positive, then our x value must also be negative (a negative divided by a negative makes a positive). Let's use the second pythagorean identity to solve for cot.

1 + cot^2 = $(\frac{-\sqrt{6}}{2})^{2}$

1 + cot^2 = 6/4

cot^2 = 2/4

cot = $\frac{\sqrt{2}}{2}$

tan = $\sqrt{2}$

Next, we can use the third trigonometric identity to solve for sec. Remember that we can get tan from cot, and cos from sec. And, from what we determined in the beginning, sec/cos will be negative.

$(\frac{2}{\sqrt{2}})^2$ + 1 = sec^2

4/2 + 1 = sec^2

2 + 1 = sec^2

sec^2 = 3

sec = - $\sqrt{3}$

cos = $\frac{-\sqrt{3}}{3}$

Hope this helps!! :)

3 0

2 years ago

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In January, you deposit $150 into your checking account. Every month after that, you deposit 10% more than you did the month bef

Reptile [31]

This is a geometric sequence of the form:

a(n)=ar^(n-1) in this case a=150, r=1.1

a(n)=150(1.1)^(n-1) August is the eighth month so

a(8)=150(1.1)^7

a(8)=292.307565

So you deposited $292.31 (to the nearest cent) in August.

5 0

2 years ago

For the month of January in a certain city, 42% of the days are cloudy. Also in the month of January in the same city, 41% o

zalisa [80]

Answer: 0.9762

Step-by-step explanation:

Let A be the event that days are cloudy and B be the event that days are rainy for January month .

Given : The probability that the days are cloudy = $P(A)=0.42$

The probability that the days are cloudy and rainy = $P(A\cap B)=0.41$

Now, the conditional probability that a randomly selected day in January will be rainy if it is cloudy is given by :-

$P(B|A)=\dfrac{P(B\cap A)}{P(A)}\\\\\Rightarrow\ P(B|A)=\dfrac{0.41}{0.42}=0.97619047619\approx0.9762$

Hence, the probability that a randomly selected day in January will be rainy if it is cloudy = 0.9762

7 0

3 years ago