Is this equation an identity?<br><br> z = -10 − z

Please help asap! Worth 15 points

sdas [7]

Answer: the second one

Step-by-step explanation:

3 0

3 years ago

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find two positive even consecutive integers such that a square of the smaller integer is 10 more than the larger integer

nalin [4]

Answer:

the two positive consecutive integers are 4 and 6.

Step-by-step explanation:

Let the smaller integer be s; then s^2 = (s + 2) + 10.

Simplifying, s^2 - s - 2 - 10 = 0, or

s^2 - s - 12 = 0.

Solve this by factoring: (s - 4)(s + 3) = 0.

Then s = 4 and s = -3.

If the first even integer is 4, the next is 6. We omit s = -3 because it's not even.

The smaller integer is 4. Does this satisfy the equation s^2 = (s + 2) + 10?

4^2 = (4 + 2) + 10 True or False?

16 = 6 + 10 = 16.

True.

So the two positive consecutive integers are 4 and 6.

6 0

3 years ago

Solve for c.<br><br> 5c+16.5=13.5+10c

Kruka [31]

Answer:

c = 3/5

Step-by-step explanation:

5c+16.5=13.5+10c

Subtract 5c from each side

5c-5c+16.5=13.5+10c-5c

16.5 = 13.5 +5c

Subtract 13.5 from each side

16.5 -13.5 =13.5-13.5 +5c

3 = 5c

Divide by 5

3/5 = 5c/5

3/5 =c

4 0

3 years ago

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Solve. 3(x+1)-2x=-6

RideAnS [48]

D.) x = - 9

C.) x = 0

Hope that helps, Good luck! (:

4 0

3 years ago

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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Is this equation an identity? z = -10 − z