Find the measures of the supplementary angles if their measures are in ratio of 11:25

2 1/2 cases of soda to split between 5 families

Charra [1.4K]

Answer:

OK SO A Which polynomial is prime?

7x2 – 35x + 2x – 10

9x3 + 11x2 + 3x – 33

10x3 – 15x2 + 8x – 12

12x4 + 42x2 + 4x2 + 14

Step-by-step explanation:

Which polynomial is prime?

7x2 – 35x + 2x – 10

9x3 + 11x2 + 3x – 33

10x3 – 15x2 + 8x – 12

12x4 + 42x2 + 4x2 + 14

5 0

3 years ago

What is t+1/5=t+7/8? What is the value of t? There should be 2 values and the slashes are fractions

muminat

There is NO solution.

Cancel t on both sides (1/5 = 7/8).

Since 1/5 = 7/8, that doesn't make any sense so that is inapplicable ( So NO answer).

4 0

3 years ago

to solve the system of equations below, Becca isolated x^2 in the firs equation and then substituted it into the second equation

bija089 [108]

The resulting equation if Becca isolated x² in the first equation and then substituted it into the second equation is (9-y²) / 25 - y²/36 = 1

<h3>Equation</h3>

x² + y² = 9

x²/25 - y²/36 = 1

From (1)

x² = 9 - y²

substitute x² = 9 - y² into (2)

x²/25 - y²/36 = 1

(9-y²) / 25 - y²/36 = 1

Therefore, the resulting equation is option C; (9-y²) / 25 - y²/36 = 1

Learn more about equation:

brainly.com/question/4172455

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4 0

1 year 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