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3 years ago

How many solutions are there to each nonlinear system of equations? How do you know?

Mathematics

1 answer:

Natali5045456 [20]3 years ago

6 0

Answer:

0, 1, 2

Step-by-step explanation:

The solution to a system of equations given graphically is at the intersection of the individual graphs.

Graph (1)

There is no intersection between the 2 graphs so no solution

Graph (2)

There is one point of intersection between the 2 graphs so 1 solution

Graph (3)

There are two points of intersection between the 2 graphs so 2 solutions

You might be interested in

The ratio of the profit, material cost and production labour of an article is 5:7:13.

IRINA_888 [86]

Answer:

The cost of producing the article is -3500 (Negative cost)

Step-by-step explanation:

The given parameters are

The ratio of the profit to material cost to production labor = 5:7:13

The amount of the material cost = 840 + Labor cost

Let the total cost = X

Therefore, we have;

The fraction of the total cost that is material cost = 7/(5 + 7 + 13) = 7/25

Therefore, the material cost = 7/25 × X

The fraction of the total cost that is labor cost = 13/(5 + 7 + 13) = 13/25

Therefore, the labor cost = 13/25 × X

However, the amount of the material cost = 840 + Labor cost, which gives;

7/25 × X = 13/25 × X + 840

7/25 × X - 13/25 × X = 840

-6/25 × X= 840

X = 840/(-6/25) = 840×(-25/6) = -3500

The cost of producing the article = -3500.

8 0

3 years ago

The pH of a substance added to a citrate buffer, y, depending on the ratio of citric acid to sodium citrate, x, can be modeled u

Charra [1.4K]

Answer:

Yes it's A. 0.79

Step-by-step explanation:

You plug in 6.5 into the y-value since it asks to find the ratio,x, if the pHis 6.5. Then you can solve using a calculator to get 0.79432, or 0.79

5 0

3 years ago

Read 2 more answers

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.

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

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.

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

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