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3 years ago

When you graph a system of two linear equations, which outcome is NOT possible?

Mathematics

1 answer:

Bumek [7]3 years ago

4 0

A. The lines do not intersect, so there are no solutions.

This is true, if the line's never intersect and don't have a common/same point, there are no solutions.

B. The lines intersect at one point, so the system has one solution.

This is true because they have 1 same point so they have 1 solution

C. The lines intersect in two different points, so the system has two solutions.

This is not possible, straight lines can only intersect 1 time. If the lines are curved, then they can intersect more than once

D. The graphs of the two equations are the same line, so there are infinitely many solutions.

This is true because they will <u>always</u> have the same points, so they will have an infinite number of solutions

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Does anyone know how to solve this?

Bezzdna [24]

Use this version of the Law of Cosines to find side b:

b^2 = a^2 + c^2 − 2ac cos(B)

We want side b.

b^2 = (41)^2 + (20)^2 - 2(41)(20)cos(36°)

After finding b, you can use the Law of Sines to find angles A and C or use other forms of the Law of Cosines to find angles A and C.

Try it....

6 0

3 years ago

What number is 35% of 600

emmainna [20.7K]

35% of 600 = .35 x 600 = 210

4 0

3 years ago

There are three power plants [X, Y, Z] that at any given time each one either generates electricity or idles. Event A is that pl

insens350 [35]

We're told that

$P(A\cap B)=0.15$

$P(A\cup B)^C=0.06\implies P(A\cup B)=0.94$

$P(B\mid A)=P(B^C\mid A)=0.5$

where the last fact is due to the law of total probability:

$P(A)=P(A\cap B)+P(A\cap B^C)$

$\implies P(A)=P(B\mid A)P(A)+P(B^C\mid A)P(A)$

$\implies 1=P(B\mid A)+P(B^C\mid A)$

so that $B\mid A$ and $B^C\mid A$ are complementary.

By definition of conditional probability, we have

$P(B\mid A)=P(B^C\mid A)$

$\implies\dfrac{P(A\cap B)}{P(A)}=\dfrac{P(A\cap B^C)}{P(A)}$

$\implies P(A\cap B)=P(A\cap B^C)$

We make use of the addition rule and complementary probabilities to rewrite this as

$P(A\cap B)=P(A\cap B^C)$

$\implies P(A)+P(B)-P(A\cup B)=P(A)+P(B^C)-P(A\cup B^C)$

$\implies P(B)-[1-P(A\cup B)^C]=[1-P(B)]-P(A\cup B^C)$

$\implies2P(B)=2-[P(A\cup B)^C+P(A\cup B^C)]$

$\implies2P(B)=[1-P(A\cup B)^C]+[1-P(A\cup B^C)]$

$\implies2P(B)=P(A\cup B)+P(A\cup B^C)^C$

$\implies2P(B)=P(A\cup B)+P(A^C\cap B)\quad(*)$

By the law of total probability,

$P(B)=P(A\cap B)+P(A^C\cap B)$

$\implies P(A^C\cap B)=P(B)-P(A\cap B)$

and substituting this into $(*)$ gives

$2P(B)=P(A\cup B)+[P(B)-P(A\cap B)]$

$\implies P(B)=P(A\cup B)-P(A\cap B)$

$\implies P(B)=0.94-0.15=\boxed{0.79}$

8 0

3 years ago

Paige use the equation y=kx to determine she would need 259 beads to create 7 necklaces. how many beads would she use per neckla

frozen [14]

Divide 259 by 7 (259/7)= 37 beads per necklace. If you wanted to check just multiply 7 and 37 (37x7)=259

5 0

3 years ago

Read 2 more answers

Special right triangles

Korolek [52]

Answer:

see down

Step-by-step explanation:

$cos 60 = \frac{x}{6\sqrt{3} } \\x = 6\sqrt{3} * cos60\\x = 6\sqrt{3} * \frac{1}{2} \\x = 3\sqrt{3}$

5 0

3 years ago