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3 years ago

Please help me out with these questions! Brainliest + 50 Points

Mathematics

2 answers:

11Alexandr11 [23.1K]3 years ago

8 0

Step-by-step explanation:

1) B

2) A

3) C

4) C

yeahhh that's about right

sveticcg [70]3 years ago

4 0

Answer:

1) B

2) B

3) C

4) D

5) A (I think )

Step-by-step explanation:

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If two lines are parallel and one has a slope of 8, what is the slope of the other line

tresset_1 [31]

Answer: 8

Parallel lines always have the same slope.

[I attached a picture that might help show you]

The lines do NOT Intersect.

6 0

2 years ago

You purchase a lottery ticket for $2 from a grocery store. There are a total of 100,000 tickets in circulation. The grand prize

FrozenT [24]

I'm pretty sure its B but I would double check before you put the answer in :)

8 0

2 years ago

35 solve the following syste

blondinia [14]

Part A: The solution is $(-0.923,5.692)$

Part B: The point $(3,7)$ is not in the solution set.

Explanation:

Part A: The given inequalities are $3 x+4 y>20$ and $x$

The solution can be determined by solving the two inequalities by substitution method.

Changing inequalities to equality, we have,

$x=3 y-18$ and $3 x+4 y=20$

Let us substitute $x=3 y-18$ in the equation $3 x+4 y=20$ , we get,

$3 (3y-18)+4 y=20$

$9y-54+4y=20$

$13y=74$

$y=5.692$

Substituting $y=5.692$ in $x=3 y-18$ , we get,

$x=3 (5.692)-18$

$=17.076-18$

$x=-0.923$

Thus, the solution set is $(-0.923,5.692)$

Part B: Now, we shall determine whether the point $(3,7)$ is in the solution set.

Let us substitute the point $(3,7)$ in the inequalities $3 x+4 y>20$ and $x$ , we get,

$3 (3)+4 (7)>20$

$9+28>20$

$37>20$

Also, substituting $(3,7)$ in $x$ , we get,

$3$

Since, the point $(3,7)$ does not satisfy one of the inequality $x$ , the solution set does not contain the point $(3,7)$

Thus, the point $(3,7)$ is not in the solution set.

8 0

2 years ago

PLS HELP ILL MARK BRAINLIST !!

andreev551 [17]

Answer:

Last option it should be

6 0

2 years ago

Read 2 more answers

Please help me now ! Thank you

jarptica [38.1K]

First we'll do two basic steps. Step 1 is to subtract 18 from both sides. After that, divide both sides by 2 to get x^2 all by itself. Let's do those two steps now

2x^2+18 = 10
2x^2+18-18 = 10-18 <<--- step 1
2x^2 = -8
(2x^2)/2 = -8/2 <<--- step 2
x^2 = -4

At this point, it should be fairly clear there are no solutions. How can we tell? By remembering that x^2 is never negative as long as x is real.

Using the rule that negative times negative is a positive value, it is impossible to square a real numbered value and get a negative result.

For example
2^2 = 2*2 = 4
8^2 = 8*8 = 64
(-10)^2 = (-10)*(-10) = 100
(-14)^2 = (-14)*(-14) = 196

No matter what value we pick, the result is positive. The only exception is that 0^2 = 0 is neither positive nor negative.

So x^2 = -4 has no real solutions. Taking the square root of both sides leads to

x^2 = -4
sqrt(x^2) = sqrt(-4)
|x| = sqrt(4)*sqrt(-1)
|x| = 2*i
x = 2i or x = -2i
which are complex non-real values

5 0

3 years ago