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3 years ago

What is the solution to this system of linear equations?

Mathematics

1 answer:

OLEGan [10]3 years ago

3 0

Answer: To solve a system of linear equations graphically we graph both equations in the same coordinate system. The solution to the system will be in the point where the two lines intersect. The two lines intersect in (-3, -4) which is the solution to this system of equations.

Step-by-step explanation: hope this helps

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3-X²=-7 How do I solve

irga5000 [103]

Answer:

x = 3.162

Step-by-step explanation:

Your first step is to get x by itself by subtracting 3 from both sides.

3- $x^{2}$ = -7

- $x^{2}$ = -10

Now multiply by both sidex by -1 to make x positive.

$x^{2}$ = 10

now take the square root of both sides.

$\sqrt {x^{2}$ = $\sqrt{10}$

x = $\sqrt{10}$

x = 3.162

8 0

3 years ago

Solve for x: 22x + 54 = -20 + 60x please help STAT :(

Doss [256]

I think that x = 37/19

22x+54=−20+60x

Step 1: Simplify both sides of the equation.

22x+54=60x−20

Step 2: Subtract 60x from both sides.

22x+54−60x=60x−20−60x

−38x+54=−20

Step 3: Subtract 54 from both sides.

−38x+54−54=−20−54

−38x=−74

Step 4: Divide both sides by -38.

−38x/ −38 = −74/−38

x=37/19

8 0

3 years ago

Read 2 more answers

This pyramid has the same base as the prism and its height is three times the height of the prism. What is the ratio of the volu

oee [108]

Answer:c

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Helppp pleaseeee!!!!

Svetradugi [14.3K]

Answer:

the answer is b :)

Step-by-step explanation:

4 0

2 years ago

The usher at a wedding asked each of the 80 guests whether they were a friend of the bride or of the groom. Find the probability

yawa3891 [41]

Complete Question

The usher at a wedding asked each of the 80 guests whether they were a friend of the bride or of the groom. The results are: 59 for Bride, 50 for Groom, 30 for both. Find the probability that a randomly selected person from this sample was a friend of the bride OR of the groom.

Answer:

0.9875

Step-by-step explanation:

Total Number of Guests which forms the Sample Space, n(S)=80

Let the Event (a friend of the bride) =A

Let the Event (a friend of the groom) =B

n(A) =59

n(B)=50

Friends of both bride and groom, $n(A \cap B)=30$

Therefore:

$n(A \cup B)=n(A)+n(B)-n(A \cap B)\\n(A \cup B)=59+50-30\\n(A \cup B)=79$

The number of Guests who was a friend of the bride OR of the groom = 79

Therefore:

The probability that a randomly selected person from this sample was a friend of the bride OR of the groom.

$P(A \cup B) =\dfrac{n(A \cup B) }{n(S)} \\\\=\dfrac{79 }{80}\\\\=0.9875$

4 0

3 years ago