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svp [43]
3 years ago
5

Solve using elimination -3-X=0 -3y=-3-2x

Mathematics
1 answer:
ryzh [129]3 years ago
7 0
<h2>The solution is (x, y) = (-3, -1)</h2>

<em><u>Solution:</u></em>

<em><u>Given, system of equations are:</u></em>

-3 - x = 0 --------- eqn 1

-3y = -3 - 2x -------- eqn 2

From eqn 1,

- 3 - x = 0

<h3>x = -3</h3>

<em><u>Substitute x = -3 in eqn 2</u></em>

-3y = - 3 - 2(-3)

-3y = -3 + 6

-3y = 3

<h3>y = -1</h3>

Thus the solution is (x, y) = (-3, -1)

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20x^2 -11x -4 = 0 what are the solutions for this equation
mezya [45]
X= 4/5, - 1/4
Hope it helps
8 0
3 years ago
In ΔABC (m∠C = 90°), the points D and E are the points where the angle bisectors of ∠A and ∠B intersect respectively sides BC an
Airida [17]

This is a little long, but it gets you there.

  1. ΔEBH ≅ ΔEBC . . . . HA theorem
  2. EH ≅ EC . . . . . . . . . CPCTC
  3. ∠ECH ≅ ∠EHC . . . base angles of isosceles ΔEHC
  4. ΔAHE ~ ΔDGB ~ ΔACB . . . . AA similarity
  5. ∠AEH ≅ ∠ABC . . . corresponding angles of similar triangle
  6. ∠AEH = ∠ECH + ∠EHC = 2∠ECH . . . external angle is equal to the sum of opposite internal angles (of ΔECH)
  7. ΔDAC ≅ ΔDAG . . . HA theorem
  8. DC ≅ DG . . . . . . . . . CPCTC
  9. ∠DCG ≅ ∠DGC . . . base angles of isosceles ΔDGC
  10. ∠BDG ≅ ∠BAC . . . .corresponding angles of similar triangles
  11. ∠BDG = ∠DCG + ∠DGC = 2∠DCG . . . external angle is equal to the sum of opposite internal angles (of ΔDCG)
  12. ∠BAC + ∠ACB + ∠ABC = 180° . . . . sum of angles of a triangle
  13. (∠BAC)/2 + (∠ACB)/2 + (∠ABC)/2 = 90° . . . . division property of equality (divide equation of 12 by 2)
  14. ∠DCG + 45° + ∠ECH = 90° . . . . substitute (∠BAC)/2 = (∠BDG)/2 = ∠DCG (from 10 and 11); substitute (∠ABC)/2 = (∠AEH)/2 = ∠ECH (from 5 and 6)
  15. This equation represents the sum of angles at point C: ∠DCG + ∠HCG + ∠ECH = 90°, ∴ ∠HCG = 45° . . . . subtraction property of equality, transitive property of equality. (Subtract ∠DCG+∠ECH from both equations (14 and 15).)
5 0
3 years ago
Which is not a step in finding the scale factor of dilation
mariarad [96]

1)To find a Scale factor of Dilation, about the origin

We have

Pre-mage Image

(x, y) k(x,y)

For example

Pre-image Image

(2,4) 2(2,4) = (4,8)

2)When it's not about the origin then we have to count from the Projection Point

Having said this, ex

Which is not a step

a)

We can divide the x value of the image over the pre-image, not the way around.

In the example, I've given if we divide the pre-image over the image value we would have found a scale factor of 1/2. In the example, The scale factor was the inverse: 2

5 0
1 year ago
A player of the National Basketball Association’s Portland Trail Blazers is the best free-throw shooter on the team, making 94%
g100num [7]

Answer:

The data for the probabilities are shown in the table below.

- A represents the probability of making the two shots for each of the best and worst shooter on the Portland Trail Blazers' team

- B represents the probability of making at least one shot for each of the best and worst shooter on the Portland Trail Blazers' team.

- C represents the probability of not making any of the two shots for each of the best and worst shooter on the Portland Trail Blazers' team.

N | Best ||| Worst

A | 0.8836 | 0.3136

B | 0.9964 | 0.8064

C | 0.0036 | 0.1936

It becomes evident why fouling the worst shooter on the team is a better tactic. The probabilities of the best shooter making the basket over the range of those two free shots are way better than the chances for the worst shooter.

Step-by-step explanation:

Part 1

Probability of the best shooter of the National Basketball Association’s Portland Trail Blazers making a shot = P(B) = 94% = 0.94

Probability that he doesn't make a shot = P(B') = 1 - 0.94 = 0.06

a) Probability that the best shooter on the team makes the two shots awarded = P(B) × P(B) = 0.94 × 0.94 = 0.8836

b) Probability that the best shooter on the team makes at least one shot.

This is a sum of probabilities that he makes only one shot and that he makes two shots.

Probability that he makes only one shot

= P(B) × P(B') + P(B') + P(B)

= (0.94 × 0.06) + (0.06 × 0.94) = 0.1128

Probability that he makes two shots = 0.8836 (already calculated in part a)

Probability that he makes at least one shot = 0.1128 + 0.8836 = 0.9964

c) Probability that the best shooter on the team makes none of the two shots = P(B') × P(B') = 0.06 × 0.06 = 0.0036

d) If the worst shooter on the team, whose success rate is 56% is now fouled to take the two shots.

Probability of the worst shooter on the team making a shot = P(W) = 56% = 0.56

Probability that the worst shooter on the team misses a shot = P(W') = 1 - 0.56 = 0.44

Part 2

a) Probability that the worst shooter on the team makes the two shots = P(W) × P(W)

= 0.56 × 0.56 = 0.3136

b) Probability that the worst shooter on the team makes at least one shot.

This is a sum of probabilities that he makes only one shot and that he makes two shots.

Probability that he makes only one shot

= P(W) × P(W') + P(W') + P(W)

= (0.56 × 0.44) + (0.44 × 0.56) = 0.4928

Probability that he makes two shots = 0.3136 (already calculated in part a)

Probability that he makes at least one shot = 0.4928 + 0.3136 = 0.8064

c) Probability that the worst shooter makes none of the two shots = P(W') × P(W') = 0.06 × 0.06 = 0.1936

From the probabilities obtained

N | Best ||| Worst

A | 0.8836 | 0.3136

B | 0.9964 | 0.8064

C | 0.0036 | 0.1936

It becomes evident why fouling the worst shooter on the team is a better tactic. The probabilities of the best shooter making the basket over the range of those two free shots are way better than the chances for the worst shooter.

Hope this Helps!!!

8 0
3 years ago
Read 2 more answers
Write greatest unit fractions that are repeating decimals. Then express each fraction as a decimal.
LekaFEV [45]
Our repeating decimal would be= 0.5555….
Let’s give 0.5555….. repeating decimal a variable of x:
x= 0.5555…. or 0.5(move 1 point to the right)
So ,10x = 5.5555 or 5.5

Now, Let’s do subtraction:
10x = 5.5555… or 5.5  
<u>-   x =  0.5555… or 0.5</u>
 9x  = 5.0            or 5.0

To get a fraction, provide a denominator the same with the numerator of variable x which is 9. Then divide the difference:
<span><u>9x</u> = <u>5.0</u>  
9        9</span> <span> <span><span> <span> <span>Therefore, x = <u>5
</u>                                     9</span> </span> </span> </span></span>


7 0
3 years ago
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