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vladimir1956 [14]
2 years ago
6

A system of equations is shown below

Mathematics
1 answer:
KatRina [158]2 years ago
7 0

Answer:

D Multiply equation B by 5 and equation A by 4 and add the results together

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(1,3), m = - 3/4 using that as a fraction write an equation in point-slope form
sukhopar [10]
Y=mx+b   m=-3/4  (1,3)
3=-3/4(1)+b
3=-3/4+b
b=3 3/4

y=-3/4+3 3 3/4
7 0
2 years ago
Multiply<br> (x-6)(4x + 3)<br> A. 47-32-18<br> B. 4) - 24%-18<br> C. 47 - 21%-18<br> D. 4/2 - 18
PilotLPTM [1.2K]

Answer:

4x² - 21x - 18

Step-by-step explanation:

Use FOIL.

(x-6)(4x+3)\\\\x*4x = 4x^2\\\\x * 3 = 3x\\\\-6 * 4x= -24x\\\\-6 * 3 = -18\\\\\\4x^2+3x-24x-18\\\\\boxed{4x^2-21x-18}

Hope this helps.

5 0
2 years ago
Read 2 more answers
What is the compliment of 41°
kozerog [31]

Answer: 49°

Step-by-step explanation:

I'm guessing you meant complement?  Complementary angles add up to 90°.  90° - 41° = 49°

3 0
3 years ago
Name the marked angle in 2 different ways.<br> Q<br> P<br> O<br> N
Alexxandr [17]

Answer:

PQO and OQP

Step-by-step explanation:

PQO: lines OQ and PQ make the angle.

OQP: Not really sure why it asks for two names. You can probably just flip PQO to OQP and that might work. N is not part of the angle so you can't have an answer with that letter.

4 0
1 year ago
The height of a punted football can be modeled with the quadratic function 0.01x^2 + 1.18x+2.
Fantom [35]

The function of the path of the punted ball is a quadratic function which

follows the path of a parabola.

The correct responses are;

Part A: The coordinates of the vertex is \underline {(59, \, 36.81)}

Part B: The maximum height of the punt is <u>36.81 ft.</u>

Part C: The defensive player must reach up to <u>7.65 feet</u> to block the punt.

Part D: The distance down the field the ball will go without being blocked is approximately <u>119.67 ft.</u>

<u />

Reasons:

The function for the height of the punted ball is; h = -0.01·x² + 1.18·x + 2

Assumption; The distances are feet.

Part A: By completing the square, we have;

f(x) = -0.01·x² + 1.18·x + 2

100·f(x) = -x² + 118·x  + 200

-100·f(x) = x² - 118·x  - 200

x² - 118·x + (118/2)²= 200 + (118/2)²

(x - 59)² = 200 + (59)² = 3681

(x - 59)² - 3681

At the vertex, -3281 = -100·f(x)

∴ f(x) at the vertex = -3681/-100 = 36.81

\mathrm{\underline{Coordinate \ of \ the \ vertex = (59, \, 36.81)}}

Part B: The maximum height is given by the y-value at the vertex = 36.81 ft.

Part C: When <em>x</em> = 5, we have;

h = -0.01·x² + 1.18·x + 2

h = -0.01 × 5² + 1.18 × 5 + 2 = 7.65

The defensive player must reach up to 7.65 feet to block the punt

Part D: The distance the ball will go before it hits the ground is given by

the function, for the height as follows;

h = -0.01·x² + 1.18·x + 2 = 0

From the completing the square method, above, we get;

-0.01·x² + 1.18·x + 2 = 0

x² - 118·x  - 200 = 0

x² - 118·x + (118/2)²= 200 + (118/2)²

x² - 118·x + (59)²= 200 + (59)² = 3681

(x - 59)² = 3681

x - 59 = ±√3681

x = 59 ± √3681

x = 59 + √3681 ≈ 119.67

The distance down the field the ball will go without being blocked, x ≈ <u>119.67 ft.</u>

<u />

Learn more here:

brainly.com/question/24136952

5 0
2 years ago
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