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Karo-lina-s [1.5K]
1 year ago
14

5. Write an equation, for the line whose slope is - and passes through the point (4,-7).

Mathematics
1 answer:
Vlad1618 [11]1 year ago
6 0

the general form of the linear equation is

y=mx+b

where m is the slope and b an initial point

we have m just need find b, we can replace the slope and the point and solve b

\begin{gathered} (-7)=(-1)(4)+b \\ -7=-4+b \\ b=-7+4 \\ b=-3 \end{gathered}

so the linear equation is

y=-x-3

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(-3,0),(5,-2) find the slope
Mashcka [7]

Answer:

-4

Step-by-step explanation:

Point slope form: y - y = m (x - x)

5 - (-3) = m (-2 - 0)

5 + 3 = m (-2)

8 = m (-2)

m = 8/(-2)

m=4

3 0
3 years ago
There are 64 pretzels in a 16-ounce bag of chocolate-covered pretzels.
Shkiper50 [21]

Answer:

64 pretzels in a 16 oz bag

1 oz will contain 64/16 pretzels = 4

5 oz will therefore contain 4 x 5 = 20 pretzels

Step-by-step explanation:

See here for a deeper explanation! ^^

Hopefully it helps :)

brainly.com/question/200543

8 0
3 years ago
Read 2 more answers
What are the solutions to the following system of equations? y=x^2+12x+30 and 8x-y=10
Ierofanga [76]
Note that the 2nd equation can be re-written as y=8x-10.
According to the second equation, y=x^2+12x+30.
Equate these two equations to eliminate y:

8x-10 = x^2+12x+30

Group all terms together on the right side.  To do this, add -8x+10 to both sides.  Then 0 = x^2 +4x +40.  You must now solve this quadratic equation for x, if possible.  I found that this equation has NO REAL SOLUTIONS, so we must conclude that the given system of equations has NO REAL SOLUTIONS.

If you have a graphing calculator, please graph 8x-10  and  x^2+12x+30 on the same screen.  You will see two separate graphs that do NOT intersect.  This is another way in which to see / conclude that there is NO REAL SOLUTION to this system of equations.
6 0
3 years ago
16. Two years of local phone service costs $883, including
Evgen [1.6K]

Answer:

The monthly fee would be $23

Step-by-step explanation:

hope it helps mark as brainllest if help!!!

3 0
2 years 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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