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kenny6666 [7]
3 years ago
12

Explain how the graph of a quadratic function relates to the solutions of the related quadratic equation.

Mathematics
1 answer:
OleMash [197]3 years ago
8 0

Answer:

The zeros of a quadratic function are the solutions of the related quadratic equation. If the graph of a quadratic function crosses the x-axis, there will be two real number solutions. If the graph of a quadratic function just touches the x-axis, there will be one unique real number, or double root, solution. If the graph of a quadratic function does not cross the x-axis, there will be no real number solution.

Step-by-step explanation:

Did the question.

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Simplify the expression 36+18x
quester [9]
The answer is 54x because 36+18=54, but then you have the x, so your answer is 54x
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Shanice bought one star fruit for $2. How many star fruit can Mark buy if he has $20?
Karo-lina-s [1.5K]
Mark can  buy 10 star fruit because 2 times 10 equals 20.So,Shanice bought one star fruit for $2. How many star fruit can Mark buy if he has $20?He can buy 10 with $20.
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3 years ago
A particle is projected with a velocity of <img src="https://tex.z-dn.net/?f=40ms%5E-%5E1" id="TexFormula1" title="40ms^-^1" alt
Katena32 [7]

Answer:

2\sqrt{55}\text{ m/s or }\approx 14.8\text{m/s}

Step-by-step explanation:

The vertical component of the initial launch can be found using basic trigonometry. In any right triangle, the sine of an angle is equal to its opposite side divided by the hypotenuse. Let the vertical component at launch be y. The magnitude of 40\text{ m/s} will be the hypotenuse, and the relevant angle is the angle to the horizontal at launch. Since we're given that the angle of elevation is 60^{\circ}, we have:

\sin 60^{\circ}=\frac{y}{40},\\y=40\sin 60^{\circ},\\y=20\sqrt{3}(Recall that \sin 60^{\circ}=\frac{\sqrt{3}}{2})

Now that we've found the vertical component of the velocity and launch, we can use kinematics equation v_f^2=v_i^2+2a\Delta y to solve this problem, where v_f/v_i is final and initial velocity, respectively, a is acceleration, and \Delta y is distance travelled. The only acceleration is acceleration due to gravity, which is approximately 9.8\:\mathrm{m/s^2}. However, since the projectile is moving up and gravity is pulling down, acceleration should be negative, represent the direction of the acceleration.

What we know:

  • v_i=20\sqrt{3}\text{ m/s}
  • a=-9.8\:\mathrm{m/s^2}
  • \Delta y =50\text{ m}

Solving for v_f:

v_f^2=(20\sqrt{3})^2+2(-9.8)(50),\\v_f^2=1200-980,\\v_f^2=220,\\v_f=\sqrt{220}=\boxed{2\sqrt{55}\text{ m/s}}

3 0
3 years ago
Divide 33 photos into two groups so the ratio is 4 to 7
Advocard [28]
4 to 7
or
4+7= 12
4x+7x=33
x=3
so,
(4×3)+(7×3)=33
12+21=33
or
12 and 21.
hope it helped
8 0
3 years ago
Is each line parallel, perpendicular, or neither parallel nor perpendicular to a line whose slope is -6?
babymother [125]

Answer:

Line M - neither

Line N - parallel

Line P - perpendicular

Line Q - neither

Step-by-step explanation:

If a line is perpendicular to another, the slope will be the opposite, e.g. -6, opposite slope, -6.

If a line is parallel to another, the slope will be the exact same, e.g. -6, same slope, 1/6.

3 0
3 years ago
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