Answer:
I'm afraid that none of them are, as the x-intercepts of the equation are 0 and 9 respectively
however, D. is the closest to it, and it would be the correct graph if there were parentheses around
Step-by-step explanation:
For what values of m dose the graph of y=3x^2+7x+m have two x-intercepts?
1 answer:
Answer:
Step-by-step explanation:
x-intercepts are for y = 0.
Put y = 0 to the equation y = 3x² + 7x + m.
<h3>3x² + 7x + m = 0</h3>Calculate the discriminant of quadratic equation ax² + bx + c = 0:
Δ = b² - 4ac
if Δ < 0, then an equation has no solution
if Δ = 0, then an equation has one solution
if Δ > 0, then an equation has two solution.
<h3>3x² + 7x + m = 0</h3>a = 3, b = 7, c = m
Δ = 7² - 4(3)(m) = 49 - 12m
Two x-intercepts for Δ > 0.
Solve the inequality:
<em>subtract 49 from both sides</em>
<em>change the signs</em>
<em>divide both sides by 12</em>
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Answer:
6 square units
Step-by-step explanation:
The attachment shows the figure redrawn to scale, along with helpful semicircle O with diameter EB. Trig relations can be used to find the area of triangle ADB. The relation between an inscribed angle and its corresponding central angle is also helpful.
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<h3>angles</h3>Any right triangle can be inscribed in a semicircle whose diameter is the hypotenuse. Accordingly, we have constructed semicircle O whose center is the midpoint of EB. Hence its radius is EB/2 = 4/2 = 2.
Then angle EBD is an inscribed angle, hence half the measure of central angle EOD. Since ∠EOD is double the measure of ∠EBD, it is congruent to ∠EBC, which is also double the measure of ∠EBD. We have called the measure of this double angle θ. Its complement is the angle marked β.
<h3>segments</h3>The fact that angles EOD and EBC are congruent means that segments OD and BC are parallel. Segment BC is the altitude of ΔADB with respect to base AD.
The tangent ratio is ...
Tan = Opposite/Adjacent
Applying that to angle β, we have
tan(β) = OD/AD [eqn 1]
tan(β) = BC/AC [eqn 2]
These two relations can be solved for the measures of AD and BC, which we need in order to find the area of ΔADB.
<h3>area</h3>Equation [eqn 1] tells us ...
AD = OD/tan(β)
Equation [eqn 2] tells us ...
BC = AC·tan(β)
The area of ΔADB is ...
Area = 1/2bh
Area = 1/2(AD)(BC) = 1/2(OD/tan(β))·(AC·tan(β))
Area = 1/2(OD)(AC) . . . . . . factors tan(β) cancel
The length of AC is given as 6, and we know OD = 2, so the area is ...
Area = 1/2(2)(6) = 6
The area of triangle ADB is 6 square units.
