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3 years ago

What are the roots of the quadratic equation X² + 4X + 4. How many roots are there.

1 answer:

3 0

If we set the equation equal to 0, we can factor it to find its roots:

x² + 4x + 4 = 0
(x + 2)(x + 2) = 0
x = -2

This graph has one root, a double root, at -2. This means that a single point, which must be the vertex of the parabola, touches the x-axis at (-2, 0)

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This box can be packed with 48 unit cubes

AURORKA [14]

Answer:

Please clarify the question so I can answer :)

Step-by-step explanation:

8 0

3 years ago

A student is given that point P(a, b) lies on the terminal ray of angle Theta, which is between StartFraction 3 pi Over 2 EndFra

Harman [31]

Answer:

A.

The student made an error in step 3 because a is positive in Quadrant IV; therefore,

$cos\theta = \frac{a\sqrt{a^2 + b^2}}{a^2 + b^2}$

Step-by-step explanation:

Given

$P\ (a,b)$

$r = \± \sqrt{(a)^2 + (b)^2}$

$cos\theta = \frac{-a}{\sqrt{a^2 + b^2}} = -\frac{\sqrt{a^2 + b^2}}{a^2 + b^2}$

Required

Where and which error did the student make

Given that the angle is in the 4th quadrant;

The value of r is positive, a is positive but b is negative;

Hence;

$r = \sqrt{(a)^2 + (b)^2}$

Since a belongs to the x axis and b belongs to the y axis;

$cos\theta$ is calculated as thus

$cos\theta = \frac{a}{r}$

Substitute $r = \sqrt{(a)^2 + (b)^2}$

$cos\theta = \frac{a}{\sqrt{(a)^2 + (b)^2}}$

$cos\theta = \frac{a}{\sqrt{a^2 + b^2}}$

Rationalize the denominator

$cos\theta = \frac{a}{\sqrt{a^2 + b^2}} * \frac{\sqrt{a^2 + b^2}}{\sqrt{a^2 + b^2}}$

$cos\theta = \frac{a\sqrt{a^2 + b^2}}{a^2 + b^2}$

So, from the list of given options;

The student's mistake is that a is positive in quadrant iv and his error is in step 3

3 0

3 years ago

Match the graph to the related table.

lana66690 [7]

Answer:

Option C

Step-by-step explanation:

Option c is right answer

3 0

2 years ago

Read 2 more answers

In the diagram of circle O, what is the measure of ? 27° 54° 108° 120°

vesna_86 [32]

Answer:

Option (2).

Step-by-step explanation:

This question is incomplete; here is the complete question and find the figure attached.

In the diagram of a circle O, what is the measure of ∠ABC?

27°

54°

108°

120°

m(minor arc $\widehat {AC}$ ) = 126°

m(major arc $\widehat {AC}$ ) = 234°

By the intersecting tangents theorem,

If the two tangents of a circle intersect each other outside the circle, measure of angle formed between them is half the difference of the measures of the intercepted arcs.

m∠ABC = $\frac{1}{2}[m(\text {major arc{AC})}-m(\text{minor arc} {AC})]$