Algebra 2 . PHOTO INCLUDED

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

2 years ago

What is the length of DE?

g100num [7]

Answer:

Step-by-step explanation:

Yes

Theorem 8.3: If two angles are complementary to the same angle, then these two angles are congruent.

∠A and ∠B are complementary, and ∠C and ∠B are complementary.

Given: ∠A and ∠B are complementary, and ∠C and ∠B are complementary.

Prove: ∠A ~= ∠C.

Statements Reasons

1. ∠A and ∠B are complementary, and ∠C and ∠B are complementary. Given

2. m∠A + m∠B = 90º , m∠C + m∠B = 90º Definition of complementary

3. m∠A = 90 º - m∠B, m∠C = 90º - m∠B Subtraction property of equality

4. m∠A = m∠C Substitution (step 3)

5. ∠A ~= ∠C Definition of ~=

6 0

2 years ago

Draw a number line to represent the inequality. 82 x

natima [27]

We can’t draw on this.....we can only type. But if I where to it would be

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7 0

2 years ago

Using the Pythagorean Theorem, which two examples are right triangles?

eimsori [14]

Answer:

2 and 4

Step-by-step explanation:

7 0

3 years ago

Dora says that of all the possible rectangles with the same area, the rectangle with the largest perimeter will have two side le

I am Lyosha [343]

Answer:

maybe

Step-by-step explanation:

Dora is apparently assuming the dimensions are integers. In that case she is correct.

If the dimensions are unconstrained, the perimeter will be largest when a pair of opposite sides will be the smallest measure allowed.

For some perimeter P and side length x, the area is ...

A = x(P/2 -x)

Conversely, the perimeter for a given area is ...

P = 2(A/x +x)

This gets very large when x gets very small, so Dora is correct in saying that the side lengths that are as small as they can be will result in the largest perimeter. We have no way of telling if her assumption of integer side lengths is appropriate. If it is not, her statement makes no sense.

8 0

2 years ago