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2 years ago

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Triangle ABC, with A = (-2,3) B = (-4,1) and C = (-3,5), is rotated 90- clockwise about the origin to make Triangle A'B'C'. What

are the y-coordinates of B'?

1 answer:

nikdorinn [45]2 years ago

6 0

Answer:

Step-by-step explanation:

-1

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What is the value of b2 - 4ac for the following equation?<br><br> 2x2 + 3x = -1

andreyandreev [35.5K]

The value of b^2-4ac is known as the discriminant of a quadratic function, and can tell you how many roots exist of this function depending on what it is equal to.

Start by moving the -1 to the other side, as we need this function to equal zero.

2x^2 + 3x + 1 = 0

This is now the standard form ax^2 + bx + c = 0. Plug each value that corresponds into the discriminant equation.

b^2-4ac
(3)^2 - 4(2)(1)
9 - 8
1

The value of the discriminant is 1, meaning that two real roots exist for the function described.

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2 years ago

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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:

<em>A.</em>

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

<em /> $cos\theta = \frac{a\sqrt{a^2 + b^2}}{a^2 + b^2}$

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;

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

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2 years ago

1) A projectile is thrown upward so that its distance above the ground after t seconds is

leonid [27]

Answer:

After 22 seconds the projectile reach its maximum height of 4,840 units

Step-by-step explanation:

we have

$h(t)=-10t^{2}+440t$

This is a vertical parabola downward (because the leading coefficient is negative)

The vertex is a maximum

Find out the coordinates of the vertex

Convert the quadratic equation in vertex form

Factor -10

$h(t)=-10(t^{2}-44t)$

Complete the square

$h(t)=-10(t^{2}-44t+22^2)+(10)(22^2)$

$h(t)=-10(t^{2}-44t+22^2)+4,840$

Rewrite as perfect squares

$h(t)=-10(t-22)^{2}+4,840$

The vertex is the point (22,4,840)

therefore

After 22 seconds the projectile reach its maximum height of 4,840 units

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2 years ago

Chris found some colorful leaves under one tree. Under another tree, he

Lynna [10]

Answer:

x+y-3+6

or

x+y+3

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2 years ago

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Delvig [45]

Answer:

It's Same perimeter only

Step-by-step explanation:

It can't be the other two because they don't share the same area

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