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dimulka [17.4K]

3 years ago

10

Simplify. t^-6 * t^2

1 answer:

Deffense [45]3 years ago

8 0

<span>t^-6 * t^2
= t^-4

hope it helps</span>

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Substitution method- y=-x+4 and y=3x

kumpel [21]

Substitue y=3x into the other equation to get 3x=-x+4. Solve this to get 4x=4 therefore x=1.
knowing this, y=3

5 0

3 years ago

Two planes that do not intersect are parallel

frez [133]

True. If they never intersect then they must have the same slope and are parallel

6 0

3 years ago

Solve: x^4-34x^2=-225

dangina [55]

X
4
−34x
2
+225=0

2 Factor
x
4
−
34
x
2
+
225
x
4
−34x
2
+225.
(
x
2
−
25
)
(
x
2
−
9
)
=
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8 0

3 years ago

The diagram below shows the graph of the quadratic function f (x) = ax² + 8x + c. calculate the value for each of the following.

Alexxandr [17]

Answer:

see explanation

Step-by-step explanation:

The y- intercept is at (0, 6) ⇒ c = 6

The x- coordinate of the vertex is at the midpoint of the zeros, that is

m = $\frac{-3-1}{2}$ = $\frac{-4}{2}$ = - 2

Since the zeros are x = - 3 and x = - 1 then the factors are (x + 3), (x + 1) and

f(x) = a(x + 3)(x + 1) = a(x² + 4x + 3)

Comparing with f(x) = ax² + 8x + 6 ⇒ a = 2

Thus

f(x) = 2x² + 8x + 6

Substitute m = - 2 into the equation for value of n

n = 2(- 2)² + 8(- 2) + 6 = 8 - 16 + 6 = - 2

Hence

c = 6, m = - 2, a = 2, n = - 2

3 0

3 years ago

Read 2 more answers

Find the obtuse angle between the following pair of straight lines.

ioda

Given:

$y=-2$

$y=\sqrt{3}x-1$

To find:

The obtuse angle between the given pair of straight lines.

Solution:

The slope intercept form of a line is

$y=mx+b$ ...(i)

where, m is slope and b is y-intercept.

The given equations are

$y=0x-2$

$y=\sqrt{3}x-1$

On comparing these equations with (i), we get

$m_1=0$

$m_2=\sqrt{3}$

Angle between two lines whose slopes are $m_1\text{ and }m_2$ is

$\tan \theta=\left|\dfrac{m_2-m_1}{1+m_1m_2}\right|$

Putting $m_1=0$ and $m_2=\sqrt{3}$ , we get

$\tan \theta=\left|\dfrac{\sqrt{3}-0}{1+(0)(\sqrt{3})}\right|$

$\tan \theta=\left|\dfrac{\sqrt{3}}{1+0}\right|$

$\tan \theta=\pm \sqrt{3}$

Now,

$\tan \theta= \sqrt{3}$ and $\tan \theta=-\sqrt{3}$

$\tan \theta= \tan 60^\circ$ and $\tan \theta=\tan (180^\circ-60^\circ)$

$\theta= 60^\circ$ and $\theta=120^\circ$

$120>90$ , so it is an obtuse angle and $60$ , so it is an acute angle.

Therefore, the obtuse angle between the given pair of straight lines is 120°.

6 0

3 years ago