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3 years ago

10

What is the range of y equals log 8 x

1 answer:

azamat3 years ago

4 0

Answer:

(-∞, ∞)

Step-by-step explanation:

The range of any log function is <em>all real numbers</em>.

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80% of what number is 64

Thepotemich [5.8K]

Make an equation.

80% = 0.8

'of' is multiplication

'what number' is 'x'

'is' is an equal sign

0.8 * x = 64

0.8x = 64

Divide 0.8 to both sides:

x = 80

5 0

3 years ago

Find the area of a parallelogram with the given vertices. P(–5, –2), Q(7, –2), R(–2, 8), S(10, 8)

beks73 [17]

Area of a parallelogram = bh

First, find b (base):

b = 12

Then find h (height):

h = 10

Multiply:

120

The answer is D) 120 units^2

4 0

3 years ago

Compute: 3.5 x 2.4 x 0.01 <br><br><br><br> 0.84<br><br><br> 0.084<br><br><br> 0.0084

Georgia [21]

The answer is 0.084.

8 0

4 years ago

A phone company charges 12 cents per minute of call. If Gerardo made a call that took 75 minutes using this plan, how much did h

sukhopar [10]

Answer:

900 cents, 9 dollars

Step-by-step explanation:

4 0

3 years ago

Help!! 50 points and brainliest!

Viktor [21]

Answer:

Second choice:

$x=2t$

$y=4t^2+4t-3$

Fifth choice:

$x=t+1$

$y=t^2+4t$

Step-by-step explanation:

Let's look at choice 1.

$x=t+1$

$y=t^2+2t$

I'm going to subtract 1 on both sides for the first equation giving me $x-1=t$ . I will replace the $t$ in the second equation with this substitution from equation 1.

$y=(x-1)^2+2(x-1)$

Expand using the distributive property and the identity $(u+v)^2=u^2+2uv+v^2$ :

$y=(x^2-2x+1)+(2x-2)$

$y=x^2+(-2x+2x)+(1-2)$

$y=x^2+0+-1$

$y=x^2$

So this not the desired result.

Let's look at choice 2.

$x=2t$

$y=4t^2+4t-3$

Solve the first equation for $t$ by dividing both sides by 2:

$t=\frac{x}{2}$ .

Let's plug this into equation 2:

$y=4(\frac{x}{2})^2+4(\frac{x}{2})-3$

$y=4(\frac{x^2}{4})+2x-3$

$y=x^2+2x-3$

This is the desired result.

Choice 3:

$x=t-3$

$y=t^2+2t$

Solve the first equation for $t$ by adding 3 on both sides:

$x+3=t$ .

Plug into second equation:

$y=(x+3)^2+2(x+3)$

Expanding using the distributive property and the earlier identity mentioned to expand the binomial square:

$y=(x^2+6x+9)+(2x+6)$

$y=(x^2)+(6x+2x)+(9+6)$

$y=x^2+8x+15$

Not the desired result.

Choice 4:

$x=t^2$

$y=2t-3$

I'm going to solve the bottom equation for $t$ since I don't want to deal with square roots.

Add 3 on both sides:

$y+3=2t$

Divide both sides by 2:

$\frac{y+3}{2}=t$

Plug into equation 1:

$x=(\frac{y+3}{2})^2$

This is not the desired result because the $y$ variable will be squared now instead of the $x$ variable.

Choice 5:

$x=t+1$

$y=t^2+4t$

Solve the first equation for $t$ by subtracting 1 on both sides:

$x-1=t$ .

Plug into equation 2:

$y=(x-1)^2+4(x-1)$

Distribute and use the binomial square identity used earlier:

$y=(x^2-2x+1)+(4x-4)$

$y=(x^2)+(-2x+4x)+(1-4)$

$y=x^2+2x+-3$

$y=x^2+2x-3$ .

This is the desired result.

3 0

4 years ago

Read 2 more answers