Ask question

Ask question

All categories

Dahasolnce [82]

3 years ago

10

What is the area of this parallelogram?

2 answers:

Liono4ka [1.6K]3 years ago

5 0

What are the measurements?

natka813 [3]3 years ago

4 0

What is the parallelogram? Can’t help if I don’t have a picture

You might be interested in

If h(z)=13z2–30z+27, use synthetic division to find h(3).

Blababa [14]

Answer:

h=13z-30+ 27/z

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

A country's population in 1991 was 136 million. In 2000 it was 141 million. Estimate the population in 2016 using the exponentia

xeze [42]

Answer: The population will be about 150 million in the year 2016.

First, let's start by using the given equation. Fill in the values that you know and solve for the missing piece.

Our equation is:

141 = 136e^(9k) I used 9 because 2000 is 9 years past 1991.

(141/136) = e^9k Divide both sides by 136

ln (141/138) = 9k Take the natural log of both sides, the e is removed.

0.004 = k Find the natural log of 141/136 and divide it by 9.

Now, we have a new equation. Just plug in 25 for x and you will get 150. We used 26 because 2016 is 25 years past the starting year of 1991.

5 0

3 years ago

Simplify the expression:<br> 2f+-1+-8+-4+-3f^2

mamaluj [8]

$2f+-1+-8+-4+-3f^2 = 2f-1 -8-4-3f^2 \\ \\ =\boxed{-3f^2+2f-13}$

5 0

3 years ago

15 1/3 - 5 7/9 = 10 1/3 - 7/9

evablogger [386]

Answer:

true

Step-by-step explanation:

7 0

3 years ago

H(x) = 2^x<br>Solve the equation h^-1 (x) =0.5

Ludmilka [50]

Answer:

The answer is

<h2> $\sqrt{2} \: \: or \: \: 1.414$ </h2>

Step-by-step explanation:

<h3> $h(x) = {2}^{x}$ </h3>

To solve the equation we must first find h-¹(x)

To find h-¹(x) equate h(x) to y

That's

<h3> $y = {2}^{x}$ </h3>

<u>Next interchange the terms </u>

x becomes y and y becomes x

That's

<h3> ${2}^{y} = x$ </h3>

<u>Make y the subject</u>

Take logarithm to base 2 to both sides

That's

<h3> $log_{2}( {2} )^{y} = log_{2}x \\ y log_{2}2 = log_{2}(x)$ </h3>

But

<h3> $log_{2}(2) = 1$ </h3><h3> $y = log_{2}(x)$ </h3>

So we have

<h3> ${h}^{ - 1} (x) = log_{2}(x)$ </h3>

Now we can solve the equation

We have

<h3> $log_{2}(x) = 0.5$ </h3>

Convert the logarithm into exponential form using the fact that

<h3> $log_{a}(x) = b \: \: \cong \: \: x = {a}^{b}$ </h3>

So we have

<h3> $x = {2}^{0.5}$ </h3>

But

<h3> ${2}^{0.5} = \sqrt{2}$ </h3>

So we have the final answer as

<h3> $\sqrt{2} \: \: or \: \: 1.414$ </h3>

Hope this helps you

8 0

3 years ago