All categories

1 year ago

Can you solve... 4(-6g-10h)+3h-4(-3h-10g)

Mathematics

1 answer:

andrezito [222]1 year ago

5 0

Answer:

16g - 28h

Step-by-step explanation:

4 ( -6g - 10h ) + 3h - 4 ( -3h - 10g) = -24g - 40h + 3h +12h + 40g

= 16g - 28h

This is pretty simple, all you need to do is distribute the values from the parenthesis and get simplify the variables

You might be interested in

In a rhombus whose side measures 20 and the smaller angle is 38°. Find the length of the larger diagonal, to the nearest tenth.

Lemur [1.5K]

A rhombus diagonals form right angles. The diagonals also cut in half the angles

use cosine: cos38 = $\frac{a}{20\\\\}$

a = 20cos38

a=15.8

4 0

2 years ago

Express this to single logarithm

zzz [600]

Answer: $\log_{2}\left(\frac{q^2\sqrt{m}}{n^3}\right)$

We have something in the form log(x/y) where x = q^2*sqrt(m) and y = n^3. The log is base 2.

===========================================================

Explanation:

It seems strange how the first two logs you wrote are base 2, but the third one is not. I'll assume that you meant to say it's also base 2. Because base 2 is fundamental to computing, logs of this nature are often referred to as binary logarithms.

I'm going to use these three log rules, which apply to any base.

log(A) + log(B) = log(A*B)
log(A) - log(B) = log(A/B)
B*log(A) = log(A^B)

From there, we can then say the following:

$\frac{1}{2}\log_{2}\left(m\right)-3\log_{2}\left(n\right)+2\log_{2}\left(q\right)\\\\\log_{2}\left(m^{1/2}\right)-\log_{2}\left(n^3\right)+\log_{2}\left(q^2\right) \ \text{ .... use log rule 3}\\\\\log_{2}\left(\sqrt{m}\right)+\log_{2}\left(q^2\right)-\log_{2}\left(n^3\right)\\\\\log_{2}\left(\sqrt{m}*q^2\right)-\log_{2}\left(n^3\right) \ \text{ .... use log rule 1}\\\\\log_{2}\left(\frac{q^2\sqrt{m}}{n^3}\right) \ \text{ .... use log rule 2}$

8 0

2 years ago

A box in a college bookstore contains books, and each book in the box is a history book, an english book or a science book. If 1

just olya [345]

B- 1/2 the books are science books. You have to add the two fractions and subtract that from one.

8 0

2 years ago

Please help!<br><br> Fill in the missing number.

liq [111]

Answer:

Step-by-step explanation:

$\left[\begin{array}{cc}5&9\\-6&9\end{array}\right] +6\left[\begin{array}{cc}-5&2\\7&8\end{array}\right]$

Multiply the second matrix by 6.

$\left[\begin{array}{cc}5&9\\-6&9\end{array}\right] +\left[\begin{array}{cc}-30&12\\42&48\end{array}\right]$

Add the corresponding cells in each matrix.

$\left[\begin{array}{cc}5-30&9+12\\-6+42&9+48\end{array}\right]$

$\left[\begin{array}{cc}-25&21\\36&57\end{array}\right]$

3 0

2 years ago

Are my answers correct? Precalculus!!

SVETLANKA909090 [29]

Answer:

Unfortunately, your answer is not right.

Step-by-step explanation:

The functions whose graphs do not have asymptotes are the power and the root.

The power function has no asymptote, its domain and rank are all the real.

To verify that the power function does not have an asymptote, let us make the following analysis:

The function $y = x ^ n$ , when x approaches infinity, where does y tend? Of course it tends to infinity as well, therefore it has no horizontal asymptotes (and neither vertical nor oblique)

With respect to the function $y= \frac{1}{x}$ we can verify that if it has asymptote horizontal in y = 0. Since when x approaches infinity the function is closer to the value 0.

For example: 1/2 = 0.5; 1/1000 = 0.001; 1/100000 = 0.00001 and so on. As "x" grows "y" approaches zero

Also, when x approaches 0, the function approaches infinity, in other words, when x tends to 0 y tends to infinity. For example: 1 / 0.5 = 2; 1 / 0.1 = 10; 1 / 0.01 = 100 and so on. This means that the function also has an asymptote at x = 0

6 0

3 years ago