So sig figs are non zero so

500 has 1 sig fig, 5
but 500. has 3 sig figs because it has a decimal

2.3 has 2 sig figs
2.03 has 3 sig figs so to make 232.76 to 2 sig figs you would put it into scientific notation and get 2.3 times 10^2 and thats 2 sig figs (the 10^2 doesn't count as a sig fig)

5 0

3 years ago

What is the slope of the line graph show below?

bonufazy [111]

Answer:

The answer is -2

Step-by-step explanation:

slope formula is m= (y2 -y1)/(x2-x1)

m=(3-1)/(0-1)=

m=2/-1 =

m=-2

6 0

3 years ago

Read 2 more answers

△PQR has two known interior angles of 34°, and 80°. △RST has two known interior angles of 24°, and 90°. What can be determined a

Sergeeva-Olga [200]

In order for two triangles to be similar, there must be 2 pairs of congruent angles. Since two angles of triangle PQR have different measures from two angles of triangle RST, it is impossible to have two pairs of congruent angles, so the triangles cannot be similar.

The answer is in bold type below.

All interior angles must be given to determine similarity.
The triangles are not similar.
Similarity cannot be determined from the given information.
The triangles are similar.

3 0

3 years ago

Read 2 more answers

Plz help urgent !!!! will give the brainliest !!

Mkey [24]

Answer:

$X = \begin{bmatrix}1&3\\ 2&4\end{bmatrix}$

Step-by-step explanation:

The question we have at hand is, in other words,

$\begin{bmatrix}4&2\\ \:1&1\end{bmatrix}\left(X\right)=\begin{bmatrix}8&20\\ \:3&7\end{bmatrix}$ - where we have to solve for the value of $X$

If we have to isolate $X$ here, then we would have to take the inverse of the following matrix ...

$\begin{bmatrix}4&2\\ \:1&1\end{bmatrix}$ ... so that it should be as follows ... $\begin{bmatrix}4&2\\ \:1&1\end{bmatrix}^{-1}$

Therefore, we can conclude that the equation as to solve for " $X$ " will be the following,

$X=\begin{bmatrix}4&2\\ 1&1\end{bmatrix}^{-1}\begin{bmatrix}8&20\\ 3&7\end{bmatrix}$ - First find the 2 x 2 matrix inverse of the first portion,

$\begin{bmatrix}4&2\\ 1&1\end{bmatrix}^{-1}$ = $\frac{1}{\det \begin{pmatrix}4&2\\ 1&1\end{pmatrix}}\begin{pmatrix}1&-2\\ -1&4\end{pmatrix}$ = $\frac{1}{2}\begin{bmatrix}1&-2\\ -1&4\end{bmatrix}$ = $\begin{bmatrix}\frac{1}{2}&-1\\ -\frac{1}{2}&2\end{bmatrix}$

At this point we have to multiply the rows of the first matrix by the rows of the second matrix,

$X = \begin{bmatrix}\frac{1}{2}&-1\\ -\frac{1}{2}&2\end{bmatrix}\begin{bmatrix}8&20\\ 3&7\end{bmatrix}$ ,

$X = \begin{pmatrix}\frac{1}{2}\cdot \:8+\left(-1\right)\cdot \:3&\frac{1}{2}\cdot \:20+\left(-1\right)\cdot \:7\\ \left(-\frac{1}{2}\right)\cdot \:8+2\cdot \:3&\left(-\frac{1}{2}\right)\cdot \:20+2\cdot \:7\end{pmatrix}$ - Simplifying this, we should get ...

$\begin{bmatrix}1&3\\ 2&4\end{bmatrix}$ ... which is our solution.

7 0

3 years ago