Someone wants to save $9 per day for 6 days. He/She then plans to spend $9 per day for another 6 days

tankabanditka [31]

Answer:

D. 22.5 sq. un.

Step-by-step explanation:

W divides BU so that there is a ratio of 1;2 meaning that there are 3 parts.

That means that 3 parts is BUV; we know that one of those parts is 4.5 sq. un. so 4.5*3=13.5 sq. un. - Also 4.5*2=9 to get WUV

V is the midpoint of the line BS so that means that BUV and UVS are equal

we know that BUV is 13.5 so we multiply that by 2 to get the whole triangle

=27

The last step is to get rid of BVW which we know is 4.5

27-4.5=22.5

7 0

2 years ago

Please help me with this question

azamat

It is 90 your welcome

7 0

2 years ago

Can someone please help me with this math question .

Alona [7]

No, Yes, No, Yes

using the laws of exponents

(1)

( $4^{3}$ )³ = $4^{9}$

4³ × 4³ = $4^{6}$

$4^{9}$ ≠ $4^{6}$ → No

(2)

( $3^{4}$ )^4 = $3^{16}$

$3^{8}$ × $3^{8}$ = $3^{16}$ → Yes

(3)

$18^{8}$ = (6 × 3 )^8 = $6^{8}$ × $3^{8}$ ≠ $6^{4}$ × $3^{4}$ → No

(4)

$20^{3}$ = (4 × 5 )³ = 4³ × 5³ → Yes

5 0

2 years ago

What is the equation of the line with a slope of -3 and passes through the point (-6,-3)

Taya2010 [7]

Answer:

f(x) = -3x - 21

Step-by-step explanation:

Graphed this equation and it passes through (-6, -3), also known as Point A.

7 0

2 years ago

Plz help<br>urgent !!!!<br>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