Find the measure of the graph

The measures of two consecutive angles between the parallel sides of a trapezoid are 5:7. The other two consecutive angle measur

DiKsa [7]

Answer:

Hi how are you doing today Jasmine

4 0

3 years ago

A savings account earns 7.2 (APR) compounded monthly, so it earns 0.6% each month. Cooper has $1500 in the account now. if he de

iragen [17]

Answer-

Cooper will receive $12.00 at the end of the month.

Solution-

Cooper has $1500 in the account now and he deposits an additional $500 at the beginning of the month.

So the total principal becomes 1500+500 = $2000

Given here,

APR = annual percentage rate = 7.2%

But as we have to calculate the monthly interest, so monthly interest rate would be $=\dfrac{7.2}{12}=0.6\%$

Time period = 1 month

So, the interest after 1 month will be,

$\text{interest}=\dfrac{\text{Principal}\times \text{Rate of interest}\times \text{Time period}}{100}$

$=\dfrac{2000\times 0.6\times 1}{100}$

$=\$12.00$

Therefore, Cooper will receive $12.00 at the end of the month.

7 0

3 years ago

Read 2 more answers

In Exercises 40-43, for what value(s) of k, if any, will the systems have (a) no solution, (b) a unique solution, and (c) infini

svet-max [94.6K]

Answer:

If k = −1 then the system has no solutions.

If k = 2 then the system has infinitely many solutions.

The system cannot have unique solution.

Step-by-step explanation:

We have the following system of equations

$x - 2y +3z = 2\\x + y + z = k\\2x - y + 4z = k^2$

The augmented matrix is

$\left[\begin{array}{cccc}1&-2&3&2\\1&1&1&k\\2&-1&4&k^2\end{array}\right]$

The reduction of this matrix to row-echelon form is outlined below.

$R_2\rightarrow R_2-R_1$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\2&-1&4&k^2\end{array}\right]$

$R_3\rightarrow R_3-2R_1$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&3&-2&k^2-4\end{array}\right]$

$R_3\rightarrow R_3-R_2$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&0&0&k^2-k-2\end{array}\right]$

The last row determines, if there are solutions or not. To be consistent, we must have k such that

$k^2-k-2=0$

$\left(k+1\right)\left(k-2\right)=0\\k=-1,\:k=2$

Case k = −1:

$\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-1-2\\0&0&0&(-1)^2-(-1)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-3\\0&0&0&-2\end{array}\right]$

If k = −1 then the last equation becomes 0 = −2 which is impossible.Therefore, the system has no solutions.

Case k = 2:

$\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&2-2\\0&0&0&(2)^2-(2)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&0\\0&0&0&0\end{array}\right]$

This gives the infinite many solution.

5 0

3 years ago

^

love history [14]

Answer:

B. x < -1

Step-by-step explanation:

Hello!

Let's put them in order from greatest to least:

$\{-2,-5,-9,-12\}$

Given our possible inequalities, we need to find the inequality that contains all of these values.

The inequality that works is x < -1, as all values are less than -1.

Therefore, the answer is B. x < -1.

4 0

2 years ago

Elaine Mellon earns $17.80 per hour as an electric billing specialist for a legal clinic. How many hours per week must she work

Nuetrik [128]

Answer:

17.80/676.4= $38

Step-by-step explanation:

8 0

3 years ago