All categories

3 years ago

5/13 t = -9 Thanks! Study work!

Mathematics

2 answers:

yaroslaw [1]3 years ago

8 0

T=-177/5.........
........

Triss [41]3 years ago

7 0

Hi there..

5/13 t = -9
Solve for T
t = 5/13 * (-9)
t= -177/5

Have a good night !

You might be interested in

Find x to the nearest hundredth place.

Arada [10]

Answer:

C. 24.27 cm

Step-by-step explanation:

sin 65° = opposite ÷ hypotenuse

sin 65° = 22/x

multiply both sides of the equation by x

(sin 65°)x = 22

divide both sides of the equation by sin 65°

x = 22 ÷ sin 65°

punch in 65 sin into your calculator

x = 22 ÷ 0.90631

x = 24.27 cm

7 0

2 years ago

Will mark brainliest if correct :)

Finger [1]

Answer:

d = 12

c = 4

so I think d > c

Step-by-step explanation:

1 x 12 = 12

0.75 x 4 = 3

12 + 3 = 15

8 0

2 years ago

If Rodney traded €228.75 for $314.52, when did he make his exchange?

never [62]

The answer is C Thursday

4 0

2 years ago

Read 2 more answers

Perform the indicated row operations, then write the new matrix.

Studentka2010 [4]

The matrix is not properly formatted.

However, I'm able to rearrange the question as:

$\left[\begin{array}{ccc}1&1&1|-1\\-2&3&5|3\\3&2&4|1\end{array}\right]$

Operations:

$2R_1 + R_2 ->R_2$

$-3R_1 +R_3 ->R_3$

Please note that the above may not reflect the original question. However, you should be able to implement my steps in your question.

Answer:

$\left[\begin{array}{ccc}1&1&1|-1\\0&5&7|1\\0&-1&1|4\end{array}\right]$

Step-by-step explanation:

The first operation:

$2R_1 + R_2 ->R_2$

This means that the new second row (R2) is derived by:

Multiplying the first row (R1) by 2; add this to the second row

The row 1 elements are:

$\left[\begin{array}{ccc}1&1&1|-1\end{array}\right]$

Multiply by 2

$2 * \left[\begin{array}{ccc}1&1&1|-1\end{array}\right] = \left[\begin{array}{ccc}2&2&2|-2\end{array}\right]$

Add to row 2 elements are: $\left[\begin{array}{ccc}-2&3&5|3\end{array}\right]$

$\left[\begin{array}{ccc}2&2&2|-2\end{array}\right] + \left[\begin{array}{ccc}-2&3&5|3\end{array}\right]$

$\left[\begin{array}{ccc}0&5&7|1\end{array}\right]$

The second operation:

$-3R_1 +R_3 ->R_3$

This means that the new third row (R3) is derived by:

Multiplying the first row (R1) by -3; add this to the third row

The row 1 elements are:

$\left[\begin{array}{ccc}1&1&1|-1\end{array}\right]$

Multiply by -3

$-3 * \left[\begin{array}{ccc}1&1&1|-1\end{array}\right] = \left[\begin{array}{ccc}-3&-3&-3|3\end{array}\right]$

Add to row 2 elements are: $\left[\begin{array}{ccc}3&2&4|1\end{array}\right]$

$\left[\begin{array}{ccc}-3&-3&-3|3\end{array}\right] + \left[\begin{array}{ccc}3&2&4|1\end{array}\right]$

$\left[\begin{array}{ccc}0&-1&1|4\end{array}\right]$

Hence, the new matrix is:

$\left[\begin{array}{ccc}1&1&1|-1\\0&5&7|1\\0&-1&1|4\end{array}\right]$

3 0

2 years ago

Read 2 more answers

Can teachers see how much time you spend on Delta Math?

NARA [144]

Answer:

Yes

Step-by-step explanation:

Short answer- yes.

Long answer- The Student Last Action feature allows teachers to see the last date and time that a student did something on DeltaMath. DeltaMath has several tools that help teachers manage student accounts. Once a student's account is created, a teacher can add the student to one or more classes.

7 0

1 year ago