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2 years ago

Ann, Mara, Jenny , Tina, and Sue are sisters. Two of the five sisters must help their father

1 answer:

8 0

Imagine taking from the group of five sisters to fill two slots. To fill the first slot, there are 5 different sisters that could do it. To fill the second slot, there would only be four left to fill the slot since one is in the first slot.

Therefore, there are 5 * 4, or 20 ways the girls can fill 2 slots. Hope this helps! :)

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5 less than the product of -2 and a number is no more than 15.

vladimir1956 [14]

Answer:

-2x - 5 (lesser than or equal to) 15

6 0

2 years ago

Slope 0; through (-5, -1)<br> with steps please.

posledela

Step-by-step explanation:

0 slope means it has in x.

so the equation is y=-1, because that is the coord for the points'y axsis

3 0

3 years ago

A hiker averages 1.4 mph downhill on a mountain trail and 0.8 mph uphill. Find the total time spent hiking in terms of the dista

nadya68 [22]

Answer:

V=d/t.

V=(1.4)(0.8)÷8m/s

=0.14

Step-by-step explanation:

bali yung divide ay guhit sa baba

3 0

3 years ago

Please help me! I need the anwser asap

sasho [114]

What is the question you are trying to ask

3 0

2 years ago

Solve the matrix equation for a, b, c, and d. [1 2] [a b] [6 5][3 4] [c d]= [19 8]

Anit [1.1K]

Answer:

The answer is " $\bold{\left[\begin{array}{cc}a&b\\c&d\end{array}\right] = \left[\begin{array}{cc}7&-2\\ -\frac{1}{2}&\frac{7}{2}\end{array}\right]}$ ".

Step-by-step explanation:

$\bold{\left[\begin{array}{cc}1&2\\3&4\end{array}\right] \left[\begin{array}{cc}a&b\\c&d\end{array}\right] = \left[\begin{array}{cc}6&5\\ 19&8\end{array}\right]}$

Solve the L.H.S part:

$\left[\begin{array}{cc}1&2\\3&4\end{array}\right] \left[\begin{array}{cc}a&b\\c&d\end{array}\right]\\\\\\\left[\begin{array}{cc}a+2c&b+2d\\3a+4c&3b+4d\end{array}\right]$

After calculating the L.H.S part compare the value with R.H.S:

$\left[\begin{array}{cc}a+2c&b+2d\\3a+4c&3b+4d\end{array}\right]= \left[\begin{array}{cc}6&5\\ 19&8\end{array}\right]} \\\\$

$\to a+2c =6....(i)\\\\\to b+2d =5....(ii)\\\\\to 3a+4c =19....(iii)\\\\\to 3b+4d = 8 ....(iv)\\\\$

In equation (i) multiply by 3 and subtract by equation (iii):

$\to 3a+6c=18\\\to 3a+4c=19\\\\\text{subtract}... \\\\\to 2c = -1\\\\\to c= - \frac{1}{2}$

put the value of c in equation (i):

$\to a+ 2 (- \frac{1}{2})=6\\\\\to a- 2 \times \frac{1}{2}=6\\\\\to a- 1=6\\\\\to a =6 +1\\\\\to a = 7\\$

In equation (ii) multiply by 3 then subtract by equation (iv):

$\to 3b+6d=15\\\to 3b+4d=8\\\\\text{subtract...}\\\\\to 2d = 7\\\\\to d= \frac{7}{2}\\$

put the value of d in equation (iv):

$\to 3b+4 (\frac{7}{2})=8\\\\\to 3b+4 \times \frac{7}{2}=8\\\\\to 3b+14=8\\\\\to 3b =8-14\\\\\to 3b = -6\\\\\to b= \frac{-6}{3}\\\\\to b= -2$

The final answer is " $\bold{\left[\begin{array}{cc}a&b\\c&d\end{array}\right] = \left[\begin{array}{cc}7&-2\\ -\frac{1}{2}&\frac{7}{2}\end{array}\right]}$ ".

4 0

3 years ago