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Rufina [12.5K]
3 years ago
10

How much is 1 watt in volts

Mathematics
1 answer:
olganol [36]3 years ago
4 0

The answer to your is 100 volts

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Arjun can type 40 words per minute.Dalia can type 55 words per minute.If Arjun and Dalia each type for 30 minutes,about how many
Citrus2011 [14]
♥ Solve:
(Arjun) 40*30=1200
(Dalia) 55*30=1650
Now subtract
1650-1200=450.
That means that Dalia will type 450 minutes more. 
Final answer: 450 

8 0
3 years ago
Read 2 more answers
If 50% of a number is 150 and 30% of the same number is 90, find 80% of that number.
Alexandra [31]

Answer:

240

Step-by-step explanation:

We know that 50% of the number is 150 so we can assume the number is 300(if its 50% just double that number). Next we use the equation 0.80 x 300 which gives us the answer 240.

Hope I helped

4 0
3 years ago
Read 2 more answers
Multiply to find the product.<br><br> 64.7 x 28 = ______
Crazy boy [7]

Answer:

1811.6

Step-by-step explanation:

647x28

18116

move once to the left bc of the point in 64.7

1811.6

7 0
2 years ago
Read 2 more answers
Solve the equation. -(7-4x)=9​
Ket [755]

Answer:

x=4

Step-by-step explanation:

-(7-4x)=9

-7+4x=9

4x=16

x=4

7 0
3 years ago
Read 2 more answers
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
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