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Kisachek [45]
3 years ago
11

HELP ME PLEASE! Find the equation of the line. Use exact numbers.

Mathematics
1 answer:
dangina [55]3 years ago
6 0

Answer:

The equation of the line (in slope-intercept form) is y=4x-9

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Omg my daughter and I are totally lost on this homework. Please help. It is multiplying decimals. She has to turn this in today.
Wewaii [24]

Answer:

5 x 0.6 = 5 x 6tenths, as 6/10 = 0.6

= 30 tenths

7 x 0.8 = 7 x 8tenths, as 8/10 is 0.8

= 56 tenths

Step-by-step explanation:

If you really want a more detailed explanation, just reply to this message.

5 0
3 years ago
Solve mathematics question, in photo. :)
Jlenok [28]
(0,4) is the correct answer.
7 0
3 years ago
3x-32=-7×+28. what does x =​
Vitek1552 [10]

Hello!

3x - 32 = -7x + 28

-32 - 28 = -7x - 3x

-60 = -10x

-6 = -x

6 = x

x = 6

I hope this helps you! Have a lovely day!

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3 0
3 years ago
Read 2 more answers
(No links please) There were 2 red marbles, 3 blue marbles and 5 gold marbles in a bag.
Rashid [163]

Answer:

5/9

Step-by-step explanation:

You started with 10 marbles total. You removed one marble and did not replace it, so you now have 9 marbles. The marble that you removed was NOT a gold marble, so you still have 5 gold marbles.

This means that out of the 9 marbles that you have left, 5 of them are gold.

This means you have a 5/9 chance of drawing a gold marble next.

4 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
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