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Kipish [7]
2 years ago
13

What is the value of y in the equation 3(3y − 15) = 0? 5 6 7 9

Mathematics
1 answer:
mezya [45]2 years ago
4 0
Y = 5
3(3y - 15) is also equal to 9y - 45, which means to find y, you have to divide 45 by 9, which is 5.
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I really need help with this one it's due today!!
baherus [9]

Answer:

-2/3

3/2

Step-by-step explanation:

So to have a line parallel the equation is: m1 = m2, meaning the slope (represented by m in this equation) is equal. So in the equation. 3y = - 2x + 4 you need to simplify the equation by dividing both sides by 3. Which would be like so: y = - 2/3x + 4/3, which the slope will be -2/3.

To be perpindicular you need the slope to be: m1 * m2 = - 1. So because we already know that the slope is -2/3, all we need to do is set up an equation: -2/3 * y = - 1

-2/3y = - 1

multiply both sides by 3/-2

y = 3/2

4 0
2 years ago
Rosa saved $100 to spend on vacation. For the first 3 days of her vacation she spent $20 each day then for the next 2 days, afte
Wewaii [24]

Answer:

hope this helps you

6 0
2 years ago
Using the following triangle, what is the cosine of angle B?
stealth61 [152]

Hello!

To find cosine, use the formula cos = adjacent / hypotenuse.

According to angle B, adjacent of angle B is side A, and the hypotenuse is side c because the hypotenuse is always opposite the right angle.

Therefore, the cosine of angle B is a/c.

3 0
3 years ago
Soil sells for 8 1/4 cents per pound. How much would you save if you bought a 100- pound sack for $7.45?
asambeis [7]

Answer:

$0.80

Step-by-step explanation:

If the 100 pounds was bought separately (of course in real life you will probably not)

= 100 pounds x $0.0825/pound

= $8.25

As you can see it will cost more than if you buy a 100 pound sack.

Cost Difference = $8.25 - $7.45 = $0.80

That means you will save $0.80 from it.

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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