All categories

3 years ago

Y+4=2(x+3) In slope-intercept form

Mathematics

2 answers:

Anna35 [415]3 years ago

6 0

<h2>Answer: y = m x + b . Y = 2 x + 2</h2><h2>______________________________________</h2><h3>Hope this helps!</h3>

Also can I please have Brainliest?

Only if I'm correct of course.

<em>Your welcome, Loves!</em>

Sati [7]3 years ago

3 0

Y=2x+2 this is the answer hope this helps

You might be interested in

How much does 13 quarters equal? <br><br> I just need the answer.

Vlad [161]

Answer:

325 cents

Step-by-step explanation:

8 0

3 years ago

Help me out !?!??!?!l

Vlada [557]

Answer:67 is correct D

Step-by-step explanation:use parallel, so 71+x=180-42 x=67 67+71=138, 67= to your answer

4 0

3 years ago

Abbie, Beth, Carl, Diego, and Elena are waiting in line. Use the following information to figure out which of the five is first

zlopas [31]

Abbie is not next to Elena or Carl

5 0

3 years ago

Complete the steps to find all zeroes of the function f(x) = x4 − 4x3 − 4x2 + 36x − 45. This function has roots.

xxTIMURxx [149]

Steps?

A graph shows zeros to be ±3. Factoring those out leaves the quadratic
(x-2)² +1
which has complex roots 2±i.

The function has roots -3, 3, 2-i, 2+i.

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

4 years ago