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nadezda [96]
3 years ago
8

Help me Instead of typing random answers

Mathematics
2 answers:
ElenaW [278]3 years ago
8 0

Answer:

whats the question

Step-by-step explanation:

Andrej [43]3 years ago
4 0

Answer:

42

Step-by-step explanation:

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Help guys pls 22x²+13x+1​
myrzilka [38]

Answer:

(11x+1)(2x+1)

8 0
3 years ago
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Diego is building a kitchen table and a coffee table. The legs of a kitchen table must be twice the height of a coffee table and
Montano1993 [528]
In the expression 4(x), 4 represents the number of legs, and x represents the height of the coffee table.

Altogether this expression shows the total length of all legs from the coffee table and kitchen table combined.
5 0
3 years ago
Plss help mee it’s not a or b
Luda [366]
C.
don’t actually put that
6 0
3 years ago
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Solve<br> h-w<br> Solve m = for the<br> 8<br> variable h.<br> I
Ilya [14]

Answer:

1) m = h - w/8

Subtract h from both sides.

m - h = -w/8

Subtract m from both sides.

-h = -m - w/8

Divide all terms by -1.

h = m + w/8.

2) m = h - w/8

Get rid of the denominator 8 by multiplying 8/1 to all terms.

8m = 8h - w

Add w on both sides and subtract 8m from both sides.

w = 8h - 8m

3) w = x + y/z

Get rid of the denominator z by multiplying it to all terms.

wz = xz + y

Subtract xz from both sides of the equation.

wz - xz = y OR y = wz - xz

4) w = x + y/z

Subtract x from both sides.

w - x = y/z

Get rid of the denominator by multiplying it to all terms.

wz - xz = y

Now factor the expression wz - xz.

z(w - x) = y

Divide both sides by w - x.

z = y / w - x

This is read as z equals to y divided by w minus x.

5) The area of a triangle is A = 1/2bh

First, get rid of the denominator by multiplying both sides by 2.

2A = bh

To find b, divide both sides by h.

2A/h = b

6) P = kt/v

Multiply both sides by v.

Pv = kt

Divide both sides by t.

Pv/t = k OR k = Pv/t.

Step-by-step explanation:

8 0
3 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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