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3 years ago

What do 3x+2x+y+2y-3 equals in combining like terms

Mathematics

1 answer:

viktelen [127]3 years ago

4 0

The answer is 5x+3y-3

explanation:

3x+2x+y+2y-3

= 5x+y+2y-3

=5x+3y-3

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Perimeter of rectangle is 48 inches area is 40 inches what are the length of the side

Aloiza [94]

$\bf \textit{perimeter of a rectangle}=p=side+side+side+side
\\\\
or\implies p=w+w+l+l\implies p=2w+2l
\\\\
p=2(w+l)\qquad 
\begin{cases}
w=width\\
l=length\\
------\\
p=48
\end{cases}\implies 40=2(w+l)\\\\
-----------------------------\\\\
\textit{area of it}=A=w\cdot l\qquad \begin{cases}
w=width\\
l=length\\
------\\
A=40
\end{cases}\implies 40=wl\\\\
-----------------------------\\\\$

$\bf thus\qquad 
\begin{cases}
40=2(w+l)\to \frac{40}{2}=w+l\to\frac{40}{2}-l=\boxed{w}
\\\\
40=wl\\
--------------\\
40=\left( \boxed{\frac{40}{2}-l} \right)\cdot l
\end{cases}$

solve for "l" to find its length

7 0

3 years ago

Please help

Harrizon [31]

Answer:

(-1/2, 3/4)

Step-by-step explanation:

Let's use the elimination by adding or subtracting method. Note that we have 8y in the first equation, and that we could obtain -8y in the second equation by multiplying the second equation by 2:

2(24x - 4y = -15) => 48x - 8y = -30

Now combine this result (this equation) with the first equation:

2x + 8y = 5

+48x -8y = -30

---------------------

50x = - 25

Dividing both sides by 50, to isolate x, we get

x = -25/50 = -1/2.

Now substitute -1/2 for x in the first equation and solve the resulting equation for y:

2x + 8y = 5

2(-1/2) + 8y = 5, or -1 + 8y = 5, or 8y = 6 (after having added 1 to both sides)

Dividing both sides of 8y = 6 by 8 leads to determining the value of y:

y = 6/8 = 3/4

The solution is (-1/2, 3/4).

3 0

3 years ago

5cm<br> 6cm<br> 8cm<br> What is the perimeter of the following

Murljashka [212]

Answer:

19cm

Step-by-step explanation:

lets assume its a triangle

perimeter = 5cm + 6cm +8cm

7 0

2 years ago

Find dy/dx for y= x^3 ln (cot x)

ICE Princess25 [194]

<h3>Answer</h3>

$\dfrac{dy}{dx} = 3x^2 \ln(\cot x)-x^3 \csc(x)\sec(x)$

<h3>Explanation</h3>

By the product rule $(d/dx)(f(x)g(x)) = f(x)g'(x) + g(x)f'(x)$ , we have

$\begin{aligned}\frac{dy}{dx} &= \left(x^3 \ln (\cot x) \right)' \\&= x^3\big(\ln (\cot x)\big)' + \ln (\cot x) \cdot \left(x^3\right)' \end{aligned}$

By the chain rule:

$\begin{aligned}\big(\ln (\cot x)\big)' &= \dfrac{1}{\cot x} \cdot (\cot x)' \\ &= \dfrac{1}{\cot x} \cdot -\csc^2 x\\&= -\tan (x) \csc^2(x) \\&= - \frac{\sin x}{\cos x} \cdot \frac{1}{\sin^2 x} = - \frac{1}{\cos x} \cdot \frac{1}{\sin x} \\&= -\csc(x)\sec(x)\end{aligned}$

By the power rule:

$(x^3)' = 3x^2$

thus

$\begin{aligned}\frac{dy}{dx} &= x^3\big(\ln (\cot x)\big)' + \ln (\cot x) \cdot \left(x^3\right)' \\&= x^3\big( -\csc(x)\sec(x) \big) + \ln(\cot x) \cdot (3x^2) \\&= -x^3 \csc(x)\sec(x) + 3x^2 \ln(\cot x) \\&= 3x^2 \ln(\cot x)-x^3 \csc(x)\sec(x)\end{aligned}$

Nothing to do to simplify any further, other than factoring out $x^2$ .

4 0

3 years ago

Beginning from a depth of 35 ft below the surface, a whale swims upward and jumps to a height of nearly 17 feet above the surfac

Shtirlitz [24]

Answer: there is a proplem in here

Step-by-step explanation:

6 0

3 years ago