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3 years ago

15

Create an equivalent expression: 2(3x -1) - 3x +4 Show all steps please! In detail!

1 answer:

Vsevolod [243]3 years ago

5 0

Answer:

The answer is 3x+2

Step-by-step explanation:

2(3x-1)-3x+4

I did distributive property on 2(3x-1) how I did that was I multiplied 2 times 3x which is 6x (2 times 3x=6x). Then I multiplied 2 times -1 which equals -2 (2 times -1=-2).

new expression:

6x-2-3x+4

I combined all the numbers with the variables (6x and -3x) and by subtracting both of them you get 3x (6x-3x=3x). After that you combine -2 and 4 which equals 2 (-2+4=2).

This is now the final expression (your answer):

3x+2

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Write an algebraic expression for the word phrase six times a lenght w?

pychu [463]

<em>Depending on what you trying to say...
</em>
Six times a length (w) is 6w
or
Six times a length is w, is 6l=w

<em>But I can't tell because of the typo
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Hope this helps!

5 0

3 years ago

X=3 y=5 z=-2<br> Please help me with number 10

zlopas [31]

Answer:

26

Step-by-step explanation:

2*3*5-3(-2)+5(-2)

6*5+6-10

30+6-10

36-10

26

3 0

2 years ago

Read 2 more answers

What is the unit price if 4 pounds of fruit cost $6.48

Vaselesa [24]

6.48 / 4 =1.62 is your answer

6 0

3 years ago

What is the value of p?<br><br> A) 55<br><br> B) 45<br><br> C) 35<br><br> D) 125

Margarita [4]

Answer:

35

Step-by-step explanation:

just subtract 125 and 99

7 0

3 years ago

Consider the following equation. f(x, y) = y3/x, P(1, 2), u = 1 3 2i + 5 j (a) Find the gradient of f. ∇f(x, y) = Correct: Your

BaLLatris [955]

$f(x,y)=\dfrac{y^3}x$

a. The gradient is

$\nabla f(x,y)=\dfrac{\partial f}{\partial x}\,\vec\imath+\dfrac{\partial f}{\partial y}\,\vec\jmath$

$\boxed{\nabla f(x,y)=-\dfrac{y^3}{x^2}\,\vec\imath+\dfrac{3y^2}x\,\vec\jmath}$

b. The gradient at point P(1, 2) is

$\boxed{\nabla f(1,2)=-8\,\vec\imath+12\,\vec\jmath}$

c. The derivative of $f$ at P in the direction of $\vec u$ is

$D_{\vec u}f(1,2)=\nabla f(1,2)\cdot\dfrac{\vec u}{\|\vec u\|}$

It looks like

$\vec u=\dfrac{13}2\,\vec\imath+5\,\vec\jmath$

so that

$\|\vec u\|=\sqrt{\left(\dfrac{13}2\right)^2+5^2}=\dfrac{\sqrt{269}}2$

Then

$D_{\vec u}f(1,2)=\dfrac{\left(-8\,\vec\imath+12\,\vec\jmath\right)\cdot\left(\frac{13}2\,\vec\imath+5\,\vec\jmath\right)}{\frac{\sqrt{269}}2}$

$\boxed{D_{\vec u}f(1,2)=\dfrac{16}{\sqrt{269}}}$

7 0

3 years ago