Identify the property that justifies step 1 when solving the equation

Mathematics

1 answer:

Ierofanga [76]2 years ago

3 0

Answer:

Distributive Property

Step-by-step explanation:

Answer choice a is correct because you are distributing 2 to each term in the parentheses. When you multiply 2 to each factor in the parentheses, it simplifies to 2x-2, thus making the new equation 6x+2x-2=30

You might be interested in

How to do this cause I'm very lost

Jlenok [28]

You put the 0 if it is close to 0, the 1/2 if it is closer to half, 1 if it is equal or closer to one, and the >1 if it is over one

5 0

2 years ago

Divide a 5-inch line into two parts so that one part (a) 2 1/4 inches shorter than the other; (b) 3 times the other.

Ilia_Sergeevich [38]

Part a), solve this system of equations:
x + y = 5
x - y = 2.25
2x = 7.25
x = 3.625
y = 1.375

part b), solve this system of equations:
x + y = 5
x = 3y
y = 1.25
x = 3.75

3 0

2 years ago

Read 2 more answers

Help math question derivative!

atroni [7]

Let $f(x)=\sec^{-1}x$ . Then $\sec f(x)=x$ , and differentiating both sides with respect to $x$ gives

$(\sec f(x))'=\sec f(x)\tan f(x)\,f'(x)=1$
$f'(x)=\dfrac1{\sec f(x)\tan f(x)}$

Now, when $x=\sqrt2$ , you get

$(\sec^{-1})'(\sqrt2)=f'(\sqrt2)=\dfrac1{\sec\left(\sec^{-1}\sqrt2\right)\tan\left(\sec^{-1}\sqrt2\right)}$

You have $\sec^{-1}\sqrt2=\dfrac\pi4$ , so $\sec\left(\sec^{-1}\sqrt2\right)=\sqrt2$ and $\tan\left(\sec^{-1}\sqrt2\right)=1$ . So $(\sec^{-1})'(\sqrt2)=\dfrac1{\sqrt2\times1}=\dfrac1{\sqrt2}$