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3 years ago

Solve for k. Describe the steps taken to find the solution. 8 + 10 - 6k = 6(4 - k)

Mathematics

2 answers:

BartSMP [9]3 years ago

5 0

Answer:

no solutions

Step-by-step explanation:

8 + 10 - 6k = 6(4 - k)

Distribute

8 + 10 - 6k = 24 -6k

Combine like terms

18 -6k = 24 -6k

Add 6k to each side

18 - 6k +6k = 24 - 6k+6k

18 = 24

This is never true so there are no solutions

Aliun [14]3 years ago

5 0

Hey there! I'm happy to help you out!

To solve for k, we have to isolate it on one side of the equation and put all the constants on the other side.

The first thing we need to do is get the k on the right side out of parentheses. To do this, we will use the distributive property!

8+10-6k=24-6k

Now, we will combine like terms.

18-6k=24-6k

We want to get all the ks on the left side and all the constants on the right. Let's add six k to both sides to cancel out the -6k on the right.

18=24

As you can see, our variables canceled out. This is because our equation was saying 18 minus six of something was equal to 24 minus six of something, which makes no sense because that doesn't work!

We are left with 18=24, which isn't even a true equation! Therefore, there is no solution to this equation.

I hope that this helps! Have a wonderful day!

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3 years ago

1) After a dilation, (-60, 15) is the image of (-12, 3). What are the coordinates of the image of (-2,-7) after the same dilatio

BARSIC [14]

Answer:

a) k = 5; (-10, -35)

Step-by-step explanation:

Given:

Co-ordinates:

Pre-Image = (-12,3)

After dilation

Image = (-60,15)

The dilation about the origin can be given as :

Pre-Image $(x,y)\rightarrow Image(kx,ky)$

where $k$ represents the scalar factor.

We can find value of $k$ for the given co-ordinates by finding the ratio of $x$ or $y$ co-ordinates of the image and pre-image.

$k=\frac{Image}{Pre-Image}$

For the given co-ordinates.

Pre-Image = (-12,3)

Image = (-60,15)

The value of $k=\frac{-60}{-12}=5$

or $k=\frac{15}{3}=5$

As we get $k=5$ for both ratios i.e of $x$ and $y$ co-ordinates, so we can say the image has been dilated by a factor of 5 about the origin.

To find the image of (-2,-7), after same dilation, we will multiply the co-ordinates with the scalar factor.

Pre-Image $(-2,-7)\rightarrow$ Image $((-2\times5),(-7\times 5))$

Pre-Image $(-2,-7)\rightarrow$ Image $(-10,-35)$ (Answer)

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3 years ago

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strojnjashka [21]

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