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3 years ago

I was just wondering if this was right most likely not

Mathematics

2 answers:

MariettaO [177]3 years ago

5 0

You’re right but you don’t have to write the + sign

y=4x-17

MrRa [10]3 years ago

4 0

Answer:

Yes you are correct :)

Step-by-step explanation:

Used a graphing calculator

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Kayden is a stunt driver. One time, during a gig where she escaped from a building about to explode(!), she drove at a constant

Anestetic [448]

Answer:

y = 25x - 85

Step-by-step explanation:

Every second (x) that passes by is 25 meters gained, and since she's 85 meters away from the safe distance still, the y intercept is -85.

8 0

2 years ago

Use a property to write an equivalent expression. 8 + y + 6

stepladder [879]

Answer:

14+y

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Factor completely. 150m^2nz+20mn^2c-120m^2nc-25mn^2z

Anika [276]

Answer:

The factor form is:

$5mn\left(6m-n\right)\left(5z-4c\right)$

Step-by-step explanation:

Here we have to find the factor of the expression:

$150m^2nz+20mn^2c-120m^2nc-25mn^2z$

So we need to take out the common terms, as we do for finding the greatest common factors.

Now the expression that is given can be re-written as:

$150m^2nz+20mn^2c-120m^2nc-25mn^2z\\=150mmnz+20mnnc-120mmnc-25mnnz\\=30\cdot \:5nmmz+4\cdot \:5nmnc-24\cdot \:5nmmc-5\cdot \:5nmnz$

Next, we will find the common terms, as follows;

$30\cdot \:5nmmz+4\cdot \:5nmnc-24\cdot \:5nmmc-5\cdot \:5nmnz\\=5nm\left(30mz+4nc-24mc-5nz\right)\\$

Now we will factorise the expression:

$\left(30mz+4nc-24mc-5nz\right)\\$

as follows;

$\left(30mz+4nc-24mc-5nz\right)\\=6m\left(5z-4c\right)+n\left(4c-5z\right)\\=\left(-4c+5z\right)\left(6m-n\right)\\$

So the final factor form is:

$150m^2nz+20mn^2c-120m^2nc-25mn^2z=5mn\left(6m-n\right)\left(5z-4c\right)$

8 0

3 years ago

Combine into a single logarithm. 3log(x+y)+2log(x-y)-log(x^2 +y^2)

seropon [69]

Answer:

$3\log _{10}\left(x+y\right)+2\log _{10}\left(x-y\right)-\log _{10}\left(x^2+y^2\right)=\log _{10}\left(\frac{\left(x+y\right)^3\left(x-y\right)^2}{x^2+y^2}\right)$

Step-by-step explanation:

Given the expression

$3log\left(x+y\right)+2log\left(x-y\right)-log\left(x^2\:+y^2\right)$

solving to write into a single logarithm

$3log\left(x+y\right)+2log\left(x-y\right)-log\left(x^2\:+y^2\right)$

$\mathrm{Apply\:log\:rule}:\quad \:a\log _c\left(b\right)=\log _c\left(b^a\right)$

$3\log _{10}\left(x+y\right)=\log _{10}\left(\left(x+y\right)^3\right)$

$=\log _{10}\left(\left(x+y\right)^3\right)+2\log _{10}\left(x-y\right)-\log _{10}\left(x^2+y^2\right)$

$\mathrm{Apply\:log\:rule}:\quad \:a\log _c\left(b\right)=\log _c\left(b^a\right)$

$2\log _{10}\left(x-y\right)=\log _{10}\left(\left(x-y\right)^2\right)$

$=\log _{10}\left(\left(x+y\right)^3\right)+\log _{10}\left(\left(x-y\right)^2\right)-\log _{10}\left(x^2+y^2\right)$

$\mathrm{Apply\:log\:rule}:\quad \log _c\left(a\right)+\log _c\left(b\right)=\log _c\left(ab\right)$

$\log _{10}\left(\left(x+y\right)^3\right)+\log _{10}\left(\left(x-y\right)^2\right)=\log _{10}\left(\left(x+y\right)^3\left(x-y\right)^2\right)$

$=\log _{10}\left(\left(x+y\right)^3\left(x-y\right)^2\right)-\log _{10}\left(x^2+y^2\right)$

$\mathrm{Apply\:log\:rule}:\quad \log _c\left(a\right)-\log _c\left(b\right)=\log _c\left(\frac{a}{b}\right)$

$\log _{10}\left(\left(x+y\right)^3\left(x-y\right)^2\right)-\log _{10}\left(x^2+y^2\right)=\log _{10}\left(\frac{\left(x+y\right)^3\left(x-y\right)^2}{x^2+y^2}\right)$

$=\log _{10}\left(\frac{\left(x+y\right)^3\left(x-y\right)^2}{x^2+y^2}\right)$

Thus,

$3\log _{10}\left(x+y\right)+2\log _{10}\left(x-y\right)-\log _{10}\left(x^2+y^2\right)=\log _{10}\left(\frac{\left(x+y\right)^3\left(x-y\right)^2}{x^2+y^2}\right)$

6 0

3 years ago

What is the solution to the opening sentence?

777dan777 [17]

Hello.

The answewr is: 3

Step 1: Subtract 2x from both sides.

4x−4−2x=2x+2−2x

2x−4=2

Step 2: Add 4 to both sides.

2x−4+4=2+4

2x=6

Step 3: Divide both sides by 2.

2x2=6/2

Have a nice day

7 0

4 years ago