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zimovet [89]
2 years ago
10

Which is the x-intercept of the graph of the equation y=3x–9?

Mathematics
1 answer:
Vinvika [58]2 years ago
5 0
Y=3x-9
0=3x-9
-3x=-9
x=3
X-int = 3
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Solve the equation and check your solution. (If all real numbers are solutions, enter REALS. If there is no solution, enter NO S
shepuryov [24]

Answer:

y=6/5 or 1 1/5

Step-by-step explanation:

combine like terms

5y+1= 10y-5

subtract 5y from both sides

1=5y-5

add 5 to both sides

6=5y

divide by 5 on both sides

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5 0
3 years ago
What is 3·3÷3·1²·5 and for any body who doesn't get it Ur not dumb lol
r-ruslan [8.4K]

I'll give you a hint. <u><em>This questions is about order of operations for pemdas. Its stand for: p-parenthesis, e-exponents, m-multiply, d-divide, a-add and s-subtract that came from left to right.</em></u>

First you had to calculate with exponents first.

1^2=1*1=1

3*3/3*1*5

Then you multiply and divide from left to right.

3*3=9

9/3*1*5

3*1*5

3*5

=15

Final answer: \boxed{=15}

Hope this helps!

And thank you for posting your question at here on brainly, and have a great day.

-Charlie

7 0
3 years ago
Read 2 more answers
A. Show that the vector v = ai + bj is perpendicular to the line ax + by = c by establishing that the slope of v is the negative
pishuonlain [190]

Answer:

Part A:

m_1m_2=-1

\frac{b}{a}(\frac{-a}{b})=-1\\-1=-1

Hence proved that Vector= ai + bj is perpendicular to the line ax + by = c.

Part B:

Slope of vector = \frac{b}{a}

Step-by-step explanation:

Condition for perpendicular is:

m_1m_2=-1

Part A:

Consider the vector v = ai + bj

x component of vector=a

y component of vector=b

Slope of vector=m_1=\frac{y}{x}=\frac{b}{a}

Consider the line ax + by = c:

Rearranging the equation:

ax+by=c

by=c-ax

y=\frac{-ax}{b}+\frac{c}{b}

According to general equation of line: y=mx+c

Where m is the slope

In our case the slope of above line is:

m_2=\frac{-a}{b}

According to the condition of perpendicular:

m_1m_2=-1

\frac{b}{a}(\frac{-a}{b})=-1\\-1=-1

Hence proved that Vector= ai + bj is perpendicular to the line ax + by = c.

Part B:

Slope of vector is also calculated above.

Since the slope of vector is negative reciprocal of the slope of the given line:

According to equation of line ax + by = c

y=\frac{-ax}{b}+\frac{c}{b}

According to  general equation of line: y=mx+c

Where m is the slope

Slope of given line=m=\frac{-a}{b}

negative reciprocal of the slope of the given line = \frac{b}{a}

Slope of vector = \frac{b}{a}

5 0
3 years ago
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