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olchik [2.2K]
2 years ago
10

Find the x-intercept and the y-intercept for the graph of y=3x-4

Mathematics
1 answer:
3241004551 [841]2 years ago
6 0
X intercept is when y = 0
0 = 3x - 4
4 = 3x, x = 4/3
y intercept is when x = 0
y = 3(0) - 4
y = 0-4, y = -4
Solution: x intercept: (4/3,0)
y intercept: (0,-4)
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Put answer in coordinate form: 2x + 2y = -2 and 9x - 6y = 36
zhannawk [14.2K]
Coordinate form is (x,y) using the variables respectively.

2x+2y=-2
-2y
2x= -2-2y
/2

x=-1-y

9x-6y=36
9 (-1-y)-6y=36
-9-9y-6y=36
+9.
-9y-6y=45
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/-15
y=-3

x=-1-y
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answer: (2,-3)
5 0
3 years ago
3sina-4sin4a/sin5a<br> if a=-45 degree
Ierofanga [76]

Given:

The given expression is:

\dfrac{3\sin a-4\sin 4a}{\sin 5a}

To find:

The value of the given expression at a=-45.

Solution:

We have,

\dfrac{3\sin a-4\sin 4a}{\sin 5a}

Substituting a=-45, we get

\dfrac{3\sin (-45)-4\sin [4(-45)]}{\sin [5(-45)]}

=\dfrac{-3\sin (45)+4\sin (180)}{-\sin (225)}

=\dfrac{-3\sin (45)+4\sin (180)}{-\sin (180+45)}

=\dfrac{-3\sin (45)+4\sin (180)}{\sin (45)}

On substituting \sin (180), we get,

=\dfrac{-3\sin (45)+4(0)}{\sin (45)}

=\dfrac{-3\sin (45)+0}{\sin (45)}

=\dfrac{-3\sin (45)}{\sin (45)}

=-3

Therefore, the value of the given expression at  a=-45 is -3.

8 0
2 years ago
Simplify <br> cosx ( tanx+cotx)
Soloha48 [4]
The first step to solving this is to use tan(t) = \frac{sin(t)}{cos(t)} to transform this expression.
cos(x) × ( \frac{sin(x)}{cos(x)} + cot(x) )
Using cot(t) = \frac{cos(x)}{sin(x)},, transform the expression again.
cos(x) × ( \frac{sin(x)}{cos(x)} +  \frac{cos(x)}{sin(x)} )
Next you need to write all numerators above the least common denominator (cos(x)sin(x)).
cos(x) × \frac{sin(x)^{2} + cos(x)^{2}  }{cos(x)sin(x)}
Using sin(t)² + cos(t)² = 1,, simplify the expression. 
cos(x) × \frac{1}{cos(x)sin(x)}
Reduce the expression with cos(x).
\frac{1}{sin(x)}
Lastly,, use \frac{1}{sin(t)} = csc(t) to transform the expression and find your final answer.
csc(x)
This means that the final answer to this expression is csc(x).
Let me know if you have any further questions.
:)
4 0
2 years ago
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