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3 years ago

If g(x) = 3x is graphed on a coordinate plane, what is the y-intercept of the graph? 0

Mathematics

2 answers:

madam [21]3 years ago

5 0

You have to solve 3x = 0, which is at x=0. The y intercept is the point (0,0), also known as the origin.

kvv77 [185]3 years ago

5 0

Answer:

(0,0)

Step-by-step explanation:

To find the y-intercept of a graph, you need to make x = 0 and then solve the equation for y.

In this case we have g(x) = 3x. This can also be written as y = 3x

So we're going to make x = 0 and solve for y:

$y=3x\\y=3(0)\\y=0$

Thus, the y-intercept of the graph is (0,0)

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2 years ago

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A pair of boots is 30dollars more than a pair of sandals. A pair of sandals cost s dollars each. Emily purchased 2 pairs of boot

babunello [35]

Answer: The expression representing the amount of money spent is $5x+60$ by Emily.

Step-by-step explanation:

Since we have given that

Let the cost price of per pair of sandals be 's'

Let the cost price of per pair of boots be 's+30'.

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Total cost would be

$2(s+30)+3s\\\\=2s+60+3s\\\\=5s+60$

Hence, the expression representing the amount of money spent is $5x+60$ by Emily.

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3 years ago

Pls help if pls do most important 15 number 20 number 26 number 28 number 30 number 17 number 25 number 22 number 24 number if u

Tamiku [17]

Answer:

Step-by-step explanation:

15. $\frac{cotx+1}{cotx-1} = \frac{cotx+cotx.tanx}{cotx-cotx.tanx} = \frac{cotx(1+tanx)}{cotx(1-tanx)} = \frac{1+tanx}{1-tanx} \\$

16. $\frac{1+cos\alpha }{sin\alpha } = \frac{sin\alpha }{1-cos\alpha }$

$=> (1+cos\alpha )(1-cos\alpha ) = sin^{2} \alpha \\=> 1-cos^{2}\alpha =sin^{2} \alpha \\=> sin^{2}\alpha +cos^{2}\alpha =1\\$

=> The clause is correct

17. Do the same (16)

18. $\frac{sinx}{1-cosx}+\frac{sinx}{1+cosx} = \frac{sinx(1+cosx) + sinx(1-cosx)}{(1+cosx)(1-cosx)} = \frac{sinx + sinx.cosx + sinx - sinx.cosx}{1-cos^{2}x} = \frac{2sinx}{sin^{2}x }= \frac{2}{sinx} = 2cosecx$

19.

$\frac{sinA}{1+cosA} + \frac{1+cosA}{sinA}=\frac{2}{sinA} \\=> \frac{sinA}{1+cosA} + \frac{1+cosA}{sinA} - \frac{2}{sinA} \\ = 0\\=> \frac{sinA}{1+cosA} + (\frac{1+cosA}{sinA} - \frac{2}{sinA} \\) = 0\\=> \frac{sinA}{1+cosA} + \frac{1+cosA-2}{sinA} = 0\\=> \frac{sinA}{1+cosA} +\frac{cosA-1}{sinA} = 0\\=> \frac{sin^{2} A+(cosA-1)(1+cosA)}{(1+cosA)sinA}=0\\=> \frac{sin^{2} A+cos^{2} A-1}{(1+cosA)sinA}=0\\=> \frac{0}{(1+cosA)sinA} =0\\$

=> The clause is correct

20. Do the same (18)

21. Left = $\frac{1}{cosecA+cotA} = \frac{1}{\frac{1}{sinA}+\frac{cosA}{sinA} } = \frac{1}{\frac{1+cosA}{sinA} }= \frac{sinA}{1+cosA}$

Right = $cosecA-cotA = \frac{1}{sinA}-\frac{cosA}{sinA}= \frac{1-cosA}{sinA} \\$

Same with (16) => Left = Right => The clause is correct

22. Do the same (21)

Too long, i'm so lazy :))))

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2 years ago