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docker41 [41]
3 years ago
11

(a) A straight line L, whose equation is 3y - 2x = -2 meets the x-axis at R Determine the co-ordinates of R.

Mathematics
1 answer:
gtnhenbr [62]3 years ago
3 0

Answer:

(1 , -2/3)

Step-by-step explanation:

when y =0

3×0-2x=-2

0-2x=-2

divide both side by -2

x=1

when x=0

3y-2×0=-2

3y-0=-2

divide both side by 3

y=-2/3

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Which of the following ( 15, 20, 30, 57 ) is the product of exactly three different prime numbers?
Andrei [34K]

Answer:

30=2*3*5, so it's 30

Step-by-step explanation:

15=3*5, so it's not 15

20=2*2*5, so it's not 20

30=2*3*5, so it's 30

57=19*3, so it's not 57

3 0
2 years ago
What is the derivative of g(x)=e^(x^2+2x)+3x
UNO [17]
Ok first we can split it in two : e^{x^2+2x} and 3x.

The derivative of 3x is 3.

For the first part, we use the chain rule : [f(g(x))]'=g'(x)f'(g(x)) hence (e^{x^2+2x})'=(x^2+2x)'e^{x^2+2x} (since the derivative of the exponential is itself) hence g'(x)=(2x+2)e^{x^2+2x}+3
7 0
3 years ago
Read 2 more answers
Find the area of the following parallelogram. pleasee helppp! ​
Pani-rosa [81]

Answer:

1736

Step-by-step explanation:

Area= b x h

      = 28 x 62

      = 1736 m

7 0
3 years ago
I need help with this
larisa [96]

Answer:

Not similar

Step-by-step explanation:

Even though

8/2=16/4,

10/3 does not have any relation to the rest. Still make sure if im right, but hope this helps!

6 0
3 years ago
SOLVE. integration of (1-v) /(1+v^2)
Nutka1998 [239]
\displaystyle\int\frac{1-v}{1+v^2}\,\mathrm dv=\int\frac{\mathrm dv}{1+v^2}-\int\frac v{1+v^2}\,\mathrm dv

The first integral is already standard and has an antiderivative in terms of \arctan v. For the second integral, take w=1+v^2 so that \dfrac{\mathrm dw}2=v\,\mathrm dv. Then

\displaystyle\int\frac{\mathrm dv}{1+v^2}-\int\frac v{1+v^2}\,\mathrm dv=\arctan v-\frac12\int\frac{\mathrm dw}w
=\arctan v-\dfrac12\ln|w|+C
=\arctan v-\dfrac12\ln(1+v^2)+C
4 0
3 years ago
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