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anzhelika [568]
3 years ago
11

Slope = 2/5 and passes through (-3, 1)

Mathematics
2 answers:
Nutka1998 [239]3 years ago
4 0

Answer:

y = 2/5x + 1/5

Step-by-step explanation:

y - y1 = m(x - x1)

y - 1 = 2/5(x - (-3))

y - 1 = 2/5(x + 3)

y - 1 = 2/5x + 6/5

y = 2/5x + 6/5 - 5/5 (since 1 equals to 1/1, 2/2, 3/3, etc.)

y = 2/5x + 1/5

Anettt [7]3 years ago
4 0
Y=1/5 and slope of 2/5
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4 years ago
What are the approximate values of the minimum and maximum points of f(x) = x5 − 10x3 + 9x on [-3,3]?
nika2105 [10]

Answer:

Minimum : -37 at x=2.4 and

Maximum = 37 at x=-2.4.

Step-by-step explanation:

Given:

f(x)=x^5-10x^3+9x; [-3,3]

Explanation:

In order to find minimum/maximum of a function, we need to find the first derivative of the function and then set it equal to 0 to get critical points.

Therefore,

f'(x)=5x^4-30x^2+9

Setting derivative equal to 0, we get

5x^4-30x^2+9=0

On applying quadratic formula, we get

x=2.4, -2.4, -0.7, 0.7.

So, those are critical points of the given function.

Plugging the values x=2.4, -2.4, -0.7, 0.7, -3 and 3 in above function, we get

f(2.4)=(2.4)^5-10(2.4)^3+9(2.4)= -37.01376   : Minimum.

f(-2.4)=(-2.4)^5-10(-2.4)^3+9(-2.4)= 37.01376 : Maximum.

f(0.7)=(0.7)^5-10(0.7)^3+9(0.7) = 3.03807

f(-0.7)=(-0.7)^5-10(-0.7)^3+9(-0.7) = -3.03807

f(-3)=(-3)^5-10(-3)^3+9(-3) =0

f(3)=(3)^5-10(3)^3+9(3) =0

Therefore the approximate values of the minimum and maximum points of f(x) = x^5- 10x^3+ 9x on [-3,3] are:

Minimum : -37 at x=2.4 and

Maximum = 37 at x=-2.4.


7 0
4 years ago
What’s an equivalent expression to 5(2x-1)
netineya [11]

Answer:

10x-5

Step-by-step explanation:

Because u can simplify the equation into a shorter one by simplifying the numbers in paranthesis with 5 and 5 times 2 equal 10 as 5 times 1 equal 5

6 0
3 years ago
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