3 years ago

Finding coordinates on a curve given a line parallel to the tangent of the curve.

Mathematics

1 answer:

creativ13 [48]3 years ago

4 0

I dont think you can do this without finding derivatives.

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3b+4b^2-a^3;a=3 and b=-5

Alik [6]

Equation: 3b+4b^2-a^3
Given: a=3 & b=-5

Plug 3 in for (a) and -5 in for (b):
3(-5)+4(-5)^2-(3)^3

Now follow PEMDAS:
3(-5) + 4(-5)^2 - (3)^3
-15 + 4(-5)^2 - (3)^3
-15 + 100 - (3)^3
-15 + 100 - 27
85 - 27
58
the asnwer is 58.

8 0

3 years ago

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if you found the percent of a number and the product is greater than the number, what do you know about the percent? explain.

Ket [755]

Well you can see it would be mixed number and it is not a number less than 1

5 0

4 years ago

I got all these wrong. Can someone please help!!!

anzhelika [568]

I need to get a high amount of points so pls dont delete comment

4 0

3 years ago

What is the answer for 6r^2-10r-24

FrozenT [24]

$6r^2-10r-24 =0\\ \\a=6, \ b= -10, \ c= -24 \\ \\ \Delta =b^2-4ac = (-10)^2 -4\cdot6\cdot (-24) = 100+576 = 676 \\ \\ r_{1}=\frac{-b-\sqrt{\Delta }}{2a}=\frac{10-\sqrt{676}}{2\cdot 6 }=\frac{ 10-26}{12}= \frac{-16}{12}=- \frac{4}{3} \\ \\r_{2}=\frac{-b+\sqrt{\Delta }}{2a}= \frac{10+\sqrt{676}}{2\cdot 6 }=\frac{ 10+26}{12}= \frac{36}{12}=3\\ \\ 6r^2-10r-24 = 6(r+\frac{4}{3})(r-3)=(6r+8)(r-3)\\ \\ \\ Answer : \ 6r^2-10r-24 =(6r-8)(r-3)$

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4 years ago

What is the greatest common factor of 4, 37, and 51

klio [65]

there and no common prime factors the GCF is 1 :)

4 0

3 years ago