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3 years ago

The inequality x ≥ −10 represents the possible values for x on a number line.

Mathematics

2 answers:

tatiyna3 years ago

5 0

-10 btw it uses that for the answer

Leokris [45]3 years ago

3 0

Answer:

-11 and lower

Step-by-step explanation:

it's saying that -10 is as low as the number can possibly be or equal to it.

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What is the area of the triangle 13,13,10

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Answer:

depends

Step-by-step explanation:

it depends on the lengths of the base and height since the area of a triangle is A=bh / 2

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3 years ago

Which equation could be used to solve the question below?

ratelena [41]

Answer:

Step-by-step explanation:

83+37=129, 120÷24=5

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4 years ago

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A scale drawing of a rectangular park is 5 inches wide and 7 inches long. The actual park is 280 yards long. What is the area of

jekas [21]

Answer:

area=56000 yards²

Step-by-step explanation:

first divide 280 by 7 and get 40

scale factor is 40

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200x280=56000

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4 years ago

In a candy box, there are 5 times as many candies with nuts as candies without nuts. There are 30 candies in the box. How many c

Luden [163]

Answer:

Step-by-step explanation:

because there are 5 time as many candies with nuts as candies without nuts and there is 18 so it will be 18×5=90

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3 years ago

Finding an Equation of a tangent Line in Exercise, find an equation of the tangent line to the graph of the function at the give

frutty [35]

Answer:

$y=\dfrac{3x}{e}+\dfrac{4}{e}$

this is the equation of the tangent at point (-1,1/e)

Step-by-step explanation:

to find the tangent line we need to find the derivative of the function g(x).

$g(x) =e^{x^3}$

we know that $\frac{d}{dx}(e^{f(x)})=e^{f(x)}f'(x)$

$g'(x) =e^{x^{3}}(3 x^{2})$

$g'(x) =3 x^{2} e^{x^{3}}$

this the equation of the slope of the curve at any point x and it also the slope of the tangent at any point x. hence, g'(x) can be denoted as 'm'

to find the slope at (-1,1/e) we'll use the x-coordinate of the point i.e. x = -1

$m =3 (-1)^{2} e^{(-1)^{3}}\\m =3e^{-1}\\m=\dfrac{3}{e}$

using the equation of line:

$(y-y_1)=m(x-x_1)$

we'll find the equation of the tangent line.

here (x1,y1) =(-1,1/e), and m = 3/e

$(y-\dfrac{1}{e})=\dfrac{3}{e}(x+1)\\y=\dfrac{3x}{e}+\dfrac{3}{e}+\dfrac{1}{e}\\$

$y=\dfrac{3x}{e}+\dfrac{4}{e}$

this is the equation of the tangent at point (-1,1/e)

3 0

3 years ago