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3 years ago

5th Grade Math Help!

Mathematics

2 answers:

Ede4ka [16]3 years ago

7 0

Answer:

Step-by-step explanation:

tatyana61 [14]3 years ago

4 0

Paula has 10 baseball cards

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Which interval for the graphed function has a local

Alex

Answer:

Interval for the function has local minimum of 0 is [2,4].

Step-by-step explanation:

You need to find the local minimum of 0 in the given function's graph.

In mathematics, local minimum is a point on a graph whose value is less than all other points near it.

See that, graph value 0 is lie on the x = 3 and in interval x = 2 to x = 4

So, final answer is :

Interval for the function has local minimum of 0 is [2,4].That's the final answer.

hope it helps

5 0

3 years ago

What is the measure of AB in O below?

8_murik_8 [283]

Answer:

The measure of an arc is equal to the angle opposite it

Therefore the answer is C. 100°

Hope this helps!

8 0

4 years ago

Let f(x) = 1/x^2 (a) Use the definition of the derivatve to find f'(x). (b) Find the equation of the tangent line at x=2

Verdich [7]

Answer:

(a) $f'(x)=-\frac{2}{x^3}$

(b) $y=-0.25x+0.75$

Step-by-step explanation:

The given function is

$f(x)=\frac{1}{x^2}$ .... (1)

According to the first principle of the derivative,

$f'(x)=lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

$f'(x)=lim_{h\rightarrow 0}\frac{\frac{1}{(x+h)^2}-\frac{1}{x^2}}{h}$

$f'(x)=lim_{h\rightarrow 0}\frac{\frac{x^2-(x+h)^2}{x^2(x+h)^2}}{h}$

$f'(x)=lim_{h\rightarrow 0}\frac{x^2-x^2-2xh-h^2}{hx^2(x+h)^2}$

$f'(x)=lim_{h\rightarrow 0}\frac{-2xh-h^2}{hx^2(x+h)^2}$

$f'(x)=lim_{h\rightarrow 0}\frac{-h(2x+h)}{hx^2(x+h)^2}$

Cancel out common factors.

$f'(x)=lim_{h\rightarrow 0}\frac{-(2x+h)}{x^2(x+h)^2}$

By applying limit, we get

$f'(x)=\frac{-(2x+0)}{x^2(x+0)^2}$

$f'(x)=\frac{-2x)}{x^4}$

$f'(x)=\frac{-2)}{x^3}$ .... (2)

Therefore $f'(x)=-\frac{2}{x^3}$ .

(b)

Put x=2, to find the y-coordinate of point of tangency.

$f(x)=\frac{1}{2^2}=\frac{1}{4}=0.25$

The coordinates of point of tangency are (2,0.25).

The slope of tangent at x=2 is

$m=(\frac{dy}{dx})_{x=2}=f'(x)_{x=2}$

Substitute x=2 in equation 2.

$f'(2)=\frac{-2}{(2)^3}=\frac{-2}{8}=\frac{-1}{4}=-0.25$

The slope of the tangent line at x=2 is -0.25.

The slope of tangent is -0.25 and the tangent passes through the point (2,0.25).

Using point slope form the equation of tangent is

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

$y-0.25=-0.25(x-2)$

$y-0.25=-0.25x+0.5$

$y=-0.25x+0.5+0.25$

$y=-0.25x+0.75$

Therefore the equation of the tangent line at x=2 is y=-0.25x+0.75.

5 0

4 years ago

Find the simple interest paid on a loan of $750 with an 18% rate over 2 years.

sleet_krkn [62]

Answer: $270

Step-by-step explanation:

I = prt

I = 0.18 x 750 x 2

I = 135 X 2

I = 270

6 0

4 years ago

Read 2 more answers

Find the possible values for m<1

Anuta_ua [19.1K]

0.5 that would be half of one if one is greater than the variable M

I really hope this helped ... have a great day !

4 0

3 years ago

5th Grade Math Help!​

5th Grade Math Help!