1. You will have to find the height first which could be done by using pythagoras theorem

split the 12 by half giving 6
a^2 = b^2 + c^2
13^2 = 6^2+c^2
169 = 36+c^2
c^2=169-36
c^2=133
c=sqrt133
c= 11.53

area of triangle
1/2 x b x h
1/2 x 6 x 11.53
34.59
34.5(3 sf)

5 0

3 years ago

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In 1934 there was an extreme drought and the Great Plains and the number 1,934 is the value of the nine and the hundredth place

WARRIOR [948]

It should be noted that No. the value of the 9 in the hundred place is not ten times the value of 3 in the tens place.

<h3>How to illustrate the information?</h3>

It should be noted that the value of 3 in 1934 is 30 and the value of 9 is 900.

1934 = 1000 + 900 + 30 + 4

Therefore the value of 9 will be:

= 30 × 3

Therefore, the the value of the 9 in the hundred place is 30 times the value of 3 in the tens place.

Learn more about place values on:

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5 0

1 year 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

3 years ago

I need this asap very easy!!! Please serious answers please!! I will mark brainliest if right

Lelechka [254]

Answer:

1. no

2. 36+196=232 c=441 Obtuse.

3. x is greater than 3, less than 21.

Step-by-step explanation:

5 0

2 years ago

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