All categories

3 years ago

Do we use Pythagorean theorem to find missing sides or missing angles in a right triangle

Mathematics

1 answer:

Marina86 [1]3 years ago

8 0

To find missing lengths of sides
hope this helped have a nice day

You might be interested in

I GIVE BRAINLIEST PLEASE EXPLAIN What is the square root of 121

andre [41]

The square root of 121 is 11 because 11 times 11 is 121.

6 0

3 years ago

Read 2 more answers

1. Factor the polynomials given one zero. a) f(x)=xº+3x-34 x+48; 3

Mars2501 [29]

Step-by-step explanation:

=x+3x+34x-34+48

=38x+14

5 0

2 years ago

Find the derivative of the following. please show the steps when you answer :1) f(x) = 8xe^x2) y= 5xe^x^43) f(x)= x^8+5/x4) f(t)

borishaifa [10]

Answer:

Since,

$\frac{d}{dx}x^n = nx^{n-1}$

$\frac{d}{dx}(f(x).g(x)) = f(x).\frac{d}{dx}(g(x)) + g(x).\frac{d}{dx}(f(x))$

$\frac{d}{dx}(\frac{f(x)}{g(x)})=\frac{g(x).f'(x) - f(x) g'(x)}{(g(x))^2}$

1) $y = 8x e^x$

Differentiating with respect to x,

$\frac{dy}{dx}=8( x \times e^x + e^x) = 8(xe^x + e^x) = 8e^x(x+1)$

2) $y = 5x e^{x^4}$

Differentiating w. r. t x,

$\frac{dy}{dx}=5(x\times 4x^3 e^{x^4}+e^{x^4})=5e^{x^4}(4x^4+1)$

3) $y = x^8 + \frac{5}{x^4}$

Differentiating w. r. t. x,

$\frac{dy}{dx}=8x^7 - \frac{5}{x^5}\times 4 = 8x^7 - \frac{20}{x^5}=\frac{8x^{12}-20}{x^5}$

4) $f(t) = te^{11}-6t^5$

Differentiating w. r. t. t,

$f'(t) = e^{11} - 30t^4$

5) $g(p) = p\ln(2p+3)$

Differentiating w. r. t. p,

$g'(p) = p\frac{1}{2p+3}(2) + \ln(2p+3) = \frac{2p}{2p+3}+\ln(2p+3)$

6) $z = (te^{6t}+e^{5t})^7$

Differentiating w. r. t. t,

$\frac{dz}{dt}=7(te^{6t}+e^{5t})^6 ( 6te^{6t}+e^{6t} + 5e^{5t})$

7) $w =\frac{2y + y^2}{7+y}$

Differentiating w. r. t. y,

$\frac{dw}{dy} = \frac{(7+y)(2+2y)-(2y+y^2)}{(7+y)^2} = \frac{14 + 2y + 14y +2y^2 - 2y - y^2}{(7+y)^2}=\frac{14+14y+y^2}{(7+y)^2}$

7 0

3 years ago

Prove the statement holds for all positive integers: 2 + 4 + 6 + ... + 2n = n² + n

crimeas [40]

Proving a relation for all natural numbers involves proving it for n = 1 and showing that it holds for n + 1 if it is assumed that it is true for any n.

The relation 2+4+6+...+2n = n^2+n has to be proved.

If n = 1, the right hand side is equal to 2*1 = 2 and the left hand side is equal to 1^1 + 1 = 1 + 1 = 2

Assume that the relation holds for any value of n.

2 + 4 + 6 + ... + 2n + 2(n+1) = n^2 + n + 2(n + 1)

= n^2 + n + 2n + 2

= n^2 + 2n + 1 + n + 1

= (n + 1)^2 + (n + 1)

This shows that the given relation is true for n = 1 and if it is assumed to be true for n it is also true for n + 1.

By mathematical induction the relation is true for any value of n.

6 0

3 years ago

Plz Help me 50 points to who ever answers firstt

artcher [175]

Answer:

$\boxed{\bold{57 \ Ft.}}$

Step-by-step explanation:

Assignment: Find area of rectangle without triangle

Area of whole rectangle including triangle: 66 Feet

(6 * 11 = 66)

Area of Triangle: $\bold{\frac{1}{2}Base \ \cdot \ Height }$

Height: 3

Base: 6 (Solved by using this: (11 - 2 + 3))

Half of base: 3

3 * 3 = 9

66 - 9 = 57

4 0

3 years ago