The height of a triangle is twice the length of its base. The

Mathematics

1 answer:

lions [1.4K]3 years ago

6 0

Answer:

height = 14.2 m , base = 7.1 m

Step-by-step explanation:

let the base be b then height h = 2b

The area (A) of a triangle is calculated as

A = $\frac{1}{2}$ bh

Here A = 50 , then

$\frac{1}{2}$ × b × 2b = 50

b² = 50 ( take square root of both sides )

b = $\sqrt{50}$ ≈ 7.1 ( to the nearest tenth )

Then base = 7.1 m and height = 2 × 7.1 = 14.2 m

You might be interested in

Verify the identity: tan(x+pi/2)=-cotx please!!

olya-2409 [2.1K]

First we use sin(a+b)= sinacosb+sinbcosa
and cos(a+b)=cosa cosb -sinasinb

tan(x+pi/2)= sin(x+pi/2) / cos(x+pi/2)
and sin(x+pi/2) = sinxcospi/2 + sinpi/2cosx =cosx,
cos(x+pi/2) = cosxcospi/2- sinxsinpi/2= - sinx,
 because cospi/2 =0, and sinpi/2=1

=tan(x+pi/2)= sin(x+pi/2) / cos(x+pi/2)= cosx / -sinx = -1/tanx = -cotx
from where tan(x+pi/2)=-cotx

5 0

3 years ago

Use Newton’s Method to find the solution to x^3+1=2x+3 use x_1=2 and find x_4 accurate to six decimal places. Hint use x^3-2x-2=

luda_lava [24]

Let $f(x) = x^3 - 2x - 2$ . Then differentiating, we get

$f'(x) = 3x^2 - 2$

We approximate $f(x)$ at $x_1=2$ with the tangent line,

$f(x) \approx f(x_1) + f'(x_1) (x - x_1) = 10x - 18$

The $x$ -intercept for this approximation will be our next approximation for the root,

$10x - 18 = 0 \implies x_2 = \dfrac95$

Repeat this process. Approximate $f(x)$ at $x_2 = \frac95$ .

$f(x) \approx f(x_2) + f'(x_2) (x-x_2) = \dfrac{193}{25}x - \dfrac{1708}{125}$

Then

$\dfrac{193}{25}x - \dfrac{1708}{125} = 0 \implies x_3 = \dfrac{1708}{965}$

Once more. Approximate $f(x)$ at $x_3$ .

$f(x) \approx f(x_3) + f'(x_3) (x - x_3) = \dfrac{6,889,342}{931,225}x - \dfrac{11,762,638,074}{898,632,125}$

Then

$\dfrac{6,889,342}{931,225}x - \dfrac{11,762,638,074}{898,632,125} = 0 \\\\ \implies x_4 = \dfrac{5,881,319,037}{3,324,107,515} \approx 1.769292663 \approx \boxed{1.769293}$