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3 years ago

A square is put together by two triangle the hypotenuse is 11cm and width of 8cm find the height of the triangle

Mathematics

1 answer:

Alecsey [184]3 years ago

5 0

Answer:

Step-by-step explanation:

by Pythagoras theorem ,

h^2 = p^2 + b^2

11^2 = p^2 + 8^2

p^2 = 121 - 64 = 57

p =√57 = 7.54cm.

(hope this helps can i plz have brainlist :D hehe)

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3 years ago

The equation giving a family of ellipsoids is u = (x^2)/(a^2) + (y^2)/(b^2) + (z^2)/(c^2) . Find the unit vector normal to each

Fynjy0 [20]

Answer:

$\hat{n}\ =\ \ \dfrac{\dfrac{x}{a^2}\hat{i}+\ \dfrac{y}{b^2}\hat{j}+\ \dfrac{z}{c^2}\hat{k}}{\sqrt{(\dfrac{x}{a^2})^2+(\dfrac{y}{b^2})^2+(\dfrac{z}{c^2})^2}}$

Step-by-step explanation:

Given equation of ellipsoids,

$u\ =\ \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}$

The vector normal to the given equation of ellipsoid will be given by

$\vec{n}\ =\textrm{gradient of u}$

$=\bigtriangledown u$

$=\ (\dfrac{\partial{}}{\partial{x}}\hat{i}+ \dfrac{\partial{}}{\partial{y}}\hat{j}+ \dfrac{\partial{}}{\partial{z}}\hat{k})(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2})$

$=\ \dfrac{\partial{(\dfrac{x^2}{a^2})}}{\partial{x}}\hat{i}+\dfrac{\partial{(\dfrac{y^2}{b^2})}}{\partial{y}}\hat{j}+\dfrac{\partial{(\dfrac{z^2}{c^2})}}{\partial{z}}\hat{k}$

$=\ \dfrac{2x}{a^2}\hat{i}+\ \dfrac{2y}{b^2}\hat{j}+\ \dfrac{2z}{c^2}\hat{k}$

Hence, the unit normal vector can be given by,

$\hat{n}\ =\ \dfrac{\vec{n}}{\left|\vec{n}\right|}$

$=\ \dfrac{\dfrac{2x}{a^2}\hat{i}+\ \dfrac{2y}{b^2}\hat{j}+\ \dfrac{2z}{c^2}\hat{k}}{\sqrt{(\dfrac{2x}{a^2})^2+(\dfrac{2y}{b^2})^2+(\dfrac{2z}{c^2})^2}}$

$=\ \dfrac{\dfrac{x}{a^2}\hat{i}+\ \dfrac{y}{b^2}\hat{j}+\ \dfrac{z}{c^2}\hat{k}}{\sqrt{(\dfrac{x}{a^2})^2+(\dfrac{y}{b^2})^2+(\dfrac{z}{c^2})^2}}$

Hence, the unit vector normal to each point of the given ellipsoid surface is

$\hat{n}\ =\ \ \dfrac{\dfrac{x}{a^2}\hat{i}+\ \dfrac{y}{b^2}\hat{j}+\ \dfrac{z}{c^2}\hat{k}}{\sqrt{(\dfrac{x}{a^2})^2+(\dfrac{y}{b^2})^2+(\dfrac{z}{c^2})^2}}$

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3 years ago

if you deposit $8000 into an account paying 9% annual interest compunded semi anually how long will it take for your money to do

Anna007 [38]

Answer:

7.87 years

Step-by-step explanation:

#First we determine the effective annual rate based on the 9% compounded semi annual;

$i_m=(1+i/m)^m-1\\\\=(1+0.09/2)^2-1\\\\=0.09203$

#We then use this effective rate in the compound interest formula to solve for n. Given that the principal doubles after 2 yrs:

$A=P(1+i)^n\\\\A=2P, i=i_m\\\\16000=8000(1.09203)^n\\\\2=1.09203^n\\\\n=\frac{log \ 2}{log \ 1.09203}\\\\=7.87324\approx7.87 \ yrs$

Hence, it takes 7.87 years for the principal amount to double.

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3 years ago

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Alla [95]

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3 years ago

If x2 + y2 = 14 and xy=5, find the value of (½x+½y)². <br><br>pls help

dimulka [17.4K]

Answer:

(½x+½y)²=6

Step-by-step explanation:

x^2 + y^2 = 14, xy=5

(A+B)^2=A^2 +2AB+B^2... (*)

(1/2x+1/2y)^2 =(*)

(1/2x)^2 +2(1/2x)(1/2y)+(1/2y)^2 =

1/4x^2 +1/2xy+1/4y^2=

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3 years ago

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