All categories

3 years ago

Help I'm really stuck!!!

Mathematics

2 answers:

faust18 [17]3 years ago

5 0

399.6 square meters
7•8/2=28
28•3=84

8•6.9/2=27.6

(12•8)•3=288

84+27.6+288=399.6

jek_recluse [69]3 years ago

3 0

336.6580 I know this because of u do the math u get this

You might be interested in

Determine which equations have the same solution set as StartFraction 2 Over 3 EndFraction minus x plus StartFraction 1 Over 6 E

Lina20 [59]

Answer:

A, B, F

Step-by-step explanation:

2/3 - x + 1/6 = 6x

Collect like terms

2/3 + 1/6 = 6x + x

(4+1) / 6 = 7x

5/6 = 7x

x = 5/6 ÷ 7

= 5/6 × 1/7

x = 5/42

a) 4 - 6x + 1 = 36x

4 + 1 = 36x + 6x

5 = 42x

x = 5/42

Equivalent to the last step of the simplification above

b) 5/6 - x = 6x

5/6 = 6x + x

5/6 = 7x

This is equivalent to the third step of the simplification

c) 4 - x + 1 = 6x

4 + 1 = 6x + x

5 = 7x

x = 5/7

Not equivalent to any of the steps in the simplification above

d) 5/6 + x = 6x

5/6 = 6x - x

5/6 = 5x

x = 5/6 ÷ 5

= 5/6 × 1/5

x = 5/30

Not equivalent to any of the steps in the simplification above

e) 5 = 30x

x = 5/30

Not equivalent to any of the steps in the simplification above

f) 5 = 42x

x = 5/42

Equivalent to the last step of the simplification above

4 0

3 years ago

Evaluate: ma + (t + h)

Eduardwww [97]

Answer:

Math

Step-by-step explanation:

math m-a-t-h mathhhhhhhhh

8 0

3 years ago

I’m confused on this one

Lorico [155]

The question is asking which of the following does not have side that are parallel. So which of them have lines that would never touch if they were never ending lines.

4 0

3 years ago

13. f(x) = x + 5 a. f(4) b. f(7) c. f(-3) d. f(0) e. f(2.4) f. f(2/3)

SashulF [63]

A.9
B.12
C.2
D.5
E.7.4
F.5 2/3 or 5.6666

4 0

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

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

3 0

3 years ago