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2 years ago

6

What is the answer to this annoying question plz help me my math teacher hates me so plz plz plz help me I'm only in six grade a

nd on the boarder line from passing which is not good at all

1 answer:

Sati [7]2 years ago

4 0

Gor f Is the answer cause hilary

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A glider flies 13 miles north from the airport and then 24 miles diagonally southeast. Approximately how far west does the glide

NISA [10]

Answer:

20.2 miles

Step-by-step explanation:

This can be described by the three sides of a right angled triangle. Let the distance of the glider to the airport be represented by x, applying the Pythagoras theorem:

$/hyp/^{2}$ = $/adj1/^{2}$ + $/adj2/^{2}$

$/24/^{2}$ = $/x/^{2}$ + $/13/^{2}$

576 = $x^{2}$ + 169

$x^{2}$ = 576 - 169

= 407

x = $\sqrt{407}$

= 20.1742

x = 20.2 miles

The glider has to fly 20.2 miles to return to the airport.

8 0

2 years ago

I really need help with this question

Step2247 [10]

Answer:

option (b) -3(3w-3.1)

7 0

3 years ago

Read 2 more answers

Find the Total Surface area of a Pyramid. Its base is a regular hexagon 4cm on a side. its height is 8 cm. Round off to nearest

oksano4ka [1.4K]

Answer:

The total surface area of a regular hexagonal pyramid is 146.2 cm².

Step-by-step explanation:

The base is a regular hexagon 4cm on a side. its height is 8 cm.

The total surface area of a pyramid with regular hexagon base

$A=\frac{3\sqrt{3}}{2}a^2+3a\sqrt{h^2+\frac{3a^2}{4}$

Where, a represents the base edge and h represents the height of the pyramid.

The area of given regular hexagonal pyramid is

$A=\frac{3\sqrt{3}}{2}(4)^2+3(4)\sqrt{(8)^2+\frac{3(4)^2}{4}$

$A=\frac{3\sqrt{3}}{2}(16)+12\sqrt{64+12}$

$A=24\sqrt{3}+12\sqrt{76}$

$A=146.182794027$

$A=146.2$

The total surface area of a regular hexagonal pyramid is 146.2 cm².

8 0

3 years ago

[36 + (3 x 2)] ÷ 6<br> can any one tell me the ans

Brilliant_brown [7]

Answer:

6+x

Step-by-step explanation:

hope this helps

3 0

2 years ago

Find a linear second-order differential equation f(x, y, y', y'') = 0 for which y = c1x + c2x3 is a two-parameter family of solu

Alisiya [41]

Let $y=C_1x+C_2x^3=C_1y_1+C_2y_2$ . Then $y_1$ and $y_2$ are two fundamental, linearly independent solution that satisfy

$f(x,y_1,{y_1}',{y_1}'')=0$
$f(x,y_2,{y_2}',{y_2}'')=0$

Note that ${y_1}'=1$ , so that $x{y_1}'-y_1=0$ . Adding $y''$ doesn't change this, since ${y_1}''=0$ .

So if we suppose

$f(x,y,y',y'')=y''+xy'-y=0$

then substituting $y=y_2$ would give

$6x+x(3x^2)-x^3=6x+2x^3\neq0$

To make sure everything cancels out, multiply the second degree term by $-\dfrac{x^2}3$ , so that

$f(x,y,y',y'')=-\dfrac{x^2}3y''+xy'-y$

Then if $y=y_1+y_2$ , we get

$-\dfrac{x^2}3(0+6x)+x(1+3x^2)-(x+x^3)=-2x^3+x+3x^3-x-x^3=0$

as desired. So one possible ODE would be

$-\dfrac{x^2}3y''+xy'-y=0\iff x^2y''-3xy'+3y=0$

(See "Euler-Cauchy equation" for more info)

6 0

3 years ago