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3 years ago

Which of the h-values satisfy the following inequality?

Mathematics

2 answers:

olasank [31]3 years ago

6 0

Answer:

Step-by-step explanation:

ElenaW [278]3 years ago

6 0

The answer I got for this question was 12

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Which shapes have at least one pair of parallel sides?

elena-s [515]

Parallelograms and trapezoids always do.

(Note: "Parallelograms" includes rectangles, squares, and rhombussesses.)

Any polygon with more than 4 sides <u>can</u> have at least one pair of parallel sides,
but it's not guaranteed.

3 0

3 years ago

A road is

Airida [17]

9514 1404 393

Answer:

25/42 mi

Step-by-step explanation:

5/6 of 5/7 of a mile is ...

(5/6)(5/7 mi) = 25/42 mi

4 0

3 years ago

A shop has a sale of

maw [93]

Answer:

19.80

Step-by-step explanation:

use the is/of x %/100 method.

33/100 x x/60

1980/100x = 19.80$

5 0

2 years ago

What is the square root of 189

valina [46]

Answer:

$3 \sqrt{21}$

Step-by-step explanation:

$\sqrt{189} = 3 \sqrt{21}$

7 0

3 years ago

Solve the differential. This was in the 2016 VCE Specialist Maths Paper 1 and i'm a bit stuck

Nimfa-mama [501]

$\sqrt{2 - x^{2}} \cdot \frac{dy}{dx} = \frac{1}{2 - y}$
$\frac{dy}{dx} = \frac{1}{(2 - y)\sqrt{2 - x^{2}}}$

Now, isolate the variables, so you can integrate.
$(2 - y)dy = \frac{dx}{\sqrt{2 - x^{2}}}$
$\int (2 - y)\,dy = \int\frac{dx}{\sqrt{2 - x^{2}}}$
$2y - \frac{y^{2}}{2} = sin^{-1}\frac{x}{\sqrt{2}} + \frac{1}{2}C$

$4y - y^{2} = 2sin^{-1}\frac{x}{\sqrt{2}} + C$
$y^{2} - 4y = -2sin^{-1}\frac{x}{\sqrt{2}} - C$
$(y - 2)^{2} - 4 = -2sin^{-1}\frac{x}{\sqrt{2}} - C$
$(y - 2)^{2} = 4 - 2sin^{-1}\frac{x}{\sqrt{2}} - C$

$y - 2 = \pm\sqrt{4 - 2sin^{-1}\frac{x}{\sqrt{2}} - C}$
$y = 2 \pm\sqrt{4 - 2sin^{-1}\frac{x}{\sqrt{2}} - C}$

At x = 1, y = 0.
$0 = 2 \pm\sqrt{4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C}$
$-2 = \pm\sqrt{4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C}$

$4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C > 0$
$\therefore 2 = \sqrt{4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C}$

$4 = 4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C$
$0 = -2sin^{-1}\frac{1}{\sqrt{2}} - C$
$C = -2sin^{-1}\frac{1}{\sqrt{2}} = -2\frac{\pi}{4}$
$C = -\frac{\pi}{2}$

$\therefore y = 2 - \sqrt{4 + \frac{\pi}{2} - 2sin^{-1}\frac{x}{\sqrt{2}}}$

6 0

3 years ago