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2 years ago

Find the slope of the line containing the points (1, 2) and (6, 12).

Mathematics

2 answers:

Blababa [14]2 years ago

6 0

Answer:

slope is x= 2

Step-by-step explanation:

slope is second y minus first y over second x minus first x, plug in the numbers and you get 10/5 simply, you get 2

Sholpan [36]2 years ago

5 0

Answer:6

Step-by-step explanation:

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Please I need help with this question. No 22. Is very important. Thank you.

kvv77 [185]

Answer:

C. $132m^3$

Step-by-step explanation:

Height = 4

Length = 6

Breadth = 7

Note : We were asked to find volume of the cuboid

Since, it has a base in the shape of a right angle triangle , we need to find the Area of the trapezium before calculating the total volume.

Step 1

Find the area of the trapezium

Area of a trapezium = $\frac{1}{2} (a+b)h$

$\frac{1}{2} (4+7)*4$

$\frac{1}{2} (11)4\\$

$\frac{1}{2} *44\\= 22$

Step 2

Calculate the volume of the cuboid

Volume of a cuboid =length× breadth × height

In this case , the volume will be our length × Area of a trapezium

V= 6 × 22

V = $132m^3$

6 0

3 years ago

What adds to get 3 but multiplies to get negative 18

Colt1911 [192]

6 and -3

6 + -3 = 3

6 x -3 = -18

Hope that helps!

3 0

3 years ago

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kramer

Answer:

Step-by-step explanation :

$\sqrt{ \frac{243}{3} }$

$= \sqrt{81}$

$= \sqrt{9 \times 9}$

= 9(Ans)

5 0

3 years ago

Read 2 more answers

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

Which is larger 3kL or 30000L

dolphi86 [110]

30000L is larger than 3kL

7 0

2 years ago