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2 years ago

Which pair of angles must be supplementary? O 1 and 6 O 2 and 5 O 5 and 4 O Z6 and2

Mathematics

1 answer:

jeka57 [31]2 years ago

7 0

Where’s the image? There has to be an image showing the angles to see which pairs of angles are supplementary.

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If a solution has 24g of sugar in 130ml of water, what is its concentration?

Jobisdone [24]

Answer: concentration = 15.58%

Step-by-step explanation:

Given: mass of sugar = 24 g = 24 ml [ ∵ 1 g = 1 ml]

Volume of solution = Quantity of water + quantity of solute

= 130 +24 ml

= 154 ml

Concentration = $\dfrac{mass\ of\ solute}{Volume\ of\ solution}\times100$

$=\dfrac{24}{154}\times 100\\\\\approx15.58\%$

Hence, the concentration =15.58%

3 0

2 years ago

One bottle contains 341 ml of juice. Céline buys 2 cases of 24 bottles of juice. Nicole buys 3 cases of 12 bottles of juice. Who

Helga [31]

Answer: Celine, 4092 ml

Step-by-step explanation:

Given

1 bottle contains 341 ml Juice

Celine buys 2 cases of 24 bottles i.e. a total of 48 bottles

Nicole buys 3 cases of 12 bottles i.e. a total of 36 bottles

Amount of Juice Celine has

$\Rightarrow 48\times 341=16,368\ ml$

Amount of Juice Nicole has

$\Rightarrow 36\times 341=12,276\ ml$

Thus, Celine has more juice than Nicole

$\Rightarrow 16,368-12,276=4092\ ml$

Celine has 4092 ml more than Nicole

7 0

2 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

2 years ago

lol Barney has 16 1 yards of fabric. To make an elf costume, he needs 5 5 2 yards of fabric. How many costumes can Barney make?

ioda

Well i happen to have that question and it is that he has 16 1/5 fabric and need 5 2/5 to make an elf costume. how many costumes can he make? Well so basically all you have to do is 16 1/5 divided by 5 2/5, which is: 3. The answer is 3 hope this helped!

7 0

2 years ago

The ratio of 9/5 is equal to

Nutka1998 [239]

The answer is 1 and 4/5. Just divide and put the remainder as the numerator and the numerator stays the same

6 0

2 years ago

Read 2 more answers