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3 years ago

Which statement correctly compares the function shown on this graph with

Mathematics

1 answer:

bija089 [108]3 years ago

7 0

Answer:

The function shown on the graph has a smaller rate of change but a higher starting point.

Step-by-step explanation:

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Plz help! It is a Grade! I will do brainsiest

ikadub [295]

Answer: I can only see the big boxes so will answer them

5/6 - 1/3

Multiply the two denominators together get 18

Then multiply the bottom denominator with the opposite numerator

The sum has now changed to:

15/18 - 6/18

15 - 6 = 9

9/18 can then be simplified to 1/2

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3 years ago

7.81.za

ohaa [14]

Answer:

what is another point Sally could have plotted

7 0

3 years ago

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Find the image of A(4,-3) under the transformation r x=2

vlabodo [156]

The is number 3 or c. I hope this helps.

8 0

3 years ago

Santa's sleigh covered a total distance of 5797 kilometres averaging a speed of 38

kogti [31]

Answer:

152.6 hours

Step-by-step explanation:

Distance = Speed × Time

Time = Distance/Speed

Santa's sleigh covered a total distance of 5797 kilometres averaging a speed of 38 km/h in snow storm conditions

Hence: Time Santa spent in snow storms is calculated as:

5797km ÷ 38 km/h

= 152.55263158 hours

Approximately = 152.6 hours

5 0

3 years ago

What is the following sum? Assume x > 0 and y > 0 sqrt x^2y^2+2 sqrt x^3y^4+xy sqrt y

Gwar [14]

Answer:

$xy(1+2y\sqrt{x}+\sqrt{y})$

Step-by-step explanation:

Given expression,

$\sqrt{x^2y^2}+2\sqrt{x^3y^4}+xy\sqrt{y}$

$=(x^2y^2)^\frac{1}{2} + 2(x^3y^4)^\frac{1}{2} + xy\sqrt{y}$

$\because (\sqrt{x}=x^\frac{1}{2})$

$=(x^2)^\frac{1}{2} (y^2)^\frac{1}{2} + 2(x^3)^\frac{1}{2} (y^4)^\frac{1}{2} + xy\sqrt{y}$

$(\because (ab)^n=a^n b^n)$

$=x^{2\times \frac{1}{2}} y^{2\times \frac{1}{2}} + 2(x^{3\times \frac{1}{2}})(y^{4\times \frac{1}{2}})+xy\sqrt{y}$

$\because (a^n)^m=a^{mn}$

$=x^1 y^1 + 2x^{1\frac{1}{2}} y^2 + xy\sqrt{y}$

$=xy+2x.(x)^\frac{1}{2} y^2 + xy\sqrt{y}$

$=xy+2xy^2\sqrt{x}+xy\sqrt{y}$

$=xy(1+2y\sqrt{x}+\sqrt{y})$

8 0

3 years ago

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