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3 years ago

Having trouble with how to solve questions like this:

1 answer:

8 0

In this question there is a small trick that needs to be understood. Basically the ratio in the question is given as a decimal. It can be expressed as a ratio as well.
Ratio of students that voted for winner to the total number of voters = 95:100
The total number of students that voted = 60
Let us assume the common ratio to be = x
then
100x = 60
x = 60/100
= 0.6
Then
The number of votes that the winner got = 95 * 0.6
= 57
So 57 students voted for the winning candidate.

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When leaving a town, a car accelerates from 30 kmh-1 to 60 kmh-1 in 5 s. Assuming the

aleksandr82 [10.1K]

Answer:

B .62.5 m

Step-by-step explanation:

convert Kmh-1 in to ms-1

30 Kmh-1 = (30×1000) ÷3600 = 8.3ms-1

60 Kmh-1 = (60×1000) ÷3600 = 16.6 ms-1

acceleration = (16.6 - 8.3) ÷ 5 = 1.66 ms-2

V^2 = U^2 +2aS

(16.6)^2 = (8.3)^2 + 2×1.66 ×S

S = 62.5 m

6 0

3 years ago

This is really confusing pls help! :)

stellarik [79]

Number of cubes increases by 3 for each term.

an= 3n+1

4 0

2 years ago

Read 2 more answers

Which toy pack has a cost of $2.50 per toy?

defon

YoYo. 15/6 = 2.5 so each YoYo costs $2.50.

5 0

3 years ago

Read 2 more answers

Match the expressions with their equivalent simplified expressions.

Tasya [4]

Answer:

$\sqrt[4]{\frac{16x^6y^4}{81x^2y^8}}\rightarrow\frac{2x}{3y}\\\sqrt[4]{\frac{81x^2y^{10}}{81x^6y^6}} \rightarrow\frac{3y}{2x}\\\sqrt[3]{\frac{64x^8y^7}{125x^2y^{10}}}\rightarrow\frac{4x^2}{5y}\\\sqrt[5]{\frac{243x^{17}y^{16}}{32x^7y^{21}}}\rightarrow\frac{3x^2}{2y}\\\sqrt[5]{\frac{32x^{12}y^{15}}{243x^7y^{10}}} \rightarrow\frac{2xy}{3}\\\sqrt[4]{\frac{16x^{10}y^{9}}{256x^2y^{17}}}\rightarrow\frac{x}{2y}$

Step-by-step explanation:

$\sqrt[4]{\frac{16x^6y^4}{81x^2y^8}} =\sqrt[4]{\frac{(2^4)(x^{6-2})(y^{4-8})}{(3^4)}} =\sqrt[4]{\frac{2^4x^4y^{-4}}{3^4}} =\frac{2xy^{-1}}{3}=\frac{2x}{3y}$

$\sqrt[4]{\frac{81x^2y^{10}}{81x^6y^6}} =\sqrt[4]{\frac{(3^4)(x^{2-6})(y^{10-6})}{(2^4)}} =\sqrt[4]{\frac{3^4x^{-4}y^{4}}{2^4}} =\frac{3x^{-1}y^1}{3}=\frac{3y}{2x}$

$\sqrt[3]{\frac{64x^8y^7}{125x^2y^{10}}} =\sqrt[3]{\frac{(4^3)(x^{8-2})(y^{7-10})}{(5^3)}} =\sqrt[3]{\frac{4^3x^6y^{-3}}{5^3}} =\frac{4x^2y^{-1}}{5}=\frac{4x^2}{5y}$

$\sqrt[5]{\frac{243x^{17}y^{16}}{32x^7y^{21}}} =\sqrt[5]{\frac{(3^5)(x^{17-7})(y^{16-21})}{(2^5)}} =\sqrt[5]{\frac{3^5x^{10}y^{-5}}{2^5}} =\frac{3x^2y^{-1}}{2}=\frac{3x^2}{2y}$

$\sqrt[5]{\frac{32x^{12}y^{15}}{243x^7y^{10}}} =\sqrt[5]{\frac{(2^5)(x^{12-7})(y^{15-10})}{(3^5)}} =\sqrt[5]{\frac{2^5x^{5}y^{5}}{3^5}} =\frac{2x^1y^{1}}{3}=\frac{2xy}{3}$

$\sqrt[4]{\frac{16x^{10}y^{9}}{256x^2y^{17}}} =\sqrt[4]{\frac{(2^4)(x^{10-2})(y^{9-17})}{(4^4)}} =\sqrt[4]{\frac{2^4x^{8}y^{-8}}{4^4}} =\frac{2x^{1}y^{-1}}{4}=\frac{x}{2y}$

Thus,

3 0

3 years ago

1) Write a system of equations to describe the situation below, solve using any method, and

Aloiza [94]

This situation can be represented by

y = 33x + 223

y = 32x + 239

I still like substitution, so let’s plug in 33x + 223 into the second equation

33x + 223 = 32x + 239

Move the terms into appropriate sides

33x - 32x = 239 - 223

Combine like terms

x = 16

Then plug in x for any equation

y = 32(16) + 239

y = 751

Erin will have to swim 16 laps for a total of 751 meters

7 0

3 years ago