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evablogger [386]
2 years ago
8

JOSIE MAKES FRUIT PUNCH BY MIXING FRUIT JUICE AND LEMONADE IN THE RATIO 1:3 SHE NEEDS 24 LITRES OF PUNCH FOR A PARTY

Mathematics
1 answer:
mario62 [17]2 years ago
3 0


the answer is as following:

Fruit juice:6

Lemonade:18

the answer has to equal 1/3 so the answer is 6 for fruit juice and 18 for lemonade

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Mike just hopped on the edge of a merry-go-round. What are his linear and angular speeds if the diameter of the merry-go-round i
dsp73

Step-by-step explanation:

Given that,

The diameter of the merry-go-round, d = 14 feet

Time taken, t = 6 seconds

Radius, r = 7 feet

The linear speed of the merry-go-round is given by :

v=\dfrac{2\pi r}{t}\\\\v=\dfrac{2\pi \times 7}{6}\\\\=7.33\ m/s

Also,

v=r\omega

Where

\omega is the angular speed

So,

\omega=\dfrac{v}{r}\\\\\omega=\dfrac{7.33}{7}\\\\=1.04\ rad/s

Hence, his linear and angular speeds are 7.33 m/s and 1.04 rad/s.

8 0
3 years ago
Help please like hurry!!
adelina 88 [10]
The answer is a 36
This is because if you take the area of one of the three triangles that make up the trapezoid you do 1/2 base times height and get 12 and you add all three together thus getting 36 square inches
5 0
2 years ago
Read 2 more answers
If f(x) = x2 – 1 and g(x) = 2x – 3, what is the domain
Ainat [17]
Begin by finding the lowest point the quadratic equation can be, the vertex; 

x²-1= is just a translation down of the graph x²

vertex; (0, -1) and since the graph of x² would extend to infinity beyond that point, we can say {x| x≥0} for domain and {y| y≥-1}. 

For the linear equation, it is possible to have all x and y values, therefore range and domain belong to all real numbers. 

Hope I helped :) 
5 0
3 years ago
U
elena55 [62]

Answer:

option 3, 45

Step-by-step explanation:

3x - 9° + 30° + 24° = 180° (angle sum property of a triangle)

3x - 9° + 54° = 180°

3x + 45° = 180°

3x = 180° - 45°

3x = 135°

x = 135/3

x = 45°

therefore, option 3 is the correct option

8 0
2 years ago
A ball is thrown into the air by a baby alien on a planet in the system of Alpha Centauri with a velocity of 30 ft/s. Its height
Crank

Answer:

a) h = 0.1: \bar v = -11\,\frac{ft}{s}, h = 0.01: \bar v = -10.1\,\frac{ft}{s}, h = 0.001: \bar v = -10\,\frac{ft}{s}, b) The instantaneous velocity of the ball when t = 2\,s is -10 feet per second.

Step-by-step explanation:

a) We know that y = 30\cdot t -10\cdot t^{2} describes the position of the ball, measured in feet, in time, measured in seconds, and the average velocity (\bar v), measured in feet per second, can be done by means of the following definition:

\bar v = \frac{y(2+h)-y(2)}{h}

Where:

y(2) - Position of the ball evaluated at t = 2\,s, measured in feet.

y(2+h) - Position of the ball evaluated at t =(2+h)\,s, measured in feet.

h - Change interval, measured in seconds.

Now, we obtained different average velocities by means of different change intervals:

h = 0.1\,s

y(2) = 30\cdot (2) - 10\cdot (2)^{2}

y (2) = 20\,ft

y(2.1) = 30\cdot (2.1)-10\cdot (2.1)^{2}

y(2.1) = 18.9\,ft

\bar v = \frac{18.9\,ft-20\,ft}{0.1\,s}

\bar v = -11\,\frac{ft}{s}

h = 0.01\,s

y(2) = 30\cdot (2) - 10\cdot (2)^{2}

y (2) = 20\,ft

y(2.01) = 30\cdot (2.01)-10\cdot (2.01)^{2}

y(2.01) = 19.899\,ft

\bar v = \frac{19.899\,ft-20\,ft}{0.01\,s}

\bar v = -10.1\,\frac{ft}{s}

h = 0.001\,s

y(2) = 30\cdot (2) - 10\cdot (2)^{2}

y (2) = 20\,ft

y(2.001) = 30\cdot (2.001)-10\cdot (2.001)^{2}

y(2.001) = 19.99\,ft

\bar v = \frac{19.99\,ft-20\,ft}{0.001\,s}

\bar v = -10\,\frac{ft}{s}

b) The instantaneous velocity when t = 2\,s can be obtained by using the following limit:

v(t) = \lim_{h \to 0} \frac{x(t+h)-x(t)}{h}

v(t) =  \lim_{h \to 0} \frac{30\cdot (t+h)-10\cdot (t+h)^{2}-30\cdot t +10\cdot t^{2}}{h}

v(t) =  \lim_{h \to 0} \frac{30\cdot t +30\cdot h -10\cdot (t^{2}+2\cdot t\cdot h +h^{2})-30\cdot t +10\cdot t^{2}}{h}

v(t) =  \lim_{h \to 0} \frac{30\cdot t +30\cdot h-10\cdot t^{2}-20\cdot t \cdot h-10\cdot h^{2}-30\cdot t +10\cdot t^{2}}{h}

v(t) =  \lim_{h \to 0} \frac{30\cdot h-20\cdot t\cdot h-10\cdot h^{2}}{h}

v(t) =  \lim_{h \to 0} 30-20\cdot t-10\cdot h

v(t) = 30\cdot  \lim_{h \to 0} 1 - 20\cdot t \cdot  \lim_{h \to 0} 1 - 10\cdot  \lim_{h \to 0} h

v(t) = 30-20\cdot t

And we finally evaluate the instantaneous velocity at t = 2\,s:

v(2) = 30-20\cdot (2)

v(2) = -10\,\frac{ft}{s}

The instantaneous velocity of the ball when t = 2\,s is -10 feet per second.

8 0
3 years ago
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