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LUCKY_DIMON [66]
3 years ago
5

1 over 2z = 3 over 4. Which of the following equals z in this equation? 3 over 8 2 over 3 1 over 1 over 2 2 over 2 over 3

Mathematics
2 answers:
Yanka [14]3 years ago
5 0
The answer is

C. 1 1/2

100% Verified

Hope This Helps!
katovenus [111]3 years ago
5 0

Answer:

The answer is

C. 1 1/2

Step-by-step explanation:

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How do you do question b?
notsponge [240]

Part (b)

We use the result of part (a) and plug in (x,y) = (0,0). This is directly from the initial condition y(0) = 0.

\arcsin(4y) = x^2 + C\\\\\arcsin(4*0) = (0)^2 + C\\\\\arcsin(0) = C\\\\0 = C\\\\C = 0\\\\

-----------------

This means,

\arcsin(4y) = x^2 + C\\\\\arcsin(4y) = x^2 + 0\\\\\arcsin(4y) = x^2\\\\4y = \sin(x^2)\\\\y = \frac{1}{4}\sin(x^2)\\\\

is the solution with the initial condition y(0) = 0.

6 0
3 years ago
In Miss perrons Class 75% of the students are boys. there are 18 boys in the class what is the total number of students in Miss
Vinvika [58]

Answer:

24 is the total number of students in Miss. Perrons class.

Step-by-step explanation:

you take 18 and divide by 75% (0.75) and you get 24

8 0
3 years ago
Plz ans with steps... Thx!!
Ganezh [65]

Answer:

16

Step-by-step explanation:

Let the 1st part of your answer be x , so the 2nd part will be 40-x . From the given information, we can write the equation: (1/4)x = (3/8) × (40-x) . We can simplify this into (1/4)x = (120-3x)/8 ; 8x = 480-12x ; 8x+12x = 480 ; 20x = 480 ; x = 480/20; x = 24

Therefore, the 1st part = 24

Plug this into your 40-x equation to get: 40 - 24 = 16

3 0
3 years ago
What is the scale factor of 8in=200mi?
BigorU [14]
200 / 8  =  25

scale factor is 1:25  or 0.04
8 0
3 years ago
Read 2 more answers
F the range of the function f(x) = 7x – 2.7 is {14.1, 30.9, 41.4, 58.9, 68}, what is its domain?
grigory [225]
You need to solve  5 equations  
1. 7x - 2.1 = 14.1
2. 7x - 2.1 = 30.9  and so on


1. 7x - 2.1 = 14.1
7x = 16.2
x = 2.314     so first number in the domain is  2.314

You can calculate the other 4 in the same way
7 0
3 years ago
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