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shutvik [7]
2 years ago
11

Can someone double check these and tell me whats the first one ill mark brianlest

Mathematics
1 answer:
HACTEHA [7]2 years ago
7 0

Answer:

The radio for the 1st : 1:3

1:3

1:3

The 2nd is : C, È, B

The 3rd is : $7.00

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Y=6x-4 y=4x+2 substitution pls help plsplslsl​
tiny-mole [99]

Answer:

Solve for the first variable in one of the equations, then substitute the result into the other equation.

Point Form:  ( 3 , 14 )

Equation Form:  x  =  3 , y  =  14

Step-by-step explanation:

4 0
3 years ago
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What is the input-output table for the function f(x) = 3x^2-x+4
mojhsa [17]

Answer:

Step-by-step explanation:

The input-output table can be made by putting value of x and finding the value of f(x)

f(x) = 3x^2-x+4

f(0) = 3(0)^2-0+4 = 0-0+4 = 4

f(1) = 3(1)^2-1+4 = 3-1+4 = 2+4 =6

f(2) = 3(2)^2-2+4 = 3(4)-2+4 = 12-2+4 = 10+4 = 14

f(3) = 3(3)^2-3+4 = 3(9)-3+4 = 27-3+4 = 24+4 = 28

So put value of x and find f(x) and fill the input-output table.

x  f(x)

0   4

1    6

2   14

3    28

3 0
3 years ago
The position of an object along a vertical line is given by s(t) = −t3 + 3t2 + 7t + 4, where s is measured in feet and t is meas
saw5 [17]

Answer:

The maximum velocity of the object in the time interval [0, 4] is 10 ft/s.

Step-by-step explanation:

Given : The position of an object along a vertical line is given by s(t) = -t^3+3t^2+7t +4, where s is measured in feet and t is measured in seconds.

To find : What is the maximum velocity of the object in the time interval [0, 4]?

Solution :

The velocity is rate of change of distance w.r.t time.

Distance in terms of t is given by,

s(t) = -t^3+3t^2+7t +4

Derivate w.r.t. time,

v(t)=s'(t) = -3t^2+6t+7

It is a quadratic function so its maximum is at vertex of the function.

The x point of the function is given by,

x=-\frac{b}{2a}

Where, a=-3, b=6 and c=7

t=-\frac{6}{2(-3)}

t=-\frac{6}{-6}

t=1

As 1 lie between interval [0,4]

Substitute t=1 in the function,

v(t)= -3(1)^2+6(1)+7

v(t)= -3+6+7

v(1)=10

Th maximum velocity is 10 ft/s.

Therefore, the maximum velocity of the object in the time interval [0, 4] is 10 ft/s.

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3 years ago
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riadik2000 [5.3K]
E. 74/116=0.637 hope this helps
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Ugggggggg idk soooooooo yeah
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