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PIT_PIT [208]
3 years ago
12

Solve for z:2^3z+9=8^2z+1

Mathematics
2 answers:
Anika [276]3 years ago
4 0

Answer:

Hope this helps

Step-by-step explanation:

Sholpan [36]3 years ago
3 0

Answer:

z = 0.143

Step-by-step explanation:

2³z+9 = 8²z+1

1. Subtract 1 from both sides

2³z+8 = 8²z

2. Solve exponents

8z+8 = 64z

3. Subtract 8z from 64z

8 = 64z-8z

8 = 56z

4. Divide 8 by 56 in order to isolate z

8/56 = z

0.143 ≈ z

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For the straight line defined by the points (2, 55) and (4, 91), determine the slope and y-intercept. do not round the answers.
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Hello : let  A(2,55)    B(4,91)<span>
the slope is :   (YB - YA)/(XB -XA)
(91-55)/(4-2)  = 36/2 =18
an equation is : y-55 = 18(x-2)</span>

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3 years ago
The sum of two numbers is 240. If one number is twice the other number, find the two numbers.
Deffense [45]

Answer:

The sum of two numbers is 240. The larger number is 6 less than twice the smaller. Find the numbers.

----------

Let the smaller be "x" ; Larger is "2x-6"

EQUATION:

x + 2x-6 = 240

3x= 246

x = 82 (smaller)

2x-6 = 158 (larger)

Step-by-step explanation:

7 0
3 years ago
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Describe two different ways that you could find the product 8 x 997 using mental math. Find the product and explain why your met
jolli1 [7]
One method is to round 997 up to 1000, multiply by 8, and then subtract 8 times 3. This would give you the solution of 7976.

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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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Answer:

The answer is X = 0

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3 years ago
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