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ludmilkaskok [199]
2 years ago
11

Solve this equation using a common base

Mathematics
1 answer:
GrogVix [38]2 years ago
6 0

Answer:

Step-by-step explanation:

1. 3x - 7 = 5 - x

   4x = 12

    x = 3

2. 6x + 4 = 2

   6x = -2

   x = -1/3

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Easy!!!
aksik [14]

Step-by-step explanation:

f(g(10))

This is called a composite function.  It's when you plug one function into another.

First, find g(10):

g(x) = √(x-9)

g(10) = √(10-9)

g(10) = √1

g(10) = 1

Then plug that into f(x):

f(x) = -9x - 9

f(g(10)) = -9 g(10) - 9

f(g(10)) = -9 (1) - 9

f(g(10)) = -9 - 9

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7 0
3 years ago
Plz help advanced calculus ​
love history [14]

Answer:

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6 0
2 years ago
Solve for x thanks .
OleMash [197]

Answer:

x = 3.87

Step-by-step explanation:

Using the right angle altitude theorem, we know that all three triangles are congruent, so the lengths of corresponding sides of the triangles are in proportion..

AD/DB = DB/DC

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8 0
2 years ago
An object is propelled upward from the top of a 300 foot building. The path that the object takes as it falls to the ground can
serg [7]

Answer:

As per the statement:

The path that the object takes as it falls to the ground can be modeled by:

h =-16t^2 + 80t + 300

where

h is the height of the objects and

t is the time (in seconds)

At t = 0 , h = 300 ft

When the objects hit the ground, h = 0

then;

-16t^2+80t+300=0

For a quadratic equation: ax^2+bx+c=0         ......[1]

the solution for the equation is given by:

x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}

On comparing the given equation with [1] we have;  

a = -16 ,b = 80 and c = 300

then;

t= \frac{-80\pm \sqrt{(80)^2-4(-16)(300)}}{2(-16)}

t= \frac{-80\pm \sqrt{6400+19200}}{-32}

t= \frac{-80\pm \sqrt{25600}}{-32}  

Simplify:

t = -\frac{5}{2} = -2.5 sec and t = \frac{15}{2} = 7.5 sec

Time can't be in negative;

therefore, the time it took the object to hit the ground is 7.5 sec

8 0
3 years ago
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