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Dmitriy789 [7]
3 years ago
15

Using the change of base formula, evaluate log312 (round to the nearest hundredth).

Mathematics
1 answer:
Svetach [21]3 years ago
5 0

Answer:

2.26

Explanation:

Given: log base 3 and 12

Using the base rule, we can write it as

= log 12 / log 3

Now we can find the value of log 12 and log 3 using a calculator.

= 1.07918124 /0.477121754

=2.26                

The answer is 2.26

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W/4 -4=3<br><br> Solve the eqution pls!
Tanzania [10]

Answer:

W=28

Step-by-step explanation:

(W/4)-4=3

(W/4)=7

W=28

4 0
3 years ago
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8) Four times the square of a positive number is eight times that number. Find the number. PLS HELP ME ASAP!​
zvonat [6]

Answer:

x=2

Step-by-step explanation:

Again, trial and error.

6 0
3 years ago
Factor –8x3 – 2x2 – 12x – 3 by grouping. What is the resulting expression?
xenn [34]
Easier if you first factor out the " - " sign.  Then you have:

–8x3 – 2x2 – 12x – 3    = -1(<span>8x3 + 2x2 + 12x +3).
</span>
Start with 12 + 3.  This factors to 3(4x+1).  Next, focus on 8x^3+2x^2.  This factors to 2x^2*(4x+1).  So, the binomial factor 4x+1 is common to the first two terms and the second two terms.  We get:  (4x+1) (  2x^2  + 3).

Now put the "-" sign back in, and we get   - (4x+1) ( 2x^2  + 3)    (answer)

Please note:  Use "^" to denote exponentiation.   2x2 is meaningless; it should be 2x^2.
3 0
3 years ago
Read 2 more answers
3x - 1 = -4x-8 how do you solve this equation with variables on both side
Nesterboy [21]

Answer:

x = -1

Step-by-step explanation:

4 0
2 years ago
(Hehe...Please ignore the markings)
inysia [295]

Answer:

See below:

Step-by-step explanation:

Problem 1:

Multiply Equation 1 by 4, keep Equation 2 the same.

x+y=8, multiply each term by 4:

4*x=4x, 4*y=4y, 8*4=32

so, the equivalent system is: 4x+4y=32 and x-y=2

Solve the system of equations:

x-y=2 becomes x=y+2

plug into 4x+4y=32 to solve for y

4(y+2)+4y=32---> 4y+8+4y=32--->8y=24---> y=3

Plug into x-y=2---> x-3=2---> x=5

Problem 1 Answer:

Equivalent system: 4x+4y=32, x-y=2; solution: x=5, y=3

Problem 2:

Keep Equation 1 the same. Add 1 and 2.

To add an equation, add the left sides together, and then the rights.

so: x+y=8 + x-y=2 gives us: 2x=10

solve for x ---> 2x/2=10/2--->x=5

plug x into x+y=8--->5+y=8--->y=3

Problem 2 answer:

Equivalent system: x+y=8, 2x=10; solution:x=5, y=3

Problem 3:

Subtract Equation 2 from 1, and keep 2 the same.

To subtract an equation, subtract the left sides, then the rights. We are subtracting 1<em> from </em>2, so its 2-1.

x-y=2 - x+y=8 gives us: -2y=-6

Solve for y by dividing by -2-->-2y/-2=-6/-2---> y=3

Plug into x-y=2---> x-3=2---> x=5

Problem 3 answer:

Equivalent system: -2y=-6, x-y=2; solution: x=5, y=3

Problem 4:

Multiply the sum of Equation 1 and 2 by a factor of 3. Keep equation 2 the same.

First we add 1 and 2: (we did this earlier) ---> 2x=10 ---> now we multiply it all by 3---> 2x*(3)=10*(3)---> this gives us: 6x=30---> now divide by 6 to solve for x: 6x/6=30/6 gives us: x=5

Now, solve for y by plugging x into equation 2: x-y=2---> 5-y=2--->y=3

Problem 4 answer:

Equivalent system: 6x=30, x-y=2; solution: x=5, y=3

______

Quick Tip: One thing inherent of Equivalent systems is that they have the same set of solutions. Thus, we know the systems are equivalent when they have the same set of solutions for x and y. Moreover, you don't need to solve every time after you attempt to find an equivalent system, instead, just plug in the values found in problem 1 to each new set of equations to test if they are equivalent.

If we find x=5 and y=3 for x+y=8 and x-y=2, then all we have to do is plug them in to 6x=30 and -2y=-6 to see if they are equivalent.

6(5)=30 ---> true

-2(3)=-6 ---> true

3 0
3 years ago
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