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arlik [135]
3 years ago
14

Please Solve for x Log15 1 =x

Mathematics
1 answer:
Fiesta28 [93]3 years ago
3 0
   
\bold{\log_{15} (1) = x}\\\\
\bold{\Longrightarrow~~~15^x = 1}\\\\
\Longrightarrow~~~\boxed{\bold{x = 0}}



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YESSS love struggling 7th grade math problem ......... help me
Dimas [21]

Answer:

x = 20

ABC = 85

BCA = 81

CAB = 14

Step-by-step explanation:

I answered your other question, so basically just read the explanation there cuz I'm too lazy to write it again :P

4 0
3 years ago
I just can't figure it out!!<br> How do I figure out Kevin's height?
rosijanka [135]

Answer:

use a ruler

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
I need help with #8. I need to use the quadratic formula. I have a calculator that is programmed for the quadratic formula so al
GuDViN [60]

Answer:

  x = 14

Step-by-step explanation:

Extend line AB so that it intersects ray CE at point G. Then angles BGC and BAD are "alternate interior angles", hence congruent.

The angle at B is exterior to triangle BCG, and is equal to the sum of the interior angles at C and G:

  138 = (376 -23x) +(x^2 -8x)

Subtracting 138 and collecting terms we have ...

  x^2 -31x +238 = 0

For your calculator, a=1, b=-31, c=238.

__

<em>Additional comment</em>

You will find that the solutions to this are x = {14, 17}. You will also find that angle BCE will have corresponding values of 54° and -15°. That is, the solution x=17 is "extraneous." It is a solution to the equation, but not to the problem.

For x=14, the marked angles are A = 84°, C = 54°.

8 0
4 years ago
5x + 6y = 32
velikii [3]

Answer:

The variable y is eliminated.

Step-by-step explanation:

To add a system of equations, we add the common terms. x with x, y with y and values with values.

In this question:

5x + 6y = 32

5x – 6y = 8

Adding:

5x + 5x + 6y - 6y = 32 + 8

10x = 40

So the variable y is eliminated.

3 0
3 years ago
Whiskers the cat weighs 3 1/2 kg and Pilglio weighs 5 kg. How many times as heavy as pilglio is whiskers? How many times as heav
enyata [817]

Given:

Whiskers the cat weighs 3\dfrac{1}{2} kg and Pilglio weighs 5 kg.

To find:

How many times as heavy as pilglio is whiskers?

How many times as heavy as whiskers is pilglio?

Solution:

We have,

Weight of cat (whiskers) = 3\dfrac{1}{2} kg

                      = \dfrac{3(2)+1}{2} kg

                      = \dfrac{7}{2} kg

                      =  3.5 kg

Weight of pilglio = 5 kg

Now,

\dfrac{Whiskers}{Pilglio}=\dfrac{3.5}{5}=0.7

Therefore, whiskers is 0.7 times as heavy as pilglio.

\dfrac{Pilglio}{Whiskers}=\dfrac{5}{3.5}

\dfrac{Pilglio}{Whiskers}=1.428571...

\dfrac{Pilglio}{Whiskers}\approx 1.43

Therefore, piglio is about 1.43 times as heavy as whiskers.

5 0
3 years ago
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