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hichkok12 [17]
3 years ago
13

What would be the solution to find x?

Mathematics
2 answers:
Ulleksa [173]3 years ago
8 0
The correct answer is 12
andreyandreev [35.5K]3 years ago
7 0
The answer would be x=12
You might be interested in
2x(3-x) what is the answer need help quick
Dvinal [7]
6x-2x^2 would be the answer 
3 0
2 years ago
Mrs. Bautista invested Php2700.00 part at 8% and the rest at 11% .How musch did she invest at each rate if her total annual inco
Margarita [4]

Answer:

The answer to the question is

She invested

Php2700.00 at 8 % and

Php 20,400.00 at 11 %

Step-by-step explanation:

To solve the question we note that

Simple interest is given by \frac{P*R*T}{100} where

P= Principal, R = Rate and T = Time

If we call the first part P₁, T₁,  and R₁ and the second part

P₂, T₂,  and R₂

Then {P_1*(1+R_1*T_1}) +{P_*(1+R_2*T_2})  = 2460.00

= 2700×0.08×1 + P₂×0.11×1 = 2460  which gives

2244÷0.11 = P₂ or P₂ = Php 20,400.00

That is she invested

Php2700.00 at 8 % and

Php 20,400.00 at 11 %

5 0
2 years ago
13
il63 [147K]

The required value after simplification of the s = -16/3. None of these are correct.

Given that,
To simplify [\frac{x^{2/3}x^{-1/2}}{x\sqrt{x^3}\sqrt[3]{x}}]^2   and to find the value of s in x^s.

<h3>What is simplification?</h3>

The process in mathematics to operate and interpret the function to make the function simple or more understandable is called simplifying and the process is called simplification.

Simplification,

=[\frac{x^{2/3}x^{-1/2}}{x\sqrt{x^3}\sqrt[3]{x}}]^2\\= \frac{x^{4/3}x^{-1}}{x^2x^3*{x}^{2/3}}\\= \frac{x^{1/3}}{x^{17/3}}\\=x^{-16/3}
Comparing with x^S
s = -16/3


Thus, the required value of the s = -16/3. None of these are correct.

Learn more about simplification here: brainly.com/question/12501526

#SPJ1

4 0
1 year ago
Help please................
Alona [7]
Hey you're using Apex xD

Answer is D. 

Here's an example 

6= 3x2
Just divide. 6/2 = 3 <span />
8 0
3 years ago
Offering 40 points and brainliest for answer
Nata [24]
In fact, this problem belongs to the chemistry section.  Recall that many other sciences require mathematical calculations.  The problem will belong to Mathematics only if no knowledge of other sciences are required to solve the problem.

Solubility for the given substances is measured in grams per 100 g of water at a particular temperature (20 deg.C).
This means that the mass (assumed to be the solute) will not change the solubility, just the minimum quantity of solvent (water) will.
Thus the solubility of sodium chloride will remain L=36 g/100g H2O for any quantity of solute.  Similarly, the solubility of lead nitrate will remain as K=54 g/100 g H2O.

The reason that they remain constant is because the quantity of solvent (water) is fixed at 100 g.  Varying amount of solute will affect the quantity of solvent required, but not the solubility.
I'll leave it to you to calculate the difference between K & L.
8 0
2 years ago
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