Answer:
Step-by-step explanation:
It maybe will be ![\neq x^{2} \leq \\ \\ \int\limits^a_b {x} \, dx \int\limits^a_b {x} \, dx \sqrt{x} \\ \left \{ {{y=2} \atop {x=2}} \right. \left \{ {{y=2} \atop {x=2}} \right. x^{2} x^{2} \sqrt{x} \lim_{n \to \infty} a_n \lim_{n \to \infty} a_n \neq \sqrt{x} \sqrt[n]{x} \frac{x}{y} \frac{x}{y} \alpha \beta x_{123} \\ x^{2} \int\limits^a_b {x} \, dx x^{2}](https://tex.z-dn.net/?f=%5Cneq%20x%5E%7B2%7D%20%5Cleq%20%5C%5C%20%5C%5C%20%5Cint%5Climits%5Ea_b%20%7Bx%7D%20%5C%2C%20dx%20%5Cint%5Climits%5Ea_b%20%7Bx%7D%20%5C%2C%20dx%20%5Csqrt%7Bx%7D%20%5C%5C%20%5Cleft%20%5C%7B%20%7B%7By%3D2%7D%20%5Catop%20%7Bx%3D2%7D%7D%20%5Cright.%20%5Cleft%20%5C%7B%20%7B%7By%3D2%7D%20%5Catop%20%7Bx%3D2%7D%7D%20%5Cright.%20x%5E%7B2%7D%20x%5E%7B2%7D%20%5Csqrt%7Bx%7D%20%20%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20a_n%20%20%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20a_n%20%5Cneq%20%5Csqrt%7Bx%7D%20%5Csqrt%5Bn%5D%7Bx%7D%20%5Cfrac%7Bx%7D%7By%7D%20%5Cfrac%7Bx%7D%7By%7D%20%5Calpha%20%5Cbeta%20x_%7B123%7D%20%5C%5C%20x%5E%7B2%7D%20%5Cint%5Climits%5Ea_b%20%7Bx%7D%20%5C%2C%20dx%20x%5E%7B2%7D)
(2x+y) * 2 = 4x+2y |
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4x+2y+3
2(1)+3
2+3
<u>5</u>
I got this picture from photomath, hope it helps
Answer: w+1.3 = 2.7
Reason:
The smaller pieces (w and 1.3) combine to form the larger piece (2.7)
To find w, subtract 1.3 from both sides to get 1.4
1.4+1.3 = 2.7
M=15...................................................................
