Given 2x + 3y-2 = 0, find the value of [(10 ^ 2) ^ x] times [(10 ^ y) ^ 3] - 99

Given 2x + 3y-2 = 0, find the value of [(10 ^ 2) ^ x] times [(10 ^ y) ^ 3] - 99

2x+3y-2=0,
2x+3y=2
[(10 ^ 2) ^ x] times [(10 ^ y) ^ 3] - 99
=10^2x×10^3y-99
=10^(2x+3y)-99
=10^2-99
=100-99
=1

Given the quadratic power of X + 2x - 1 = 0, find the value of the cubic power of 2x + the quadratic power of 5x + 1999

x²＋2x=1
Then:
2x³＋5x²＋1999
=2x(x²＋2x)＋x²＋1999
=2x＋x²＋1999
=1＋1999
=2000

Let a > 0, when - 1 ≤ x ≤ 1, the minimum value of the function y = - x2 ax + B + 1 is - 4, and the maximum value is 0

Y = − x2 − ax + B + 1 = − (x + A2) 2 + A24 + B + 1; (1) if − A2 ≤− 1, i.e. a ≥ 2, the function y decreases monotonically on [- 1, 1]; the minimum value of the function is B-A = - 4; the maximum value is a + B = 0, i.e. a = 2, B = - 2; (2) if − 1 < A2 < 0, i.e. 0 < a < 2, when x = − A2, the maximum value of function y is A24 + B + 1 = 0 & nbsp; & nbsp; ① Moreover, f (- 1) = a + B, f (1) = - A + B, f (1) ＜ f (- 1), the minimum value of function y is - A + B = - 4 & nbsp; & nbsp; ②; ① ② the simultaneous solution of the two equations obtains a = 2, B = - 2, which does not conform to 0 ＜ a ＜ 2, which does not exist; to sum up, a = 2, B = - 2

Two thirds of a is equal to three quarters of B. what is the ratio of a to B

A:B=3/4:2/3=9:8

The top of a trapezoid is three times the length of the bottom. If the bottom is extended to 8 cm, the trapezoid will become a parallelogram. How many cm are the top and bottom of the trapezoid?

The bottom is 8 (3-1) = 4cm
The top and bottom are 4 × 3 = 12 cm

What is the basis for drawing an angle equal to a known angle with a ruler

SSS is the basis for drawing an angle equal to a known angle with ruler

If the coefficients of two quadratic equations of one variable are equal, why can we take the unknowns of these two equations as the two roots of another equation and use the unreachable theorem

1. A quadratic equation with one variable has real roots, which must be two
2. Substituting the value of the unknowns into the equation, the equal sign of the equation holds, which shows that the value of the unknowns is a root of the equation. Therefore, we can take the unknowns of the two equations as the two roots of the other equation, and then use the unreachable theorem

It is known that the area of the circle in the figure below is 62.8 square centimeters, so we can find the area of isosceles right triangle
One right side of a right triangle is the radius of a circle

S circle = π × R × r = 62.8
So r = √ 20
Area of isosceles right triangle = 1 / 2 × right side × another right side = 1 / 2 × 20 = 10

The image of function y = 12x − 4 is given, and the questions are answered according to the image: (1) when x takes what value, Y > 0; (2) when - 1 ≤ x ≤ 2, the value range of Y is obtained

(1) Drawing can be: from the image, when x ＞ 8, y ＞ 0; (2) first express the unknown x with y, and then calculate the range of Y according to the value range of X. from y = 12x-4 to x = 2Y + 8, that is - 1 ≤ 2Y + 8 ≤ 2, the solution is - 92 ≤ y ≤ - 3. Answer: the value range of Y is - 92 ≤ y ≤ - 3

Given z = ln (x, y), find the first and second partial derivatives of Z

Z'|X＝Y/﹙XY﹚＝1/X Z"|XX＝-1/X²
Z'|Y＝X/﹙XY﹚＝1/Y Z"|YY＝-1/Y²
Z"|XY＝Z"|YX＝0