As shown in the figure, a warehouse with an area of 130m2 is to be built. One side of the warehouse is close to the wall (the length of the wall is 16m) and a 1m wide door is opened on the side parallel to the wall. At present, there are 32m long wooden boards which can be enclosed to calculate the length and width of the warehouse

Let the length of one side of the warehouse perpendicular to the wall be X. according to the meaning of the question, (32-2x + 1) x = 130, 2x2-33x + 130 = 0, (X-10) (2x-13) = 0,  X1 = 10 or x2 = 6.5, when X1 = 10, 32-2x + 1 = 13 ＜ 16; when x2 = 6.5, 32-2x + 1 = 20 ＞ 16, which does not fit the meaning of the question. Answer: the length and width of the warehouse are 13m and 10m respectively

As shown in the figure, a rectangular sheet of iron with a length of 60cm and a width of 40cm is to cut off four equal small squares at its four corners and fold it into a cuboid water tank without a cover, so that its bottom area is 800cm square. The side length of the square is to be cut off

Let the side length of the truncated square be x cm. According to the meaning of the question, the length and width of the bottom of the rectangle are (60-2x) cm and (40-2x) cm respectively, so the bottom area of the rectangle is (60-2x) (40-2x) = 800, that is, x2-50x + 400 = 0, and the solution is X1 = 10, X2 = 40

An applied problem of quadratic equation of one variable in grade two of junior high school
What is the width of the silver edge if you want to insert the same width silver edge on the outside of an 8cm * 12cm photo frame and make the area of the silver edge equal to the area of the photo frame?
What does "^" mean?

If the width of the silver edge is x cm, then
2 * 12 * x + 2 * 8 * x + 4x ^ 2 = 8 * 12
{{{X - > - 12} (negtive root rounding), {X - > 2}}
X ^ 2 is the square of X

We know that E1 and E2 are two non collinear vectors in the plane, a = 3e1-2e2, B = - 2E1 + E2, C = 7e1-4e2. We use a and B to represent C

Let C = n × a + m × B
So, 7e1-4e2 = n × (3e1-2e2) + m × (- 2E1 + E2)
Therefore, 7e1 = n × 3e1-m × 2E1; - 4e2 = - n × 2e2 + m × E2
n=1；m=-2
So, C = a-2b

Solution equation: 1 / 19 + (1540-x) × 1 / 10 = 100

x/19+(1540-x)/10=100
-(9/190) (-1140 + x)=0
x=1140

Solving equation x - (2 / 5x + 28) = 9 / 4x-14
incorrect. Wrong list. It should be X - (2 / 5x + 28) = 4 / 9x-14

X-(2/5X+28)=9/4X-14
Then 3 / 5x + 28 = 9 / 4x-14
Then 12x + 560 = 45x-280
Then 33x = 840
The solution is x = 280 / 11

What is the value range of X satisfying log root 3 (3x + 1) > log root 3 (2x-1)?

Firstly, in log root 3 (3x + 1) and log root 3 (2x-1), both 3x + 1 and 2x-1 must be greater than 0 to get x > 1 / 2
And the root sign 3 is about 1.732 larger than 1, that is to say, the equation of log root sign 3 (x) is monotonically increasing, so in order to satisfy the inequality in the question, we should have 3x + 1 > 2x-1, and get x > - 2. Put together with the above results, we get the answer as x > 1 / 2
(theorem: when the base of log equation is larger than 1, it is monotonically increasing, and when the base is between 0 and 1, it is monotonically decreasing.)

Urgent request with 74ls161 design 24 base counter, has the circuit diagram to be better

Because it's a mobile phone, I can't give you the circuit diagram. I can give you a plan
74ls161 is a hexadecimal counter with asynchronous setting and synchronous clearing. There are two ways to form a 24 ary counter
1. Asynchronous setting method. Because it is asynchronous, it can be set directly without waiting for the clock signal. To form a 24-ary counter, two chips need to be cascaded. The first chip counts 16 times and then carries once, and the second chip counts once. When the first chip counts 8 times and the second chip counts once, it counts 24 times. At this time, the gate circuit translates the setting signal to the setting terminal
2. Synchronous clearing method. The principle is the same as the number setting method, but it is synchronous clearing and needs to wait for the clock signal to work together to clear. Therefore, after counting the first piece for 7 times and the second piece for 1 time, it is 23 times counting. At this time, the clearing signal is translated, and then wait for a clock signal. The time counting is 24 times and the clearing is just completed
If you don't know anything, please let me know
I hope my answer can help you

The radii of two different circles passing through point C (3.4) and tangent to x-axis and y-axis are R1 and R2 respectively?

Let the radius of a circle be r (r > 0), because it passes through C (3,4) and X axis, and Y axis is tangent, so the whole circle is in the first quadrant, and the coordinate of the center of the circle is (R, R). Then the equation of the circle is (X-R) ^ 2 + (y-r) ^ 2 = R ^ 2. Because the circle passes through (3,4), substituting x = 3 and y = 4 into the equation, we get the sum of (3-R) ^ 2 + (4-R) ^ 2 = R ^ 2-7r + 25 = 0, R1 + R2 =

In the expansion of binomial (1-2x) ^ 6, the sum of all coefficients is

The sum of all the coefficients is that when x = 1, (1-2) ^ 6 = 1