Solution equation: X: 6 / 5 = 25 / 42

Solution equation: X: 6 / 5 = 25 / 42

X = 25 / 42 times 6 / 5
x=5/7
Please click the "select as satisfactory answer" button below,

X / (X-30) = (x + 40) / X solves the equation,

X/(x-30)=(x+40)/X
X^2=(X-30)*(X+40)
X^2=X^2+40X-30X-1200
10X=1200
X=120

There is a rectangular stadium. If its length and width are increased by six meters, its area will be increased by 1236 square meters. What is the perimeter of the original stadium?

Let the original side lengths be x and Y respectively
Now the extra area is 6 * y + X * 6 + 36 = 1236
Then (x + y) * 6 = 1200
Then x + y = 200
The original perimeter is 2 * (x + y) = 400m

The development of the seventh grade mathematics Evaluation Handbook Suke Edition
A and B both go from a to B. A walks at a speed of 5km / h, 1.5 hours first; B rides a bicycle, and it takes 50 minutes for them to arrive at B at the same time
That's question 5 on page 79

14 km / h
Set X kilometers per hour
1.5*5+50/60=50/60X
X=14

Indefinite integral with (e ^ 2x + 5) radical

Let t = √ [e ^ (2x) + 5], then x = 1 / 2 · ln (T & # 178; - 5), DX = t / (T & # 178; - 5) DT
∫√[e^(2x)+5] dx
=∫t·t/(t²-5) dt
=∫(t²-5+5)/(t²-5) dt
=∫[1+5/(t²-5)]dt
=∫dt+1/2·∫[1/(t-5)-1/(t+5)]dt
=t+1/2·(ln|t-5|-ln|t+5|)+C
=t+1/2ln|(t-5)/(t+5)|+C
=√[e^(2x+5)]+1/2 ln|{√[e^(2x)+5]-5}/{√[e^(2x)+5]+5}|+C

Please put the following 9 letters in correct order to spell out an English word
If this sign is on my dressing room door ,what am I
_ _ _ _ _ _ _ _ _
e i r c l y e b t

celebrity
Star, celebrity

The side view of a cylinder is a rectangle with an area of 16 π square. What is the radius of the bottom of the cylinder
It's a process

The side view is a rectangle, 2 π r in length and h in width. The answer is: r = 8 / h, OK?

Please add appropriate symbols or brackets to the left of the following formula to make the equation hold: 0 = 24

(cos0+cos0+cos0+cos0)!=4!=24

If ax & # 178; + ax + 1 ≥ 0 holds for all x ∈ R, then the value range of a is

If ax & # 178; + ax + 1 ≥ 0 holds for all x ∈ R
① 1 ≥ 0 holds for all x ∈ r when a = 0
accord with
② When a ≠ 0
Satisfy a > 0, Δ = A & # 178; - 4A ≤ 0
So 0 < a ≤ 4
In conclusion, the value range of a is [0,4]
If you don't understand, please ask, I wish you a happy study!

It is proved that the square of 5x - 6x + 10 and 4 / 5 is always greater than 0

5x^2-6x+10+4/5
=5(x^2-6x/5)+10+4/5
=5(x^2-6x/5+9/25)-9/5+10+4/5
=5(x-3/5)^2+9
Because 5 (x-3 / 5) ^ 2 > = 0,
So 5 (x-3 / 5) ^ 2 + 9 > 0
That is, 5x ^ 2-6x + 10 + 4 / 5 > 0