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If the function f (x) = asinx bcosx (AB ≠ 0) has f (π / 4-x) = f (π / 4 + x) for any real number x, then the slope of the line ax-2by + C = 0 is
If f (π / 4-x) = f (π / 4 + x), then the axis of symmetry of function f (x) is x = π / 4. Since x = π / 4 is the axis of symmetry, then f (0) = f (π / 2),
Therefore, there are three aspects
－b=a =====>>>> a/b=－1
For the straight line ax-2by + C = 0, the slope k = A / (2b) = - 1 / 2, that is, the slope of the straight line is - 1 / 2
The area of triangle ABO is 18 square centimeters, Bo = 3od? What is the area of trapezoid ABCD
Triangle AOB is similar to triangle ODC
S（AOB）/（S（ODC）=（OB/OD）^2=9
S(ODC)=15/9=5/3
S(AOD)/S(AOB)=OD/OB
S(AOD)=1/3S(AOB)=5
S(OBC)/S(ODC)=OB/OD
S(0BC)=3*5/3=5
S(ABCD)=15+5/3+5+5=80/3
If a * b = B (b > A) or a (a ≥ b) is defined, then the range of function f (x) = lgx / LG2 * lgx / lg0.5 is
x> 1, f (x) = lgx / LG2 > 0
0
If the cone, bottom circumference 12.56 cm, 6 cm, cone volume is () cubic centimeter
Bottom radius = 12.56 △ 3.14 △ 2 = 2cm
Bottom area = 3.14 × 2 × 2 = 12.56 square centimeter
Volume = 12.56 × 3.6 × 1 / 3 = 15.072 cm3
① (2a + 5b) ^ 2 - (4a-7b) (4a + 7b) + (2a-5b) ^ 2 (m-n + P-Q) (m-n-p-q) cough, speed, ha ziha, thank you
①(2a+5b)^2-(4a-7b)(4a+7b)+(2a-5b)^2=4a²+20ab+25b²-16a²+49b²+4a²-20ab+25b²=99b²-8a²②(m-n+p-q)(m-n-p-q)=(m-n-q)²-p²=(m-n)²-2(m-n)q+q²-p²...
The lengths of three sides of a triangle are 15 cm, 20 cm and 25 cm respectively?
The ratio of the three sides of the triangle
=15：20：25
=3：4：5
So, this is a right triangle
The sides 15 and 20 cm long are right angles
Its area
=15×20÷2
=150 (square centimeter)
The height of the longest side of the triangle
=150×2÷25
=12 (CM)
What number is even and prime
2
Given that the perimeter of a rectangle is 10 and the length of one side is x, the functional expression of area s and X is
The other side is 5-x long
So s = (5-x) X
As shown in the figure, AB is the diameter of ⊙ o, ad is the tangent of ⊙ o, point C is on ⊙ o, and BC ∥ OD. If AB = 4, OD = 6, then the length of BC is equal to______ ．
∵ AB is the diameter, ∵ C = 90 °, ∵ ad is the tangent line of ⊙ o, ∵ Dao = 90 °, ∵ C = ∵ Dao, and ∵ BC ∥ OD, ∵ B = ∵ AOD, ∥ ABC ∥ DOA, ∥ bcoa = abod, that is, BC2 = 46, the solution is BC = 43, so the answer is: 43
As shown in the figure, m --- s --- P --- Q --- t --- n, MP: PQ: QN = 3:2:4, points s and T are the midpoint of MP and QN, and St = 11cm,
Then Mn = - cm
18cm
Let MP be 3x, PQ be 2x and QN be 4x, then st = MP / 2 + PQ + QN / 2 = 1.5x + 2x + 2x = 5.5x = 11cm
All x = 2cm, Mn = 3x + 2x + 4x = 9x = 18cm
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