Are large amounts followed by uncountable nouns and predicate verbs in singular or plural form?

Are large amounts followed by uncountable nouns and predicate verbs in singular or plural form?

Amount is an uncountable noun, a large amount of + uncountable noun, the predicate verb is singular. Large amounts of + uncountable n the predicate verb is plural eg.Large amounts of money are spent on the subject.
Depending on the amount, whether quantity is singular or plural, if it is singular, use the singular; if it is plural, use the plural
How to determine the singular and plural of the predicate verbs when the nouns modified by a quantity of and quantities of are used as subjects,
Well, you can either use a large / small quantity of or the quantity of ~ to get to the point. Look carefully ~ a large / small quantity of + plural noun + predicate verb plural a large / small quantity of + uncountable noun +
A quantity of silver which had been fused in a ladle was allowed to solidify.
The silver dissolved in the spoon solidified.
He collected a quantity of interesting information.
He collected a lot of interesting information.
To prove that f (x) = xcosx is not a periodic function
The good man helped
Generally speaking, Y1 = x
Y2 = cosx, the first is not a period, the second is a period, so the product is not a period
Let the minimum positive period of F (x) be t,
Then f (x + T) = f (x); that is, (x + T) cos (x + T) = xcosx;
Obviously, for any x, the equation is always true, so take x = - π / 2,0;
The results show that (- π / 2 + T) cos (- π / 2 + T) = 0; tcost = 0;
Only t = π / 2 satisfies the two equations; (t = 0 is not a positive number, rounding off)
So the period of F (x) can only be π / 2,
But obviously f (π / 3), f ((5 / 6) * π... Expand
Let the minimum positive period of F (x) be t,
Then f (x + T) = f (x); that is, (x + T) cos (x + T) = xcosx;
Obviously, for any x, the equation is always true, so take x = - π / 2,0;
The results show that (- π / 2 + T) cos (- π / 2 + T) = 0; tcost = 0;
Only t = π / 2 satisfies the two equations; (t = 0 is not a positive number, rounding off)
So the period of F (x) can only be π / 2,
But obviously f (π / 3), f ((5 / 6) * π) are not equal. contradiction
So f (x) = xcosx is not a periodic function
Although a periodic function multiplied by an aperiodic function is definitely an aperiodic function, it obviously does not meet the requirements of the topic. Put it away
Given that a, B and X are positive numbers and LG (BX) · LG (AX) + 1 = 0, the value range of AB is obtained
Let LGB (LGB) be (LGA + 2) + LGB (LGB) = (LGA + 2) + (LGB) = (LGA + 2) + (LGB) = (LGA + 2) + (lga-2) + (LGB) = (LGA + 2) + (LGB) = (LGA + 2) + (LGB) = (lga-2) + (LGA + 2) + (LGB) = (LGA + 2) + (lga-2) + (LGB) = (LGA + 2) + (lgb-2) + (LGA + 2) + (lga-2) + (LGA + LGB) LGB (LGA + 2) + (LGB) LGB (LGA + LGB (LGB) = (LGA + 2) + (LGB (lga-2) + (LGA + (LGA + 2) + (LGA + LGB) LGA + LGA + LGB (LGA + LGA + 2 + LGA + LGA + LG Or 0 ＜ ab ≤ 1100. The range of AB is (01100) ∪ [100, + ∞)
The odd function f (x) defined on R satisfies: when x > 0, f (x) = 201ox + log201ox, then the number of real roots of the equation f (x) = 0 on R is ()
A. 1B. 2C. 3D. 4
When x ＞ 0, Let f (x) = 0, that is 201x x = - log201x, draw the images of functions F1 (x) = 201x, F2 (x) = log201x in the same coordinate system, as shown in the right figure, we can see that there is only one intersection point between two images, that is, the equation f (x) = 0 has only one real root, ∵ f (x) is an odd function defined on R, ∵ when x ＜ 0, the equation f (x) = 0 also has a real root, and ∵ f (0) = 0, The number of real roots of the equation f (x) = 0 is 3
How to find the period of function
Sin3x * cos3x = (2sin3x * cos3x) / 2 = (sin6x) / 2, so the period T = 2 π / 6 = π / 3. Because the maximum and minimum of SiNx are 1 and - 1 respectively, when sin6x is 1, the function gets the maximum value of 1 / 2. When sin6x is - 1, the function gets the minimum value of - 1 / 2. (2) 1 / 2-sin2x-x0d constant term
In the problem of logarithmic function, the function y = LG (AX-1) decreases monotonically on (- ∞, 1), and the value range of a is obtained
The function y = LG (AX-1) decreases monotonically on (- ∞, 1), and the value range of a is obtained
I Know
Let u = ax + 1, a
As a value of X in function lgx, u (1) must satisfy the requirement of X, that is, the basic condition of x > 0 in y = lgx
Given the function f (x) = - x ^ 2 + ln (1 + 2x), let b > a > 0, it is proved that ln (a + 1) / B + 1 > (a-b) (a + B + 1)
It is proved that: ln (a + 1) > 0, B + 1 > 0, a + B + 1 > 0, that is ln (a + 1) / B + 1 > 0, and (a-b) (a + B + 1) < 0
Let g (x) = - x ^ 2 + ln (1 + x) g ′ (x) = - 2x + 1 / (1 + x) when x > 0, G ′ (x) - B ^ 2 + ln (1 + b) is transferred to ln (a + 1) / B + 1 > A ^ 2-B ^ 2 > A ^ 2-B ^ 2 + A-B, then ln (a + 1) / B + 1 > (a-b) (a + B + 1)
So I think you should change "f (x) = - x ^ 2 + ln (1 + 2x)" to "f (x) = - x ^ 2 + ln (1 + x)"
How to judge whether this function is periodic
f(x)=xcosx
Let f (x) = f (x + T) - f (x) f (x) = 0. For any x constant, f (x) = xcos (x + T) + TCOS (x + T) - xcosx = TCOS (x + T) - 2xsin (x + T / 2) sin (T / 2) let x = 0, f (0) = tcost = 0
Let the logarithmic equation LG (AX) = 2lg (x-1). When a takes a value in what range, the equation has a solution, and its solution is obtained
lg(ax)=2lg(x-1),
=> ax=(x-1)²
And ax > 0, X-1 > 0, = > a > 0, x > 1
=>X & # 178; - (a + 2) x + 1 = 0
=> Δ=(a+2)²-4≥0,
=>A ≥ 0 or a ≤ - 4
=> a>0
The solution is (a) = + 2,
X>1
=> x=[a+2+√(a²+4a)]/2 , a>0