All categories

3 years ago

The complement of 25% is 52% true or false

Mathematics

2 answers:

ki77a [65]3 years ago

8 0

Answer:

False

Step-by-step explanation:

Complementary Percentages gives 100% after adding, where 100% represents the Grand Total amount.

Let x be the complement percentage of 25 %,

Thus, by the above statement,

x + 25 % = 100 %

x = 100 % - 25 % = 75 %

So, the complement of 25 % is 75 %,

Hence, this is a false statement that the complement of 25% is 52% .

Anettt [7]3 years ago

3 0

False; it would be 75%.

You might be interested in

Which simplified expression is the difference of (0.5n+0.3) - (0.75n-0.45)

never [62]

The answer is B . your welcome

4 0

3 years ago

Two identical masses are moving down a slope one masses traveling twice as fast as the other mass how does the kinetic energy of

GaryK [48]

Answer:

One mass's kinetic energy is 4 times the other.

Step-by-step explanation:

The formula for kinetic energy is $KE=\frac{1}{2}mv^2$

Where

KE is the kinetic energy

m is the mass

v is the velocity

Now, let mass of the first be m (second is identical so mass of 2nd is m as well)
Let velocity of the first one be v. Since 2nd one is travelling twice as fast, its velocity will be 2v

KE of 1st mass is:

$\frac{1}{2}mv^2$

KE of 2nd mass is:

$\frac{1}{2}m(2v)^2\\=\frac{1}{2}m(4v^2)\\=\frac{1}{2}mv^2 * 4$

Hence, the kinetic energy of the 2nd mass is 4 times the first.

6 0

3 years ago

What is the domain of ^3√x^4+9?

navik [9.2K]

That would be c

x could be any real number.

3 0

3 years ago

Solve the equation. Round to the nearest tenth if necessary.

ASHA 777 [7]

The answer will be 30

5 0

3 years ago

Evaluate the surface integral. s xy ds s is the triangular region with vertices (1, 0, 0), (0, 8, 0), (0, 0, 8)

zmey [24]

Parameterize $S$ by

$\vec s(u,v)=(1-u)v\,\vec\imath+8uv\,\vec\jmath+8(1-v)\,\vec k$

with $0\le u\le1$ and $0\le v\le1$ .

Then the surface element is

$\mathrm dS=\|\vec s_u\times\vec s_v\|\,\mathrm du\,\mathrm dv=8\sqrt{66}\,v\,\mathrm du\,\mathrm dv$

and the surface integral is

$\displaystyle\iint_Sxy\,\mathrm dS=8\sqrt{66}\int_0^1\int_0^18(1-u)uv^3\,\mathrm du\,\mathrm dv=\boxed{8\sqrt{\dfrac{22}3}}$

4 0

3 years ago