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Anuta_ua [19.1K]
2 years ago
5

If t is 25% of u, then what percent of 4t is 2u?​

Mathematics
2 answers:
vichka [17]2 years ago
8 0
T = 0.25u
4t = 4(0.25u) =
u
Then u is 50 percent of 2u.
Hope this helps, please mark brainliest if possible :)
dexar [7]2 years ago
7 0
<h2>Howdy,</h2>

Step-by-step explanation:

t = .25u        (given)

4 t = 4 (.25u)  = u

then u = 50% of 2 u

You might be interested in
If x and y are integers, and 3x + 2y = 13, which of the following could be the value of y ?
lions [1.4K]
They are all integers so...
first lets substitute
A.3x=13
X=13/3
b.3x+2=13
3x=11
x=11/3
c.3x+4=13
3x=9
x=3
d.3x+6=13
3x=7
x=7/3

E.3x+8=13
3x=5
x=5/3

therefore the possible is letter c because it must be an integer and the definition of integer states that it is not a decimal nor a fraction

3 0
4 years ago
Read 2 more answers
3 times 5 with a two on the top
jenyasd209 [6]
Ok so it could be
1. \frac{2}{3 times 5}
2. 3 timies \frac{2}{5}


ok if 1. then
\frac{2}{3 times 5}=\frac{2}{15}

if 2.  3 timies \frac{2}{5}= \frac{6}{5}



3 0
3 years ago
I need help I’m stuck on this
arsen [322]

press ask for help i think it my help im also stumped on that

8 0
3 years ago
All vectors are in Rn. Check the true statements below:
Oduvanchick [21]

Answer:

A), B) and D) are true

Step-by-step explanation:

A) We can prove it as follows:

Proy_{cv}y=\frac{(y\cdot cv)}{||cv||^2}cv=\frac{c(y\cdot v)}{c^2||v||^2}cv=\frac{(y\cdot v)}{||v||^2}v=Proy_{v}y

B) When you compute the product Ax, the i-th component is the matrix of the i-th column of A with x, denote this by Ai x. Then, we have that ||Ax||=\sqrt{(A_1 x)^2+\cdots (A_n x)^2}. Now, the colums of A are orthonormal so we have that (Ai x)^2=x_i^2. Then ||Ax||=\sqrt{(x_1)^2+\cdots (x_n)^2}=||x||.

C) Consider S=\{(0,2),(2,0)\}\subseteq \mathbb{R}^2. This set is orthogonal because (0,2)\cdot(2,0)=0(2)+2(0)=0, but S is not orthonormal because the norm of (0,2) is 2≠1.

D) Let A be an orthogonal matrix in \mathbb{R}^n. Then the columns of A form an orthonormal set. We have that A^{-1}=A^t. To see this, note than the component b_{ij} of the product A^t A is the dot product of the i-th row of A^t and the jth row of A. But the i-th row of A^t is equal to the i-th column of A. If i≠j, this product is equal to 0 (orthogonality) and if i=j this product is equal to 1 (the columns are unit vectors), then A^t A=I    

E) Consider S={e_1,0}. S is orthogonal but is not linearly independent, because 0∈S.

In fact, every orthogonal set in R^n without zero vectors is linearly independent. Take a orthogonal set \{u_1,u_2\cdots u_p\} and suppose that there are coefficients a_i such that a_1u_1+a_2u_2\cdots a_nu_n=0. For any i, take the dot product with u_i in both sides of the equation. All product are zero except u_i·u_i=||u_i||. Then a_i||u_i||=0 then a_i=0.  

5 0
4 years ago
Help me on my math homework first and correct one gets brain.lyest :)
stira [4]
The answer is the third choice they both equal -16/77. Hope this helps:)
7 0
3 years ago
Read 2 more answers
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