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QveST [7]
3 years ago
8

If a+b=6 and a-b = 4, what is the value of 2a + 3b?

Mathematics
1 answer:
elena-s [515]3 years ago
3 0
A = 5
b = 1
(2*5) + (3*1) = 
10 + 3 = 
13
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In circle A, CD = 25 cm and KL = 25 cm. If m = 110° and m = (2x − 10)°, what is the value of x?
Alex777 [14]
X equals 60 because 2x60=120. and 120-10=110

6 0
3 years ago
Which choicest are equivalent to the fraction below check all that apply 12/32
notka56 [123]
6/16 & 3/8. Hope it helped.
5 0
3 years ago
Billy is a professional baseball player. He had 99 singles, 47 doubles, 4 triples, and 15 home runs in his first year playing. I
Maurinko [17]

Answer: 792

Step-by-step explanation:

From what I got it should be 792 because it is asking for doubles, triples, and home runs all together at the same amount that he got on his first year.

It says after 12 years

Doubles 47 times 12 years = 564

Triples 4 times 12 years = 48

and

Home runs 15 times 12 years = 180

564 + 48 = 612

612 + 180 = 792

8 0
3 years ago
Read 2 more answers
Assume V and W are​ finite-dimensional vector spaces and T is a linear transformation from V to​ W, T: Upper V right arrow Upper
scZoUnD [109]

Answer:

Thus for the vectors v_1, v_2, v_p there are scalars c_1, c_2, c_p not all zeros, such that c_1v_1 +c_2v_2+... +c_pv_p = 0. It means that the vectors v_1, v_2, v_p are linearly dependent in contradiction with the fact that the vectors form a basis for H. So the assumption that T(v_1), T(v_2),..., T(v_p) are linearly dependent is false, proving the required.  

Step-by-step explanation:

Let B = {v_1 ,v_2,..., v_p} be a basis of H, that is dim H = p and for any v ∈ H there are scalars c_1 , c_2, c_p, such that v = c_1*v_1 + c_2*v_2 +....+ C_p*V_p It follows that  

T(v) = T(c_1*v_1 + c_2v_2 + ••• + c_pV_p) = c_1T(v_1) +c_2T(v_2) + c_pT(v_p)

so T(H) is spanned by p vectors T(v_1),T(v_2), T(v_p). It is enough to prove that these vectors are linearly independent. It will imply that the vectors form a basis of T(H), and thus dim T(H) = p = dim H.  

Assume in contrary that T(v_1 ), T(v_2), T(v_p) are linearly dependent, that is there are scalars c_1, c_2, c_p not all zeros, such that  

c_1T(v_1) + c_2T(v_2) +.... + c_pT(v_p) = 0

T(c_1v_1) + T(c_2v_2) +.... + T(c_pv_p) = 0

T(c_1v_1+ c_2v_2 ... c_pv_p) = 0  

But also T(0) = 0 and since T is one-to-one, it follows that c_1v_1 + c_2v_2 +.... + c_pv_p = O.

Thus for the vectors v_1, v_2, v_p there are scalars c_1, c_2, c_p not all zeros, such that c_1v_1 +c_2v_2+... +c_pv_p = 0. It means that the vectors v_1, v_2, v_p are linearly dependent in contradiction with the fact that the vectors form a basis for H. So the assumption that T(v_1), T(v_2),..., T(v_p) are linearly dependent is false, proving the required.  

8 0
3 years ago
Each year the Student Council conducts a food drive. At the end of the drive, the members report on the items collected.
larisa86 [58]

Answer:

A)  Canned and Packet

    Grain, Meat and Vegetables

B)  Let C = canned goods, let P = packet goods

     20C + 7P

     Let G = grain, let M = meat, let V = vegetables

     7G + 3M + 17V

C)  The number of grain and packet goods are the same in each expression, as this food group has been packaged in the same way.

The other variables in both expression are different.  The meat and vegetable food groups have been separated in the second expression, whereas they are part of the same group in the first expression.

3 0
2 years ago
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