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VikaD [51]
3 years ago
10

Can some help me fast!

Mathematics
1 answer:
Mice21 [21]3 years ago
7 0

Answer:

5x +2y =38

5(6) +2y = 38

30 +2y= 38

2y= 8

y= 4

(6, 4)

5(0) +2y =38

2y= 38

y= 19

(0,19)

5(-2) +2y= 38

-10 +2y =38

2y= 48

y=24

(-2,24)

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Christopher made 3 1/4 dozen cupcakes. He added vanilla icing to 1/3 of the cupcakes and added chocolate icing to the rest. How
Arada [10]
1 dozen = 12

First you would multiply.

3.25 x 12 = 39 cupcakes total
1/3 x 12 = 4 cupcakes with vanilla icing

Then you subtract.

39 - 4 = 35 cupcakes with chocolate icing

Then to get the answer back in to terms of dozens, divide by 12.

35/12 = 2 11/12

Hope this helped
6 0
3 years ago
Read 2 more answers
Factor this trinomial.
andrew11 [14]

Answer:

C

Step-by-step explanation:

If you were to factor the equation out you would get

(x-5x)(15-3)

Simplify it to

x(1-5) -3(-5-1)

Where you would then get

(x-3)(x-5)

Use FOIL to check answer



4 0
4 years ago
Read 2 more answers
WILL GIVE BRIANIEST PLEASE HELP !!!
gregori [183]

Answer:

C 180-123

Step-by-step explanation:

This is because 180 is what triangles always equal and if one angle is 128 then the rest will be answered from this equation

3 0
3 years ago
L :V --> W is a linear transformation. Prove each of the following (a) ker L is a subspace of V. (b) range L is a subspace of
iragen [17]

Answer:

a) Assume that x,y\in\ker L, and \alpha is a scalar (a real or complex number).

<em>First. </em>Let us prove that \ker L is not empty. This is easy because L(0_V)=0_W, by linearity. Here, 0_V stands for the zero vector of V, and 0_W stands for the zero vector of W.

<em>Second.</em> Let us prove that \alpha x\in\ker L. By linearity

L(\alpha x) = \alpha L(x)=\alpha 0_W=0_W.

Then, \alpha x\in\ker L.

<em>Third. </em> Let us prove that y+ x\in\ker L. Again, by linearity

L(x+y)=L(x)+L(y) = 0_W + 0_W=0_W.

And the statement readily follows.

b) Assume that u and v are in range of L. Then, there exist x,y\in V such that L(x)=u and L(y)=v.

<em>First.</em> Let us prove that range of L is not empty. This is easy because L(0_V)=0_W, by linearity.

<em>Second.</em> Let us prove that \alpha u is on the range of L.

\alpha u = \alpha L(x) = L(\alpha x) = L(z).

Then, there exist an element z\in V such that L(z)=\alpha u. Thus \alpha u is in the range of L.

<em>Third.</em> Let us prove that u+v is in the range of L.

u+v = L(x)+L(y) = L(x+y)=L(z).

Then, there exist an element z\in V such that L(z)= u +v. Thus u +v is in the range of L.

Notice that in this second part of the problem we used the linearity in the reverse order, compared with the first part of the exercise.

Step-by-step explanation:

6 0
3 years ago
In which of these figures will the diagonals not always be perpendicular?
Harman [31]
Rectangle and square

However, I would need to know which figures are involved seeing as there are no pictures of anything to work with.
7 0
4 years ago
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