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3 years ago

Pleasse and thank u <33

Mathematics

1 answer:

liubo4ka [24]3 years ago

7 0

Answer:

4x-2x+3y+6x+6y distributive property

4x-2x+6y+3y+6y commutative property of addition

8x+9y combine like terms

Step-by-step explanation:

I hope this helps!

You might be interested in

Lila evaluates the expression using the following steps:(−7.13+63.97)+(−54.8−3.4)−71.1+(−58.2)−129.3Describe the error that Lila

mihalych1998 [28]

Lila's error was that she added -7.13 to 63.97 to get -71.1

Lila steps are as folows

(−7.13 + 63.97) + (−54.8 − 3.4) = −71.1 + (−58.2) = −129.3

In the first bracket, Lila was suppose to subtract 7.31 from 63.97 since there is a negative in front of 7.31.

So, the correct expression should be

(−7.13 + 63.97) + (−54.8 − 3.4) = 56.84 + (−58.2) = 56.84 - 58.2 = −1.36

So, Lila's error was that she added -7.13 to 63.97 to get -71.1

Learn more about addition and subtraction here:

brainly.com/question/16812794

4 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).

First. 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.

Second. Let us prove that $\alpha x\in\ker L$ . By linearity

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

Then, $\alpha x\in\ker L$ .

Third. 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 $L(y)=v$ .

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

Second. 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 $L$ .

Third. 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

What is 5 divided by 720

Sav [38]

5 divides by 720
i think it's answer is 144
you do 144 times table by 5 you got 720
so this answer is 144.

3 0

3 years ago

HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

valentinak56 [21]

The answer is 3. Hope this helped! Good luck! Let me know if you have any more questions! :)

8 0

4 years ago

Read 2 more answers

At what point do the curves r1(t) = t, 5 − t, 48 t2 and r2(s) = 8 − s, s − 3, s2 intersect?

DerKrebs [107]

Set components equal:
$t = 8-s \\ 5-t = s-3 \\ 48 t^2 = s^2$

From this we find:
$t = 8-s \\ s = 4 \sqrt{3} t$

Substitute and solve system:
$t = \frac{8}{1+4 \sqrt{3}} = 1.009 \\ \\ s = 8-t = 8 - 1.009 = 6.991$

The point where they intersect is:
$(1.009, 5-1.009, 48(1.009)^2)$

3 0

4 years ago