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babymother [125]
3 years ago
14

Here are two rectangles.

Mathematics
2 answers:
dedylja [7]3 years ago
5 0

Step-by-step explanation:

Scale Factor = Ratio of Side Lengths

= 12/8 = 9/6 = 1.5.

Vladimir79 [104]3 years ago
5 0

Answer:

It is

Scale Factor = Ratio of Side Lengths

= 12/8 = 9/6 = 1.5.

Step-by-step explanation:

Hope this helped have an amazing day!

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4. Round to the nearest 100 to<br>estimate the difference.<br>390 - 122 =<br>​
VARVARA [1.3K]

Answer:

400-100=300

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4 years ago
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QveST [7]
-3>-4 because the number line shows it going to the right side which means the number is bigger then -4 so the only bigger answer is -3 and is correct
8 0
3 years ago
Read 2 more answers
Find the vertex and the x-inter of the function given below. y = x^2 + 2x + 6
DIA [1.3K]

Answer: Vertex = (-1, 5) x-intercepts = No real solutions

Step-by-step explanation:

Vertex = -2/2= -1

-1^2 + 2(-1) +6

1 -2 +6

-1 +6

=5

Discrminats= 2^2 -4(1)(6)

4 -4(1)(6)

4-24

= -20 (It's a negative so there's no real solutions, meaning there's no xintercepts/R0XS)

7 0
3 years ago
Solve for x. 3x/2 = 15 x = 22 1/2 x = 10 1/2 x = 10 x = 1/10
Leokris [45]

Answer:

x = 10

Step-by-step explanation:

(3x)/2 = 15 is your question.

Isolate the variable (x). Note the equal sign, what you do to one side, you do to the other. Do the opposite of PEMDAS (Parenthesis, Exponent (& roots), Multiplication, Division, Addition, Subtraction).

First, multiply 2 to both sides

((3x)/2) = (15)(2)

3x = (15)(2)

3x = 30

Next, to isolate the variable (x), divide 3 from both sides

(3x)/3 = (30)/3

x = 30/3

x = 10

x = 10 is your answer

~

8 0
3 years ago
Read 2 more answers
L :V --&gt; 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
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