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3 years ago

State if the two triangles are congruent, if they are state how you know.

Mathematics

1 answer:

Mnenie [13.5K]3 years ago

4 0

Answer:

I belive it would be HL because its defenitly not SSS

Step-by-step explanation:

HL Postulate is if the hypotenuse and leg of one right triangle are congruent to the hypotenuse and leg of another right triangle, then the two triangles are congruent.

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Let g' be the group of real matricies of the form [1 x 0 1]. Is the map that sends x to this matrix an isomorphism?

aliina [53]

Yes. Conceptually, all the matrices in the group have the same structure, except for the variable component $x$ . So, each matrix is identified by its top-right coefficient, since the other three entries remain constant.

However, let's prove in a more formal way that

$\phi:\ \mathbb{R} \to G,\quad \phi(x) = \left[\begin{array}{cc}1&x\\0&1\end{array}\right]$

is an isomorphism.

First of all, it is injective: suppose $x \neq y$ . Then, you trivially have $\phi(x) \neq \phi(y)$ , because they are two different matrices:

$\phi(x) = \left[\begin{array}{cc}1&x\\0&1\end{array}\right],\quad \phi(y) = \left[\begin{array}{cc}1&y\\0&1\end{array}\right]$

Secondly, it is trivially surjective: the matrix

$\phi(x) = \left[\begin{array}{cc}1&x\\0&1\end{array}\right]$

is clearly the image of the real number x.

Finally, $\phi$ and its inverse are both homomorphisms: if we consider the usual product between matrices to be the operation for the group G and the real numbers to be an additive group, we have

$\phi (x+y) = \left[\begin{array}{cc}1&x+y\\0&1\end{array}\right] = \left[\begin{array}{cc}1&x\\0&1\end{array}\right] \cdot \left[\begin{array}{cc}1&y\\0&1\end{array}\right] = \phi(x) \cdot \phi(y)$

8 0

3 years ago

What is the answer to this

Softa [21]

Answer:

5 f

Re-order terms so that constants are on the left

. 5

5 f

solution

5 f

7 0

3 years ago

1.) What is the maximum number of zeros for the following function f(x) = x - 2x2 - 11x + 12

Lesechka [4]

Answer:

-14x+12

Step-by-step explanation

f(x)=-14

4 0

3 years ago

In a group of a hundred and fifty students attending a youth workshop in mombasa, 125 of them are fluent in kiswahili, 135 in en

jek_recluse [69]

Answer:

The probability that a student chosen at random is fluent in English or Swahili.

P(S∪E) = 1.1

Step-by-step explanation:

Step(i):-

Given total number of students n(T) = 150

Given 125 of them are fluent in Swahili

Let 'S' be the event of fluent in Swahili language

n(S) = 125

The probability that the fluent in Swahili language

$P(S) = \frac{n(S)}{n(T)} = \frac{125}{150} = 0.8333$

Let 'E' be the event of fluent in English language

n(E) = 135

The probability that the fluent in English language

$P(E) = \frac{n(E)}{n(T)} = \frac{135}{150} = 0.9$

n(E∩S) = 95

The probability that the fluent in English and Swahili

$P(SnE) = \frac{n(SnE)}{n(T)} = \frac{95}{150} = 0.633$

Step(ii):-

The probability that a student chosen at random is fluent in English or Swahili.

P(S∪E) = P(S) + P(E) - P(S∩E)

= 0.833+0.9-0.633

= 1.1

Final answer:-

The probability that a student chosen at random is fluent in English or Swahili.

P(S∪E) = 1.1

8 0

2 years ago

Y = (x + 4)2 - 1 Standard form

denpristay [2]

answer :

y = 2x + 7

i hope it helped

3 0

3 years ago

Read 2 more answers