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3 years ago

Please help

Mathematics

1 answer:

Yuri [45]3 years ago

7 0

Answer:

Yes

Step-by-step explanation:

$\angle EFT \cong \angle MKL...(1)\\(given) \\\angle EFT \cong \angle GFH... (2)\\(vertical\: \angle 's) \\\angle MKL \cong \angle JKN... (3)\\(vertical\: \angle 's)$

From equations (1), (2) & (3)

$\angle JKN \cong \angle GFH\\$

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Question 2 of 5

Ipatiy [6.2K]

It may be B—JL is longer than JK

5 0

3 years ago

B2-4b+4=0 using quadratic forumla

Ivan

Trying to factor by splitting the middle term

Factoring b2-4b+4
The first term is, b2 its coefficient is 1 .
The middle term is, -4b its coefficient is -4 .
The last term, "the constant", is +4 

Step-1 : Multiply the coefficient of the first term by the constant 1 • 4 = 4

Step-2 : Find two factors of 4 whose sum equals the coefficient of the middle term, which is -4 .

-4 + -1 = -5 -2 + -2 = -4 That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -2 and -2
 b2 - 2b - 2b - 4

Step-4 : Add up the first 2 terms, pulling out like factors :
 b • (b-2)
 Add up the last 2 terms, pulling out common factors :
 2 • (b-2)
Step-5 : Add up the four terms of step 4 :
 (b-2) • (b-2)
 Which is the desired factorization

8 0

4 years ago

Read 2 more answers

GIVING BRAINIEST IF CORRECT!! Factor x^4 - 16 completely

ioda

Answer:

$(x^2 + 4)(x + 2)(x - 2)$

Step-by-step explanation:

Hello!

We can use the difference of square method.

<h2>Difference Of Squares (DOS)</h2>

The formula for the DOS is $(a + b)(a - b) = a^2 - b^2$

It is a simple way to factor polynomials.

The criteria:

Has to begin and end with a perfect square
The operation has to be subtraction

<h3>Factor:</h3>

$x^4 - 16$

Begins with a perfect square (x² * x²) and ends with a perfect square (4 * 4)

$(x^2 + 4)(x^2 - 4)$

Warning! Watch out, there may be another DOS!

$x^2 - 4$ is another DOS

$(x^2 + 4)(x + 2)(x - 2)$

The x² + 4 is not a DOS because the operation is addition.

The final factored form is $(x^2 + 4)(x + 2)(x - 2)$

8 0

2 years ago

Find the perimeter of the quadrilateral.

levacccp [35]

Question:

Find the perimeter of the quadrilateral. if x = 2 the perimeter is ___ inched.

The complete figure of this question is attached below

Answer:

<h3>The perimeter of the quadrilateral is 129 inches</h3>

Solution:

The complete figure of this question is attached below

Given that, a quadrilateral with,

Side lengths are:

$4x^2 + 8x\ inches \\\\3x^2-5x+20\ inches \\\\7x + 30\ inches \\\\31\ inches$

The values of the side lengths when x = 2 are

$(4x^2+8x)=(4\times 2^2+8\times 2)=(4\times 4+16)=16+16=32\ inch\\\\(3x^2-5x+20)=(3\times 2^2-5\times 2+20)=(3\times 4-10+20)=12+10=22\ inch\\\\(7x+30)=(7\times 2+30)=14+30=44\ inch$

Perimeter of a quadrilateral = Sum of its sides

Perimeter of given quadrilateral = 32 + 22 + 44 + 31 = 129 inches

Thus perimeter of the quadrilateral is 129 inches

5 0

3 years ago

Help please thanks!!!!!!!!!!

sergeinik [125]

The answer is b please be thankful

6 0

3 years ago