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3 years ago

100 points! Mhanifa please help. Look at the picture attached. I will mark brainliest!

Mathematics

2 answers:

patriot [66]3 years ago

6 0

Answer:

Step-by-step explanation:

BD is the perpendicular disector of isosceles triangle DAC

AB=BC

6x=9x-27

3x=27

x=9

BC=9x-27=54

For triangle JKL and JML,

KL=ML

JL=JL

JK=JM from properties of right-angled triangle

So triangle JKL and JML are congruent

<KJL = <MJL

4x+6=3x+10

x=4

<KJL=4*4+6=22

natima [27]3 years ago

5 0

Answer: BC = 54 <kjm=44

Step-by-step explanation:

1st 0ne-

if you look at it as 2 triangles with D being the shared side- I can determine that the angles are congruent (by SSA theorem)

6x=9x-27 (by corresponding parts of congruent Theorem)

-3x=-27

x=9

BC- 9(9)-27= 81-27=54

2nd one-

J bisects KJM

4x+6=3x+10

x=4

3(4)+10= 22

4(4)+6= 22

KJM= 22+22= 44

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Jet001 [13]

Answer:

x = 3 + √6 ; x = 3 - √6 ; $x = \frac{2+3\sqrt{2}}{2}$ ; $x = \frac{2-(3)\sqrt{2}}{2}$

Step-by-step explanation:

Relation given in the question:

(x² − 6x +3)(2x² − 4x − 7) = 0

Now,

for the above relation to be true the following condition must be followed:

Either (x² − 6x +3) = 0 ............(1)

(2x² − 4x − 7) = 0 ..........(2)

now considering the equation (1)

(x² − 6x +3) = 0

the roots can be found out as:

$x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$

for the equation ax² + bx + c = 0

thus,

the roots are

$x = \frac{-(-6)\pm\sqrt{(-6)^2-4\times1\times(3)}}{2\times(1)}$

$x = \frac{6\pm\sqrt{36-12}}{2}$

$x = \frac{6+\sqrt{24}}{2}$ and, x = $x = \frac{6-\sqrt{24}}{2}$

$x = \frac{6+2\sqrt{6}}{2}$ and, x = $x = \frac{6-2\sqrt{6}}{2}$

x = 3 + √6 and x = 3 - √6

similarly for (2x² − 4x − 7) = 0.

we have

the roots are

$x = \frac{-(-4)\pm\sqrt{(-4)^2-4\times2\times(-7)}}{2\times(2)}$

$x = \frac{4\pm\sqrt{16+56}}{4}$

$x = \frac{4+\sqrt{72}}{4}$ and, x = $x = \frac{4-\sqrt{72}}{4}$

$x = \frac{4+\sqrt{2^2\times3^2\times2}}{2}$ and, x = $x = \frac{4-\sqrt{2^2\times3^2\times2}}{4}$

$x = \frac{4+(2\times3)\sqrt{2}}{2}$ and, x = $x = \frac{4-(2\times3)\sqrt{2}}{4}$

$x = \frac{2+3\sqrt{2}}{2}$ and, $x = \frac{2-(3)\sqrt{2}}{2}$

Hence, the possible roots are

x = 3 + √6 ; x = 3 - √6 ; $x = \frac{2+3\sqrt{2}}{2}$ ; $x = \frac{2-(3)\sqrt{2}}{2}$

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3 years ago

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Virty [35]

Answer:

5/6

Step-by-step explanation:

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3 0

3 years ago

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andriy [413]

Answer:

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3 years ago

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Lubov Fominskaja [6]

Answer:

75 minutes

Step-by-step explanation:

Assuming that Clare will not take a break throughout the session

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Cross multiplying to find the value of x

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Making x the subject

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