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olga_2 [115]
3 years ago
13

I need help with number 5

Mathematics
2 answers:
Tju [1.3M]3 years ago
7 0

Answer:

B. BD = 9

Step-by-step explanation:

Simply set the 2 equations equal to each other (both are congruent due to perpendicular bisector):

6x + 3 = 3x = 6

3x + 3 = 6

3x = 3

x = 1

Then substitute 1 for <em>x</em> in BD

6(1) + 3

6 + 3

BD = 9

disa [49]3 years ago
4 0
I think the answer is B
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X squared+ y squared = 2 y = 2x squared – 3 Which of the following describes the system?
cluponka [151]

Answer:

x=-1,1,-\sqrt{\frac{7}{4} },\sqrt{\frac{7}{4}} and y=-1,\frac{1}{2}

The ordered pair solutions are (-\sqrt{\frac{7}{4}},0.5), (\sqrt{\frac{7}{4}},0.5), (-1,-1), and (1,-1).

Step-by-step explanation:

I'm assuming the system is \left \{ {x^2+y^2=2} \atop {y=2x^2-3}} \right.:

x^2+y^2=2

x^2+(2x^2-3)^2=2

x^2+(4x^4-12x^2+9)=2

x^2+4x^4-12x^2+9=2

4x^4-11x^2+9=2

4x^4-11x^2+7=0

x^4-11x^2+28=0

(x^2-7)(x^2-4)=0

(4x^2-7)(x^2-1)=0

4x^2-7=0

4x^2=7

x^2=\frac{7}{4}

x=\pm\sqrt{\frac{7}{4}}

x^2-1=0

x^2=1

x=\pm1

y=2x^2-3

y=2(\pm\sqrt{\frac{7}{4}})^2-3

y=2({\frac{7}{4}})-3

y=\frac{7}{2}-3

y=\frac{1}{2}

y=2x^2-3

y=2(\pm1)^2-3

y=2(1)-3

y=2-3

y=-1

Therefore, x=-1,1,-\sqrt{\frac{7}{4} },\sqrt{\frac{7}{4}} and y=-1,\frac{1}{2}

The ordered pair solutions are (-\sqrt{\frac{7}{4}},0.5), (\sqrt{\frac{7}{4}},0.5), (-1,-1), and (1,-1).

4 0
2 years ago
Please help I’ll mark you as brainliest if correct
Len [333]

Answer:

I think it is option B

Step-by-step explanation:

I dont know if I am right

if I am your welcome

4 0
3 years ago
Read 2 more answers
Which statement is true?
Mumz [18]
The answer is D.

|-4| < 4
   4 < 4

|-1| < 0
   1 < 0

|-6| = -6
   6 = -6

|-14| > 10
   14 > 10
6 0
2 years ago
I am giving branliest ​
Oduvanchick [21]

Answer:

eisdmejdjdj

Step-by-step explanation:

kdjeirkffn

4 0
2 years ago
Can someone please help me with this i don't know
Liula [17]
50.2/3divided by 8/7
     it will equal 1 5/16 which is C

7 0
2 years ago
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