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3 years ago

13

I need help with number 5

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