X^2+4+10=0
x=(-4(+/-)root16-40)/2, so we now know that the zeros are imaginary, because you can't square root a negative number and 16-40 is -24
so the two roots are…
-2+iroot6 and -2-iroot6

7 0

3 years ago

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HELP ME PLEASE WILL PICK BRAINLYEST

Ronch [10]

Here's one way to do it.

AB ≅ AC . . . . . . . . . . given
∠BAY ≅ ∠CAY . . . . given
AY ≅ AY . . . . . . . . . . reflexive property
ΔBAY ≅ ΔCAY . . . .. SAS congruence
XY ≅ XY . . . . . . . . . . reflexive property
∠AYB ≅ ∠AYC . . . . CPCTC
BY ≅ CY . . . . . . . . . . CPCTC
ΔXYB ≅ ΔXYC . . . .. SAS congruence

Therefore ...
∠XCY ≅ ∠XBY . . . . CPCTC

3 0

3 years ago

Please help me I would really appreciate it.

Karo-lina-s [1.5K]

Given:

$\overline{AB}\cong \overline {CD},\overline{BC}\cong \overline {DA}$

To prove:

$\Delta CDA\cong \Delta ABC$

Proof:

In $\Delta CDA\text{ and } \Delta ABC$ ,

$\overline{AB}\cong \overline {CD}$ because Given

$\overline{BC}\cong \overline {DA}$ because Given

$\overline{AC}\cong \overline {AC}$ because Reflexive property.

Corresponding sides of both triangles are congruent. So,

$\Delta CDA\cong \Delta ABC$ because SSS

Therefore, the required missing values are: Given, $\overline{BC}\cong \overline {DA}$ , Reflexive property , $\Delta CDA\cong \Delta ABC$ , SSS respectively.

5 0

3 years ago