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dmitriy555 [2]
3 years ago
5

Which step can be used when solving x^2-6x-25=0?

Mathematics
1 answer:
Anon25 [30]3 years ago
5 0
Complete the square

since a=1 for ax²+bx²+c=0
we can do
take 1/2 of the linear coefient and square it
-6/2=-3, (-3)²=9
add that to both sides

x²-3x+9-25=9
factor perfect square trinomial
(x-3)²-25=9
add 25 to both sides
(x-3)²=34
sqrt both sides, rmember to take positive and negative roots
x-3=+/-√34
add 3
x=3+/-√34

x=3+√34 or 3-√34
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Congruent triangles & proofs need answers to all the ones i got wrong. thank u!
notka56 [123]

Answer:

Step-by-step explanation:

13. Since, it is given that we have to prove ΔGDH≅ΔFEH by HL rule, thus From ΔGDH, ∠GDH=90° and GH is the hypotenuse and from ΔFEH, ∠HEF=90° and HF is the hypotenuse, thus in ΔGDH and ΔFEH,

∠GDH≅∠HEF=90°

GH≅FH,

Thus by HL the given triangles are congruent.

Option A is correct.

14. From the figure, it can be seen that both triangles have a right angle congruent and hypotenuse congruent, thus by HL rule, triangle scan be prove congruent.

Option A is correct.

15. It is given that BI is parallel to RD, thus using the alternate interior angle property, ∠B≅∠R.

Option C is correct.

16. It is given that BI is parallel to RD, and ∠BSI and ∠RSD form the vertically opposite angles, thus ∠BSI≅∠RSD by vertical angles are congruent.

Option A is correct.

17. Since Δ ABC is an isosceles triangle, thus two sides of the triangle are equal, therefore ∠A=∠B=3x+5, using the angle sum property in ΔABC, we have

3x+5+2x+3x+5=180°

⇒8x+10=180

⇒8x=170

⇒x=21.25

Thus, ∠B=3x+5=3(21.25)+5=63.75+5=68.75°

18. From ΔJKL and ΔDEF, we have

∠K=∠E (given)

∠L=∠F(given)

JL=DF

Thus, By AAS rule ΔJKL ≅ ΔDEF

As corresponding angles and corresponding sides are equal in these two triangles as compared to the other triangles.

8 0
2 years ago
If h(x)=x-9, find h(13)
Helen [10]

If you plug in h, the answer would be 13-9 which is 4.

6 0
2 years ago
Read 2 more answers
15! + 12!=
MrMuchimi

Answer:

27 sana makatulong hahahahah

6 0
2 years ago
Please help me!!!!!!!!!!!
shutvik [7]

The given parabola is y+4=12(x-7)^2. Compare with the standard equation of a parabola in vertex form, y-c=a(x-b)^2.

The vertex of this parabola is (b,c). Compare y+4=12(x-7)^2 with the standard equation y-c=a(x-b)^2.

we see c=-4,b=7. Thus, the vertex of the given parabola is (7,-4).


8 0
3 years ago
Read 2 more answers
Given cosα = −3/5, 180 < α < 270, and sinβ = 12/13, 90 < β < 180
torisob [31]

I got

-  \frac{6 3}{65}

What we know

cos a=-3/5.

sin b=12/13

Angle A interval are between 180 and 270 or third quadrant

Angle B quadrant is between 90 and 180 or second quadrant.

What we need to find

Cos(b)

Cos(a)

What we are going to apply

Sum and Difference Formulas

Basics Sine and Cosines Identies.

1. Let write out the cos(a-b) formula.

\cos(a - b)  =  \cos(a)  \cos(b)  +  \sin(a)  \sin(b)

2. Use the interval it gave us.

According to the given, Angle B must between in second quadrant.

Since sin is opposite/hypotenuse and we are given a sin b=12/13. We. are going to set up an equation using the pythagorean theorem.

.

{12}^{2}  +  {y}^{2}  =  {13}^{2}

144 +  {y}^{2}  = 169

25 =  {y}^{2}

y = 5

so our adjacent side is 5.

Cosine is adjacent/hypotenuse so our cos b=5/13.

Using the interval it gave us, Angle a must be in the third quadrant. Since cos is adjacent/hypotenuse and we are given cos a=-3/5. We are going to set up an equation using pythagorean theorem,

.

( - 3) {}^{2}  +  {x}^{2}  =  {5}^{2}

9 +  {x}^{2}  = 25

{x}^{2}  = 16

x = 4

so our opposite side is 4. sin =Opposite/Hypotenuse so our sin a =4/5.Sin is negative in the third quadrant so

sin a =-4/5.

Now use cosine difference formula

-  \frac{3}{5}  \times  \frac{5}{13}  +   - \frac {4}{5}  \times  \frac{12}{13}

- \frac{15}{65} + (  - \frac{48}{65}  )

-  \frac{63}{65}

Hope this helps

6 0
2 years ago
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