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klemol [59]
3 years ago
12

Find the coordinates of the point(s on the curve 2y=10?x2 that are closest to the origin

Mathematics
1 answer:
sergey [27]3 years ago
4 0
No it far 10 is way out there
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Slope of a parallel line with the points (7,-5) and (-5,7)
VLD [36.1K]

Answer: Negative 1  The slope of parallel lines is the same.

The attachment shows two such lines, given coordinates labeled.

Step-by-step explanation:

Find the slope of the line passing through the given points.

 rise/run  

Rise is the difference in y-values  7-(-5)  = 12

Run is the difference between x-values  -5 - 7  = - 12

The Slope is 12/-12   simplify:

slope = -1

5 0
3 years ago
PLEASE HELP ME what is the solution to the system of equations?
masha68 [24]
Times the bottom equation by -4 to cancel out the x’s.
Add the equations.
Solve for y.
Plug in y to find x.

8 0
2 years ago
Which order does it go?
gladu [14]
Subtract 4 from each side
Divide both sides by 5
Take square root of both sides
Add 3 to both sides
4 0
2 years ago
Need help on this math pls points and brainlest Math angles
iVinArrow [24]

Answer:

x = 28

m<ABC = 57°

Step-by-step explanation:

✔️(2x + 1)° + 33° = 90° (complementary angles)

Solve for x

2x + 1 + 33 = 90

Add like terms

2x + 34 = 90

2x = 90 - 34 (subtraction property of equality)

2x = 56

Divide both sides by 2

x = 28

✔️m<ABC = 2x + 1

Plug in the value of x

m<ABC = 2(28) + 1

= 56 + 1

m<ABC = 57°

5 0
2 years ago
Read 2 more answers
Segments
Kazeer [188]

Given

AB  and  CD  intersect

AC,  CB,  BD  and  AD  are congruent.

Prove that AB  is the bisector of ∠CAD and ray  CD  is the bisector of ∠ACB.

and AB  and  CD  are perpendicular.

To proof

Bisector

<em>A bisector is that which cut an angle in two equal parts.</em>

In ΔACB and ΔADB

AD = AC  ( Given )

AB = AB   ( common )

BC = DB  ( Given )

by SSS congurence property

we have

ΔACB ≅ΔADB

∠CAB =∠ DAB

∠CBA = ∠DBA

( By corresponding sides of the congurent triangle )

Thus AB is the bisector of the ∠CAD.

InΔ DAC and ΔDBC

AD = DB (Given)

AC = CB  ( Given )

CD = CD (common)

By SSS congurence property

ΔDAC≅ Δ DBC

∠  ACD =∠ BCD

∠ADC =∠BDC

( By corresponding sides of the congurent triangle )

Therefore CD is the bisector of the CAD.

In ΔBOC andΔ BOD

BO = BO ( Common )

∠BCO = ∠BDO

( As prove above ΔACB ≅ΔADB

Thus ∠ACB = ∠ADB by corresponding sides of the congurent triangle , CD is a bisector

∠BCO = ∠BDO )

 CB = DB ( given )

by SAS congurence property

ΔBOC ≅ ΔBOD

∠BOC =∠ BOD

∠BOC +∠ BOD = 180 °( Linear pair )

2∠ BOC = 180°

∠BOC = 90°

∠BOC =∠ BOD = 90°

also

In ΔCOA and ΔAOD

AO = AO ( Common )

∠ACO =∠ ADO

(  As prove above ΔACB ≅ΔADB Thus ACB = ADB by corresponding sides of congurent triangle ,CD is a bisector

thus  ∠ACO = ∠ADO )

AC =AD ( given )

by SAS congurence property

Δ COA ≅ ΔAOD

∠AOC = ∠AOD

( By corresponding angle of corresponding sides )

∠AOC + ∠AOD = 180°

2∠ AOC = 180°   ( Linear pair )

∠AOC = 90°

∠AOC = ∠AOD = 90 °

Thus AB  and  CD  are perpendicular.

Hence proved









   


 



6 0
3 years ago
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