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shusha [124]
2 years ago
6

What is the value of the expression. (-8+4)- (3-9)

Mathematics
1 answer:
Mekhanik [1.2K]2 years ago
7 0

Answer:

2

PLZ MARK BRAINLIEST

Step-by-step explanation:

(-8+4) - (3-9) = (-4) - (-6) = -4 + 6 = 2

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X+10=2 what is the value of x
allsm [11]
The answer to your question is X = -8

8 0
3 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
Using the technique in the model above, find the missing sides in this 30°-60°-90° triangle. Long = 3 Hypotenuse =
VMariaS [17]
Alright, if I have this right, first of all, your provided with only one side of the triangle, the long leg. You need to calculate the short leg, and then the hypotenuse for it to all be complete
1. Find the short leg's measurement by dividing the length of the long leg by the square root of 3 (use a calc.)
2. 3/ the square root of 3 = to 1.73
3. The length of the short leg, supposedly, is 1.73 (roughly, was rounded)
4. Find the hypotenuse by taking the length of the short leg times 2
5. 1.73*2=3.46, (roughly, was rounded)
6. Apply the measurements to the figure.
7.  The long leg is equal to 3, the hypotenuse is equal to 3.46, and the short leg is equal to 1.73
~Hope this helps!
(Would appreciate brainliest :p)

6 0
3 years ago
(1) Let {v1,v2,v3} be a set of vectors in Rn . If u is Span {v1,v2,v3}, show that 3u is in Span {v1,v2,v3}.
Evgesh-ka [11]

Answer:

(1)

Multiplying by 3 both sides of the equality you get that

3u = 3c_1v_1+3c_2v_2+3c_3v_3

3u  is in the Span of the vectors \{v_1,v_2,v_3\}.

(2)

That's not true, consider the following counter example.

v_1 = (0,0,0,1)\\v_2 = (0,0,1,0)\\v_3 = (0,1,0,0)\\v_4 = (1,0,0,0)\\u = (0,1,1,1)

u is a linear combination of v_1,v_2,v_3 but is NOT a linear combination of v_1,v_2,v_3,v_4.

Step-by-step explanation:

(1)

As the hint indicates, you know that

u = c_1 v_1 + c_2v_2+c_3v_3

Then, if you multiply both sides of the equality by 3, you get that

3u = 3c_1v_1+3c_2v_2+3c_3v_3

And that's it. 3u  is in the Span of the vectors \{v_1,v_2,v_3\}

(2)

That's not true, consider the following counter example.

v_1 = (0,0,0,1)\\v_2 = (0,0,1,0)\\v_3 = (0,1,0,0)\\v_4 = (1,0,0,0)\\u = (0,1,1,1)

u is a linear combination of v_1,v_2,v_3 but is NOT a linear combination of v_1,v_2,v_3,v_4.

4 0
3 years ago
Can you delete your posts?
DerKrebs [107]
I’d look in the help center but you can probably click on the 3 dots next to your post or contact an admin
6 0
3 years ago
Read 2 more answers
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