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Gwar [14]
2 years ago
8

If it is completely impossible to put a contract in writing, which of the following would be the best substitute?

Mathematics
1 answer:
I am Lyosha [343]2 years ago
4 0

Answer:

B

Step-by-step explanation:

Because the the contract will be in verbal form. no matter if it is videotaped or recorded or both a "videoed verbal contract" not just written. so like paper thst is notarized or not this is still the same bc the the same rules apply the contract is still valid. the rest of the options are more like what they call gentlemens agreements which would be the handshake type of thing. and no matter what any agreement someone has to witness the transaction or contract. so definitely the answer is B

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A runner completes a 400-m race in 43.9 s. In a 100-m race, he finishes in 10.4 s. In which race was his speed faster?
Sveta_85 [38]

Answer:

100m race

Step-by-step explanation:

because 10.4 x 4 because 100m and 400m right 10.4x4=41.6 and 43.9 so obviously the 100m race is faster.

5 0
3 years ago
Read 2 more answers
A boat can travel 456 miles on 114<br> gallons of gasoline. How far can it<br> travel on 67 gallons?
Ede4ka [16]

Answer:

268 miles i think

Step-by-step explanation:

if 456=114 gallons

456/114=4

so 4 miles per gallon

and then

4*67=268

Hope this helps: )

3 0
3 years ago
In this rectangular box, EF = 16, FD = 5, and DB = 30. Find AF.
AVprozaik [17]

Answer:

FA = 25.87  

Step-by-step explanation:

Given: BD = 30 , DF = 5 , EF = 16

To find: FA

Rectangular box is a Cuboid.

Figure attached.

AB = FE = 16

So,

In  ΔADB

using Pythagoras theorem,

DB^2=AB^2+AD^2\\30^2=16^2+AD^2\\900=256+AD^2\\AD^2=900-256\\AD^2=644\\AD=\sqrt{644}\\AD=2\sqrt{161}

Again using Pythagoras theorem in Δ ADF we get

FA^2=DF^2+AD^2\\FA^2=5^2+(2\sqrt{161})^2\\FA^2=25+644\\FA^2=669\\FA=\sqrt{669}\\FA=25.87

Therefore, FA = 25.87

6 0
3 years ago
Read 2 more answers
According to the table below, which of these is a possible taxable income for
enyata [817]

Answer:

I think c is the answer.

4 0
2 years ago
A. Find the length of the mid segment of an equilateral triangle with side lengths of 12.5 cm.
boyakko [2]

Answer:

A) The length of the mid-segment  is 6.25 cm

B) The length of AT = 33 units

C) The value of x is 3

Step-by-step explanation:

A) Find the length of the mid segment of an equilateral triangle with side lengths of 12.5 cm

<u>Solution:</u>

A mid segment of a triangle is a segment connecting the midpoints of two sides of a triangle. This segment has two special properties. It is always parallel to the third side, and the length of the mid segment is half the length of the third side.

This is an equilateral triangle with side lengths 12.5cm

The length of the mid segment = 1/2 the length of the third side.

The length of the mid segment = 1/2 * 12.5

The length of the mid segment = 6.25 cm

B) Given that UT is the perpendicular bisector of AB, where T is on AB, find the length of AT given AT = 3x +6 and TB = 42 - x

<u>Solution:</u>

UT is the perpendicular bisector of AB

T lies on AB

AT = BT

3x+6 = 42-x

Combine the like terms:

3x+x=42-6

4x= 36

Divide both sides by 4

4x/4 = 36/4

x= 9

Now plug the value of x in AT= 3x+6.

=3(9)+6

=27+6

=33

The length of AT = 33 units.

C) Given angle ABC has angle bisector for BD, where AB = CB, find the value of x if AD = 5x + 10 and DC = 28 - x.

<u>Solution:</u>

In Δ ABC

AB = BC

Δ ABC is an isosceles triangle

BD bisects angle ABC

AC is the opposite side of the vertex B

BD bisects the side AC at D

AD=CD

AD=5x+10

CD= 28-x

Equate both the equations:

5x+10 = 28-x

Combine the like terms:

5x+x=28-10

6x=18

Divide both sides by 6

6x/6 = 18/6

x= 3

Thus the value of x is 3....

5 0
3 years ago
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