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NNADVOKAT [17]
3 years ago
12

Best anwser gets brain select all that apply​

Mathematics
2 answers:
denpristay [2]3 years ago
7 0

7,4 ×10^ -5

.....................

iVinArrow [24]3 years ago
6 0
The second option, fourth, and the last are all scientific notation.
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The length of the base and the height of a triangle are numerically equal. Their sum is 6 less than the number of units in the a
Troyanec [42]
Base = Height, Let them be x.

( x \times x) \div 2 = (x + x) + 6 \\ x ^{2} \div 2 = 2x + 6 \\ \frac{1}{2} x^{2} - 2x - 6 = 0
Solve quadratic equation:
x=6, x=-2(rejected because base and height is positive)

6x6/2 =18
6 0
2 years ago
The distance a race car travels is given by the equation d=v0t+12at2 where v0 is the initial speed of the race car, a is the acc
Scrat [10]

Answer:

v0 + 1/2at

Step-by-step explanation: Given that the distance a race car travels is given by the equation d = v0t+12at2 where v0 is the initial speed of the race car, a is the acceleration, and t is the time travelled.

The equation for the driver's average speed s during the acceleration will be:

(v0t+12at2) / t

Since Average speed is equal to distance divided by time.

Therefore, the equation will be:

v0+1/2at

7 0
2 years ago
Hlp me plz i will give you guys 25 points
mart [117]
(A4+a2) hoped this helps
6 0
3 years ago
What is the following quotient? StartFraction RootIndex 3 StartRoot 60 EndRoot Over RootIndex 3 StartRoot 20 EndRoot EndFraction
FrozenT [24]

Answer:

\sqrt[3]{3}

Step-by-step explanation:

We are required to simplify the quotient: \dfrac{\sqrt[3]{60} }{\sqrt[3]{20}}

Since the <u>numerator and denominator both have the same root index</u>, we can therefore say:

\dfrac{\sqrt[3]{60} }{\sqrt[3]{20}} =\sqrt[3]{\dfrac{60} {20}}

=\sqrt[3]{3}

The simplified form of the given quotient is \sqrt[3]{3}.

6 0
2 years ago
Read 2 more answers
The London theater offers child tickets at 10$each and adult tickets at 20$ each.Mrs burham buys 34 tickets at a total cost of 5
kirill115 [55]

Answer:

A+C=34

10C+20A=570

4 0
2 years ago
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