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3 years ago

Find the solutions to the equation below. 2x2 + 7x + 3 = 0

Mathematics

2 answers:

Tanzania [10]3 years ago

8 0

Answer: x=-1

Step-by-step explanation: 2 times 2 is 4 plus 7x plus 3. Move 4 and 3 over and you get negative 7. Divide by 7 and x=-1.

zmey [24]3 years ago

3 0

Answer:

x = -1/2 and x = -3

Step-by-step explanation:

2x² + 7x + 3 = 0

Using AC method:

2 × 3 = 6

Factors of 6 that add up to 7: 1 and 6

Divide by 2 and reduce: 1/2, 6/2 = 3/1

The factors are (2x + 1) and (x + 3)

(2x + 1) (x + 3) = 0

2x + 1 = 0

x = -1/2

x + 3 = 0

x = -3

The solutions are x = -1/2 and x = -3.

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The designer also programs a bird with a path that can be modeled by a quadratic function. The bird starts at the vertex of the

Nady [450]

Answer:

-1.2

Step-by-step explanation:

Given that the designer also programs a bird with a path that can be modeled by a quadratic function.

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3 years ago

Find the LCM of n^3 t^2 and nt^4. A) n 4t^6 B) n 3t^6 C) n 3t^4 D) nt^2

patriot [66]

Answer:

The correct option is C.

Step-by-step explanation:

The least common multiple (LCM) of any two numbers is the smallest number that they both divide evenly into.

The given terms are $n^3t^2$ and $nt^4$ .

The factored form of each term is

$n^3t^2=n\times n\times n\times t\times t$

$nt^4=n\times t\times t\times t\times t$

To find the LCM of given numbers, multiply all factors of both terms and common factors of both terms are multiplied once.

$LCM(n^3t^2,nt^4)=n\times n\times n\times t\times t\times t\times t$

$LCM(n^3t^2,nt^4)=n^3t^4$

The LCM of given terms is $n^3t^4$ . Therefore the correct option is C.

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3 years ago

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A pile of earth removed from an excavation is a cone measuring 6ft high and 30ft across its base. How many trips will it take to

nignag [31]

Answer: 14

Step-by-step explanation:

Given

Excavation cone measures

height $h=6\ ft$

Diameter $d=30\ ft$

Truck can dump $125\ ft^3$ at a time

The volume of a cone is

$V=\dfrac{1}{3}\pi r^2h$

Putting values

$\Rightarrow V=\dfrac{1}{3}\times \pi\times 15^2\times 6\\\Rightarrow V=1413.9\approx 1414\ ft^3$

No of trips(N) required to accumulate this much volume is given by

$\Rightarrow N=\dfrac{1414}{125}=13.3\approx 14$

Therefore, 14 trips are necessary to accumulate a cone of volume $1414\ ft^3$

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