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3 years ago

7t-4(-x+.5)-12t (simplify)

1 answer:

4 0

Answer:
–5t + 4x – 2

Explanation:
7t – 4(–x + .5) – 12t
combine 7t and – 12
7 – 12 = – 5
– 4(–x +.5) – 5t
Now distribute –4:
4x – 2 – 5t
Rearrange:
4x – 5t – 2
or
5t + 4x – 2

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The answer is D

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Which diagram represents a fraction equivalent to 75%?

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The last one is the right answer

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2. Ariana has created a paper airplane for her physics class at school where all the students will be testing their planes by se

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Answer:

2. The airplane will crash into the ground after 5 seconds.

3. The domain of the function is [0,30] and range of the function is $[-\infty,900]$ .

Step-by-step explanation:

Ariana initial attempt is modeled by the function,

$h(x)=x^2-12x+35$

where, h is the height in feet and x is the time in seconds of the paper airplanes path.

If airplane crash into the ground, then $h(x)=0$ ,

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Equate each factor equal to 0.

$x=5,7$

Therefore the airplane will crash into the ground after 5 seconds.

The height of the rocket in meters is modeled by the function shown below, where t is time in seconds.

$h(t)=-4t^2+120t$

The value of t must be positive because time can not be negative.

Find the zeros of the function.

$0=-4t(t-30)$

$t=0,30$

The leading coefficient is negative, so the it is a downward parabola. Since the zeros of the function are 0 and 30, therefore the function is negative before 0 and after 30.

The height can not be negative, so the domain of the function is

$0\leq t\leq 30$

The vertex of a downward parabola is the maximum point.

Vertex of a parabola,

$(\frac{-b}{2a},h(\frac{-b}{2a}))$

$(\frac{-120}{2(-4)},h(\frac{-120}{2(-4)}))$

$(15,h(15))$

Put t=15 it in the equation.

$h(15)=-4(15)^2+120(15)=900$

$(15,900)$

The range of the function is

$[-\infty,900]$

Therefore the domain of the function is [0,30] and range of the function is $[-\infty,900]$ .

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3 years ago

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Answer:

for the triangle on top on side is 8 one side is 4 so if we use the theorem

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Side b = 17.66349

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Angle ∠B = 63° = 1.09956 rad = 7/20π

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Perimeter p = 46.4877

Semiperimeter s = 23.24385

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3 years ago

A population of crawfish in a section of the Louisiana bayou increased from 900,000 to 1.2 million between 2010 and 2011. If the

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Since the population of crawfish in a section of the Louisiana bayou increased from 900,000 to 1.2 million between 2010 and 2011. The percentage increase in population will be:

= (1.2 million - 900000) / 900000 × 100

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Since the population increases by the same percentage between 2011 and 2012, then the population in 2012 will be:

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= 133.33% × 1.2 million

= 1.3333 × 1.2 million

= 1,599,960

The population in 2012 is 1,599,960

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