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Marysya12 [62]
3 years ago
7

Help.

Mathematics
2 answers:
Zinaida [17]3 years ago
4 0

Answer:

Eat them. )llllllllllllllllllllllllll)

Tcecarenko [31]3 years ago
4 0
I would go with both ;p <3
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A group of employees is stranded on top of a 120 feet building because a fire occurred. Your goal is to throw a rope from a 60 f
Paul [167]

Step-by-step explanation:

-16t² + 64t + 60 = 120

this is a quadratic equation, and its points of f(t) = 0 can be calculated. there are usual 2 such solutions for a quadratic equation.

so, we need to bring it to a form that ends in "= 0".

-16t² + 64t - 60 = 0

that is also the same as

-4t² + 16t - 15 = 0

the formula for the solutions of such a quadratic equation is

x = (-b ± sqrt(b² - 4ac))/(2a)

a = -4

b = 16

c = -15

so,

t = (-16 ± sqrt(256 - 4×-4×-15))/(2×-4) =

= (-16 ± sqrt(256 - 240))/-8 =

= (-16 ± sqrt(16))/-8 = (-16 ± 4)/-8

t1 = (-16 + 4)/-8 = -12/-8 = 3/2

t2 = (-16 - 4)/-8 = -20/-8 = 5/2

so, it depends on what the unit of time t is. let's assume seconds.

therefore, the rope will reach the top of the building while going up after 3/2 = 1 1/2 = 1.5 seconds.

and then, as the shot rope makes a curve and comes back down again (that is why we have 2 solutions : the rope will reach the height of 120ft during its flight path twice : once while still going up, and once when coming back down again) it will be at the height of the top of the building after 5/2 = 2 1/2 = 2.5 seconds.

so, it depends if the people there can grab it at the first chance, and if the path of the rope would allow them to grab it also again on its way back down. this we don't know.

I therefore guess, that your teacher is aiming for the first solution.

4 0
3 years ago
Find the distance that the points is given on the graph ?
EleoNora [17]

From the graph, the given points are (2,-1) and (4,0). Use the distance between two points formula:

\large \boxed{ \sqrt{ {(x_2 - x_1)}^{2} +  {(y_2 - y_1)}^{2} } }

Substitute these points in the formula.

\large{ \sqrt{ {(2 - 4)}^{2}  +  {( - 1 - 0)}^{2} } } \\  \large{ \sqrt{ {( - 2)}^{2} +  {( - 1)}^{2}  } }  \\  \large{ \sqrt{4 + 1} \longrightarrow  \sqrt{5}  }

Hence, the distance is √5

Answer

  • √5

7 0
3 years ago
Jamal simplified ...
brilliants [131]
\bf \sqrt{75x^5y^8}\qquad &#10;\begin{cases}&#10;75=3\cdot 25\\&#10;\qquad 3\cdot 5^2\\&#10;x^5=x^{4+1}\\&#10;\qquad x^4x^1\\&#10;\qquad (x^2)^2x\\&#10;y^8=y^{4\cdot 2}\\&#10;\qquad (y^4)^2&#10;\end{cases}\implies \sqrt{3\cdot 5^2\cdot (x^2)^2x(y^4)^2}&#10;\\\\\\&#10;5x^2y^4\sqrt{3x}
4 0
3 years ago
12a^3b^2 +18a²b^2 – 12ab^2<br> Factor completely
AleksAgata [21]

The factorization of 12a^3b^2 +18a²b^2 – 12ab^2 is 6 a b^{2}(a+2)(2 a-1)

<u>Solution:</u>

Given, expression is 12 a^{3} b^{2}+18 a^{2} b^{2}-12 a b^{2}

We have to factorize the given expression completely.

Now, take the expression

12 a^{3} b^{2}+18 a^{2} b^{2}-12 a b^{2}

Taking b^2 as common term,

b^{2}\left(12 a^{3}+18 a^{2}-12 a\right)

Taking "a" as common term,

b^{2}\left(a\left(12 a^{2}+18 a-12\right)\right)

Taking "6" as common term,

b^{2}\left(a\left(6\left(2 a^{2}+3 a-2\right)\right)\right)

Splitting "3a" as "4a - a" we get,

b^{2}\left(a\left(6\left(2 a^{2}+4 a-a-2\right)\right)\right)

\begin{array}{l}{b^{2}(a(6(2 a(a+2)-1(a+2))))} \\\\ {b^{2}(a(6((a+2) \times(2 a-1))))} \\\\ {6 a b^{2}(a+2)(2 a-1)}\end{array}

Hence, the factored form of given expression is 6 a b^{2}(a+2)(2 a-1)

4 0
3 years ago
A statement and portions of the flowchart proof are shown. Which reason should appear in box labeled 1?
larisa [96]

Answer:

Angle Addition Postulate

Step-by-step explanation:

Given:

m\angle UVW=30^{\circ}\\ \\m\angle WVX=75^{\circ}

Prove:

m\angle UVX=105^{\circ}

Proof:

1. m\angle UVX=m\angle UVW+m\angle WVX - Angle Addition Postulate

2. m\angle UVW=30^{\circ} - given

3. m\angle WVX=75^{\circ} - given

4. m\angle UVX=30^{\circ}+75^{\circ} - Substitution Property of Equality

5.  m\angle UVX=105^{\circ} - Simplify

4 0
4 years ago
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