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Amiraneli [1.4K]
3 years ago
11

I need help with 1-3 i don’t understand that problems and its already over due

Mathematics
1 answer:
Nina [5.8K]3 years ago
7 0

Step-by-step explanation:

For a quadratic equation y = ax² + bx + c, the vertex (the maximum or minimum point) is at x = -b/(2a).

1) y = -0.5t² + 2t + 38

The maximum is at:

t = -2 / (2 × -0.5)

t = 2

The maximum height is:

y = -0.5(2)² + 2(2) + 38

y = 40

The coordinates of the vertex are (2, 40).  That means the missile reaches a maximum height of 40 km after 2 minutes.

2) y = -4.9t² + 12t + 1.6

The maximum is at:

t = -12 / (2 × -4.9)

t = 1.22

The maximum height is:

y = -4.9(1.22)² + 12(1.22) + 1.6

y = 8.95

The coordinates of the vertex are (1.22, 8.95).  That means the missile reaches a maximum height of 8.95 m after 1.22 seconds.

3) y = -0.04x² + 0.88x

The maximum is at:

x = -0.88 / (2 × -0.04)

x = 11

The maximum height is:

y = -0.04(11)² + 0.88(11)

y = 4.84

The maximum height of the tunnel is 4.84 meters.

The maximum width is when y = 0.

0 = -0.04x² + 0.88x

0 = -0.04x (x − 22)

x = 22

The maximum width is 22 feet.

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Area of given triangle is 939.15cm² and smallest altitude is 30.8cm

<h3>Solution:</h3>

We are given three sides of a triangle, Let the sides be :

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Heron's formula was founded by hero of Alexandria, for finding the area of triangle in terms of the length of its sides. Heron's formula can be written as:

\sf{   \pmb { \longrightarrow \:  \sqrt{s(s - a)(s - b)(s - c)} }}

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\begin {aligned}\quad & \quad \longmapsto  \sf  s =  \dfrac{a + b + c}{2}  \\  & \quad \longmapsto  \sf s =  \dfrac{35 + 54 + 61}{2}  \\ & \quad \longmapsto  \sf s =  \dfrac{150}{2}  \\ & \quad \longmapsto  \sf s  = 75cm \end{aligned}

Now, Area of triangle will be:

\begin{aligned}&:\implies \sf\quad \sf \:  A = \sqrt{s(s - a)(s - b)(s - c)} \\ &:\implies \sf\quad \sf \:  A = \sqrt{75(75 - 35)(75 - 54)(75 - 61)}   \\&:\implies \sf\quad \sf \:  A = \sqrt{75 \times 40 \times 21 \times 14}  \\ &:\implies \sf\quad \sf \:  A = \sqrt{5 \times 5 \times 3 \times 3 \times 2 \times 2 \times 7 \times 7 \times 2 \times 2 \times 5}  \\ &:\implies \sf\quad \sf \:  A =5 \times 3 \times 2 \times 7 \times 2 \sqrt{5}  \\ &:\implies \sf\quad \sf \:  A =420 \times 2.23 \\ &:\implies \sf\quad \sf \boxed{ \pmb{ \sf   A =939.15 {cm}^{2} }} \end{aligned}

Also, we have to find the smallest altitude, and the smallest altitude will be on the longest side. So,

\begin{aligned}&:\implies \sf\quad \sf \:  Area =939.15 \\ &:\implies \sf\quad \sf \:   \dfrac{1}{2}  \times b \times h =939.15 \\ &:\implies \sf\quad \sf \:   \dfrac{1}{2} \times 61 \times h  = 939.15 \\&:\implies \sf\quad \sf \:  h =939.15 \times  \dfrac{2}{61}   \\&:\implies \sf\quad \sf \:  h = \dfrac{1818.3}{61}  \\ &:\implies \sf\quad  \boxed{ \pmb{\sf \:  h =30.79 \: (approx)}} \end{aligned}

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