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3 years ago

Help me with this question please please

Mathematics

1 answer:

marysya [2.9K]3 years ago

5 0

The answer would be 36.

Since 17u and 19u have the same variable all you have to do is add 17 and 19.

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Find the area of the shaded region below. Give your answer in terms of T. 12m 5m

Step2247 [10]

The area of the shaded region is [(169π/4)-60]m².

The length of the rectangle is 12 m.
The breadth of the rectangle is 5 m.
The area of the rectangle = length * breadth.
The area of the rectangle = 12 * 5
The area of the rectangle = 60m²
Let h be the diagonal of the rectangle.
Using the Pythagorean theorem,
h² = 12² + 5²
h² = 144 + 25
h² = 169
h = √169
h = 13
The diagonal of the rectangle is also the diameter of the circle.
Let r be the radius of the circle.
2r = 13
r = 13/2m
The area of the circle is πr².
The area of the circle = π(13/2)²
The area of the circle = (169π/4)m²
The area of the shaded region equals the area of the circle minus the area of the rectangle.
The area of the shaded region = [(169π/4) - 60]m²

To learn more about area, visit :

brainly.com/question/27683633

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7 0

2 years ago

a person kicks a ball from the ground into the air with an initial upward velocity of 8 feet per second. what is the maximum hei

KengaRu [80]

Chech the picture below.

if the person kicked it from the ground, that means its initial height is 0.

it reaches its maximum height at the y-coordinate of its vertex, and it will hit the ground when y = 0, as you see in the picture.

$\bf ~~~~~~\textit{initial velocity}
\\\\
\begin{array}{llll}
~~~~~~\textit{in feet}
\\\\
h(t) = -16t^2+v_ot+h_o
\end{array} 
\quad 
\begin{cases}
v_o=\stackrel{8}{\textit{initial velocity of the object}}\\\\
h_o=\stackrel{0}{\textit{initial height of the object}}\\\\
h=\stackrel{}{\textit{height of the object at "t" seconds}}
\end{cases}
\\\\\\
h(t)=-16t^2+8t+0$

now let's find the y-coordinate of its vertex

$\bf \textit{vertex of a vertical parabola, using coefficients}
\\\\
h(t)=\stackrel{\stackrel{a}{\downarrow }}{-16}t^2\stackrel{\stackrel{b}{\downarrow }}{+8}t\stackrel{\stackrel{c}{\downarrow }}{+0}
\qquad \qquad 
\left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right)
\\\\\\
\left(\qquad ,~~0-\cfrac{8^2}{4(-16)} \right)\implies (\quad ,~0+1)\implies (\quad ,~1)$

when will it hit the ground?

$\bf \stackrel{h(t)}{0}=-16t^2+8t\implies 16t^2-8t=0\implies 8t(2t-1)=0
\\\\\\
8t=0\implies \stackrel{\textit{0 seconds when it took off first}}{t=0}
\\\\\\
2t-1=0\implies 2t=1\implies \stackrel{\textit{half a second later it when it came back down}}{t=\cfrac{1}{2}}$

3 0

3 years ago

What number makes the expressions equivalent? 1/2(-1.4m +0.4)=?m+0.2

ch4aika [34]

Answer:

2/4 6/12 4/8 5/10

4 0

3 years ago

Estimate the perimeter and the area of the shaded figure to the nearest whole number.

soldier1979 [14.2K]

Answer:

Step-by-step explanation:

The area and perimeter of the given shaded figure are respectively 33.12 unit² and 20.56 units.

Area and perimeter-based problem:

What information do we have?

Radius of semi-circle = 8 / 2 = 4 unit

Length of remain Rectangle = 8 unit

Width of remain Rectangle = 2 unit

Perimeter of shaded figure = πr + (l + 2b)

Perimeter of shaded figure = (3.14)(4) + [4 + (2)(2)]

Perimeter of shaded figure = 12.56 + 4 + 4

Perimeter of shaded figure = 20.56 units

Area of shaded figure = πr²/2 + lb/2

Area of shaded figure = (3.14)(4)²/2 + (8)(2)/2

Area of shaded figure = (3.14)(8) + 8

Area of shaded figure = 25.12 + 8

Area of shaded figure = 33.12 unit²

5 0

2 years ago

Read 2 more answers

Help please!! super urgent !

alekssr [168]

Answer:

Step-by-step explanation:

tan 35 = 16/x

3 0

3 years ago