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Olegator [25]
3 years ago
8

Segment BD is an altitude of triangle ABC. Find the area of the triangle.

Mathematics
2 answers:
inn [45]3 years ago
8 0
The answer is 7.5
Just count the squares and I did it so it is right.

hodyreva [135]3 years ago
3 0
Area = 1/2 * base  * height
the height is BDand the base is AC 
so its  area = 1/2 * BD * AC
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a new flagpole needs to be 54ft high and the old flagpole is 15m high, how many meters taller does the new flag need to be? roun
Elena L [17]
It would need to be 16.4592 meters higher which rounds to 16 meters rounded to the nearest tenth
7 0
2 years ago
Question 2
cluponka [151]

Answer:

4:5

Step-by-step explanation:

28:35

28/7:35/7

4:5

hopefully this helps :)

4 0
2 years ago
A baseball is thrown in a parabolic arc. It's position above the ground at a given point in time can be represented by the quadr
SCORPION-xisa [38]

Answer:

The baseball reached 9 feet high above the ground

Step-by-step explanation:

The given quadratic function representing the position of the baseball above ground is p(t) = 1/2·g·t² + v₀·t + p₀

Where;

t ≥ 0

g = -32 ft./sec²

v₀ = The initial velocity

p₀ = The initial position

Given that when the ball is thrown, we have;

The initial, straight up, velocity, v₀ = 16 ft./sec

The initial position, p₀ = 5 ft.

Substituting the above values in the quadratic function representing the position of the baseball above ground, we have;

p(t) = 1/2·(-32)·t² + 16·t + 5 = 16·t - 16·t² + 5

At the maximum point, the rate of change of the height with time = 0, therefore;

dp(t)/dt = 0 = d(16·t - 16·t² + 5)/dt = 16 - 32·t

16 - 32·t = 0

16 = 32·t

t = 16/32 = 0.5 seconds

Therefore, the time takes to reach the maximum height = 0.5 seconds

The height (maximum) reached in 0.5 seconds is given as follows;

h(t) = 16·t - 16·t² + 5, from which we have;

h(0.5) = 16 × 0.5 - 16 × (0.5)² + 5 = 9

Therefore, the height baseball reached = 9 ft. above ground

7 0
2 years ago
FURTHER MATHEMATICS Use determinants to solve the systems of equation:
maw [93]

Answer:

Step-by-step explanation:

\left\{\begin{array}{ccc}2x+1y+2z&=&13\\1x+1y-2z&=&8\\1x+2y+1z&=&11\\\end{array}\right.\\\\\\\Delta=\left| \begin{array}{ccc}2&1&2\\1&1&-2\\1&2&1\end{array}\right| =2*\left| \begin{array}{ccc}1&\frac{1}{2}&1\\1&1&-2\\1&2&1\end{array}\right| =2*\left| \begin{array}{ccc}1&\frac{1}{2}&1\\0&\frac{1}{2}&-3\\0&1&3\end{array}\right| =2*(\frac{3}{2}+3)=9\\\\

\Delta_1=\left| \begin{array}{ccc}13&1&2\\8&1&-2\\11&2&1\end{array}\right| =2*\left| \begin{array}{ccc}13&1&2\\8&1&-2\\\frac{11}{2}& 1&\frac{1}{2}\end{array}\right| =2*\left| \begin{array}{ccc}13&1&2\\-5&0&-4\\\frac{-5}{2}& 0&\frac{5}{2}\end{array}\right| \\\\=2*(-1)*(\frac{-25}{2}-\frac{20}{2}) =45\\

\Delta_2=\left| \begin{array}{ccc}2&13&2\\1&8&-2\\1&11&1\end{array}\right| \\\\\\=\left| \begin{array}{ccc}3&21&0\\3&30&0\\1&11&1\end{array}\right| \\\\\\=1*(90-63) =27\\

\Delta_3=\left| \begin{array}{ccc}2&1&13\\1&1&8\\1&2&11\end{array}\right| \\\\\\=\left| \begin{array}{ccc}0&-1&-3\\0&-1&-3\\1&2&11\end{array}\right| \\\\\\=0\\

\left\{\begin{array}{ccc}x=\dfrac{\Delta_1}{\Delta}=\dfrac{45}{9}=5\\\\y=\dfrac{\Delta_2}{\Delta}=\dfrac{27}{9}=3\\\\z=\dfrac{\Delta_3}{\Delta}=\dfrac{0}{9}=0\\\end{array}\right.

8 0
3 years ago
HELP ME PLEASEEEEEEEEEEEEEEEEE
katrin2010 [14]
The answer is just x
6 0
3 years ago
Read 2 more answers
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