The computation shows that the placw on the hill where the cannonball land is 3.75m.
<h3>How to illustrate the information?</h3>
To find where on the hill the cannonball lands
So 0.15x = 2 + 0.12x - 0.002x²
Taking the LHS expression to the right and rearranging we have:
-0.002x² + 0.12x -.0.15x + 2 = 0.
So we have -0.002x²- 0.03x + 2 = 0
I'll multiply through by -1 so we have
0.002x² + 0.03x -2 = 0.
This is a quadratic equation with two solutions x1 = 25 and x2 = -40 since x cannot be negative x = 25.
The second solution y = 0.15 * 25 = 3.75
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Complete question:
The flight of a cannonball toward a hill is described by the parabola y = 2 + 0.12x - 0.002x 2 . the hill slopes upward along a path given by y = 0.15x. where on the hill does the cannonball land?
1-3
2-6
3-9 This is an example of porportional.
1-4
2-3
3-9 There is no pattern so thin is Non-Porportional.
The weight of each cabbage is not given.
I'll assume that each cabbage weighs p pounds.
Now, we know that Yasmin bought 6 cabbage each weighing p pounds. To know the total weight, we will simply multiply the number of cabbages by the weight of each as follows:
Total weight = 6 * p = 6p pounds
Hope this helps :)
Answer: your dumb
Step-by-step explanation:
Part 1) we know that
m∠5=44° m∠11=86°
m∠2=m∠5------> by vertical angles
m∠2=44°
m∠13=m∠11------> by vertical angles
m∠13=86°
m∠12+m∠13=180°-----> supplementary angles
m∠12=180-86-----> m∠12=94°
m∠14=m∠12----> by vertical angles
m∠14=94°
m∠1=m∠11----> by corresponding angles
m∠1=86°
m∠4=m∠1----> by vertical angles
m∠4=86°
m∠2+m∠1+m∠6=180
m∠6=180-(86+44)----> 50°
m∠6=50°
m∠3=m∠6----> by vertical angles
m∠3=50°
m∠8=m∠3----> by corresponding angles
m∠8=50°
m∠8+m∠7=180°-----> supplementary angles
m∠7=180-50----> 130°
m∠7=130°
m∠10=m∠6----> by corresponding angles
m∠10=50°
m∠10+m∠9=180°-----> supplementary angles
m∠9=180-50-----> 130°
m∠9=130°
the answers Part 1) are
m∠1=86°
m∠2=44°
m∠3=50°
m∠4=86°
m∠5=44°
m∠6=50°
m∠7=130°
m∠8=50°
m∠9=130°
m∠10=50°
m∠11=86°
m∠12=94°
m∠13=86°
m∠14=94°
Part 2)
a) what is m∠TPR?
in the right triangle PTR
m∠PTR+m∠TPR+m∠TRP=180° ( the sum of internal angles of triangle is equal to 180 degrees)
m∠PTR=30°
m∠TRP=90°
so
m∠TPR=180-(90+30)----> 60°
the answer Part 2a) is
m∠TPR=60°
b) what is the length in inches of segment PR?
in the right triangle PTR
sin 30=PR/TP-----> PR=TP*sin 30-----> PR=14*(1/2)----> 7 in
the answer Part 2b) is
PR=7 in
c) what is the length in inches of segment TR?
in the right triangle PTR
cos 30=TR/PT-----> TR=PT*cos 30-----> TR=14*(√3/2)---> TR=7√3 in
the answer Part 2c) is
TR=7√3 in
d) what is the length in inches of segment PQ?
in the right triangle PQR
PR=7 in
RQ=PR-----> by angle 45°
so
RQ=7 in
applying the Pythagoras Theorem
PQ²=RQ²+PR²-----> 7²+7²-----> PQ²=98-----> PQ=√98 in---> PQ=7√2 in
the answer Part 2d) is
PQ=7√2 in
Part 3) Patrice buys a block of wax in the shape of a right rectangular prism. The dimensions of the block are 20 cm by 9 cm by 8 cm.
<span><span>(a) </span>What is the volume of the block?
volume of the prism=20*9*8-----> 1440 cm³
the answer Part 3 a) is
the volume of the block is 1440 cm³
<span>
Patrice melts the wax and creates a candle in the shape of a circular cylinder that has a diameter of 10 cm and a height of 15 cm.<span>(b) </span>To the nearest centimeter, what is the volume of the candle?
</span></span>volume of a cylinder=pi*r²*h
diameter=10 cm
radius r=10/2----> 5 cm
h=15 cm
volume of a cylinder=pi*5²*15----> 1177.5 cm³-----> 1178 cm³
the answer Part 3b) is
the volume of the candle is 1178 cm³
<span>Patrice decides to use the remaining wax to create a candle in the shape of a cube.<span>(c) </span>To the nearest centimeter, what is the length of the side of the cube?
</span>
the remaining wax=volume of the prism-volume of a cylinder
=1440-1178-----> 262 cm³
volume of a cube=b³
where b is the length side of the cube
262=b³-------b=∛262-----> b=6.40 cm-----> b=6 cm
the answer Part 3c) is
the length of the side of the cube is 6 cm
