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balandron [24]
3 years ago
6

Find the area and perimeter of the shaded figure

Mathematics
1 answer:
puteri [66]3 years ago
6 0

Answer:

Area - 6 unit^2   Perimeter - 10 units

Step-by-step explanation:

The shaded figure is 3 units long and 2 units high.

L = 3  W = 2

A = LW

A = 3 x 2 = 6 unit^2

Perimeter = 2L + 2W

P = 2(3) + 2(2)

P = 6 + 4 = 10 units

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Which equation are equivalent to Y equals2/3 X -4 when written in slope intercept form check all that apply
Neporo4naja [7]

Answer:

tg

Step-by-step explanation:

ggr

5 0
3 years ago
What is reductionism ......... ​
Sever21 [200]

Answer:

the practice of analysing and describing a complex phenomenon in terms of its simple or fundamental constituents, especially when this is said to provide a sufficient explanation

7 0
3 years ago
Read 2 more answers
At a sand and gravel plant, sand is falling off a conveyor, and onto a conical pile at a rate of 10 cubic feet per minute. The d
kherson [118]

Answer:

dh/dt=9.82x10^-3 ft/min

Step-by-step explanation:

1. You have that the rate is10 ft³/min. Then:

dV/dt=10

2. The formula for calculate the volume of a cone, is:

V=1/3(πr²h)

"r" is the radius and "h" is the height.

3. The diameter of the base of the cone is approximately 3 times the altitude. Then, the radius is:

r=diameter/2

diameter=3h

r=3h/2

4. When you susbstitute r=3h/2 into the formula V=πr²h/3, you have:

V=1/3(πr²h)

V=1/3(π(3h/2)²(h)

V=1/3(π9h²/4)(h)

V=9πh³/12

5. Therefore:

dV/dt=(9πh²/4)dh/dt

h=12

dV/dt=10

6. When you substitute the values of dV/dt and h into dV/dt=(9π(12)²/4)dh/dt, you have:

dV/dt=(9π(12)²/4)dh/dt

10=(1017.876)

7. Finally, you obtain:

dh/dt=10/1017.876

dh/dt=9.82x10^-3 ft/min

5 0
3 years ago
Which one is the answer? please help!!
Vitek1552 [10]

Answer:

Its 100 percent B

Step-by-step explanation:

first u multiply 3 which will give u 3V on the left side.

then you divide pie times radius squared which will give you 3V divided by pi radius squared equal to H which is the height

7 0
3 years ago
Find the real or imaginary solutions of each equation by factoring
anyanavicka [17]
Sum of  2 perfect cubes
a³+b³=(a+b)(x²-xy+y²)
so

x³+4³=(x+4)(x²-4x+16)
set each to zero
x+4=0
x=-4

the other one can't be solveed using conventional means
use quadratic formula
for
ax^2+bx+c=0
x=\frac{-b+/- \sqrt{b^2-4ac} }{2a}
for x²-4x+16=0
x=\frac{-(-4)+/- \sqrt{(-4)^2-4(1)(16)} }{2(1)}
x=\frac{4+/- \sqrt{16-64} }{2}
x=\frac{4+/- \sqrt{-48} }{2}
x=\frac{4+/- (\sqrt{-1})(\sqrt{48}) }{2}
x=\frac{4+/- (i)(4\sqrt{3}) }{2}
x=2+/- 2i\sqrt{3}


the roots are
x=-4 and 2+2i√3 and 2-2i√3
8 0
3 years ago
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