How many faces have an area of 6 cm2?

Mathematics

2 answers:

denis23 [38]2 years ago

8 0

Answer:

Step-by-step explanation:

There are 6 sides:

3 by 4 = 3*4 = 12, no

3 by 3 = 3*3 = 9, no

5 by 3 = 5*3 = 15, no

Triangles:

Half of 3 by 4 = 1/2 * 3* 4 = 1/2 * 12 = 6, yes

The other triangle is the same, so 2

Liono4ka [1.6K]2 years ago

3 0

Answer:

two

Step-by-step explanation:

Area of the rectangle on the left = 4 * 3 = 12 cm^2

area of the square = 3^2 = 9 cm^2

Area of the rectangle on the right = 5 * 3 = 15 cm^2

Area of the two triangles = (4 * 3)/2 = 6 cm^2

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15:59:26

Viefleur [7K]

Answer:

Step-by-step explanation: I just took the quiz.

5 0

2 years ago

37= -3+5(x + 6) slove for x

lions [1.4K]

Step 1: Simplify both sides of the equation.
37
=
−
3
+
5
(
x
+
6
)
37
=
−
3
+
(
5
)
(
x
)
+
(
5
)
(
6
)
(Distribute)
37
=
−
3
+
5
x
+
30
37
=
(
5
x
)
+
(
−
3
+
30
)
(Combine Like Terms)
37
=
5
x
+
27
37
=
5
x
+
27
Step 2: Flip the equation.
5
x
+
27
=
37
Step 3: Subtract 27 from both sides.
5
x
+
27
−
27
=
37
−
27
5
x
=
10
Step 4: Divide both sides by 5.
5
x
5
=
10
5
x
=
2

8 0

3 years ago

If a factory continuously dumps pollutants into a river at the rate of the quotient of the square root of t and 45 tons per day,

julsineya [31]

<h2>Hello!</h2>

The answer is:

The first option, the amount dumped after 5 days is 0.166 tons.

To solve the problem, we need to integrate the given expression and evaluate using the given time.

So, integrating we have:

$\int\limits^5_0 {\frac{\sqrt{t} }{45} } \, dt=\int\limits^5_0 {\frac{1}{45} (t)^{\frac{1}{2} } } \, dt\\\\\int\limits^5_0 {\frac{1}{45} (t)^{\frac{1}{2} } } \ dt=\frac{1}{45}\int\limits^5_0 {t^{\frac{1}{2} } } } \ dt\\\\\frac{1}{45}\int\limits^5_0 {t^{\frac{1}{2} } } } \ dt=(\frac{1}{45}*\frac{t^{\frac{1}{2}+1} }{\frac{1}{2} +1})/t(5)-t(0)\\\\(\frac{1}{45}*\frac{t^{\frac{1}{2}+1} }{\frac{1}{2} +1})/t(5)-t(0)=(\frac{1}{45}*\frac{t^{\frac{3}{2}} }{\frac{3}{2}})/t(5)-t(0)$

$(\frac{1}{45}*\frac{t^{\frac{3}{2}} }{\frac{3}{2}})/t(5)-t(0)=(\frac{1}{45}*\frac{2}{3}*t^{\frac{3}{2} })/t(5)-t(0)\\\\(\frac{1}{45}*\frac{2}{3}*t^{\frac{3}{2} })/t(5)-t(0)=(\frac{2}{135}*t^{\frac{3}{2}})/t(5)-t(0)\\\\(\frac{2}{135}*t^{\frac{3}{2}})/t(5)-t(0)=(\frac{2}{135}*5^{\frac{3}{2}})-(\frac{2}{135}*0^{\frac{3}{2}})\\\\(\frac{2}{135}*5^{\frac{3}{2}})-(\frac{2}{135}*0^{\frac{3}{2}})=\frac{2}{135}*11.18-0=0.1656=0.166$