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Nookie1986 [14]
3 years ago
11

5n - 12 = -3n + 4 30 points

Mathematics
2 answers:
Levart [38]3 years ago
6 0

Answer:

5n-12=-3n+4

5n+3n=4+12

8n=16

n=16/8

n=2

Step-by-step explanation:

ipn [44]3 years ago
4 0

Answer:

n = 2

Step-by-step explanation:

5n - 12 = -3n + 4

5n + 3n = 4 + 12

8n = 16

n = 2

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What acronym is used for sine , cosine and tangent
sergejj [24]

Answer:

SOHCAHTOA.

Step-by-step explanation:

acronym is an abbreviation formed from the initial letters of other words and pronounced as a word

The acronym for sin cosine and tangent is

SOHCAHTOA

Sine =Opposite over Hypotenuse

Cosine= Adjacent over Hypotenuse

Tangent= Opposite over Adjacent

8 0
3 years ago
Ms. Ortiz is painting her canoe. Two-fifths of gallon of paint covers ¼ of the canoe.
madreJ [45]

Answer: 8/5 or 1 3/5

Step-by-step explanation:

2/5 + 2/5 + 2/5 + 2/5

8/5 which is 1 3/5

6 0
3 years ago
NEED HELP FAST PLS CLICK TO SEE
Alexeev081 [22]
Y = -12 , -9

That’s the answer and good luck!
8 0
2 years ago
Please help due in a few minutes I will mark u brainliest
Goryan [66]

Answer:

y=4x-5

Step-by-step explanation:

4 0
2 years ago
Read 2 more answers
Water is being pumped into a conical tank that is 8 feet tall and has a diameter of 10 feet. If the water is being pumped in at
Deffense [45]

The rate of change of the depth of water in the tank when the tank is half

filled can be found using chain rule of differentiation.

When the tank is half filled, the depth of the water is changing at  <u>1.213 × </u>

<u>10⁻² ft.³/hour</u>.

Reasons:

The given parameter are;

Height of the conical tank, h = 8 feet

Diameter of the conical tank, d = 10 feet

Rate at which water is being pumped into the tank, = 3/5 ft.³/hr.

Required:

The rate at which the depth of the water in the tank is changing when the

tank is half full.

Solution:

The radius of the tank, r = d ÷ 2

∴ r = 10 ft. ÷ 2 = 5 ft.

Using similar triangles, we have;

\dfrac{r}{h} = \dfrac{5}{8}

The volume of the tank is therefore;

V = \mathbf{\dfrac{1}{3} \cdot \pi \cdot r^2 \cdot h}

r = \dfrac{5}{8} \times h

Therefore;

V = \dfrac{1}{3} \cdot \pi \cdot \left(  \dfrac{5}{8} \times h\right)^2 \cdot h = \dfrac{25 \cdot h^3 \cdot \pi}{192}

By chain rule of differentiation, we have;

\dfrac{dV}{dt} = \mathbf{\dfrac{dV}{dh} \cdot \dfrac{dh}{dt}}

\dfrac{dV}{dh}=\dfrac{d}{h} \left(  \dfrac{25 \cdot h^3 \cdot \pi}{192} \right) = \mathbf{\dfrac{25 \cdot h^2 \cdot \pi}{64}}

\dfrac{dV}{dt} = \dfrac{3}{5}  \ ft.^3/hour

Which gives;

\dfrac{3}{5} =  \mathbf{\dfrac{25 \cdot h^2 \cdot \pi}{64} \times \dfrac{dh}{dt}}

When the tank is half filled, we have;

V_{1/2} = \dfrac{1}{2} \times  \dfrac{1}{3} \times \pi \times 5^2 \times 8 =\mathbf{ \dfrac{25 \cdot h^3 \cdot \pi}{ 192}}

Solving gives;

h³ = 256

h = ∛256

\dfrac{3}{5} \times \dfrac{64}{25 \cdot h^2 \cdot \pi} = \dfrac{dh}{dt}

Which gives;

\dfrac{dh}{dt} = \dfrac{3}{5} \times \dfrac{64}{25 \cdot (\sqrt[3]{256}) ^2 \cdot \pi} \approx \mathbf{1.213\times 10^{-2}}

When the tank is half filled, the depth of the water is changing at  <u>1.213 × 10⁻² ft.³/hour</u>.

Learn more here:

brainly.com/question/9168560

6 0
2 years ago
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