All categories

3 years ago

3^4 + 3 x 5 = (the '^4' is an exponent)

Mathematics

2 answers:

juin [17]3 years ago

5 0

$\bold{Heya~!}$

<h2><u>Your answer will be is :</u></h2>

<u />

<u /> $\sf{= \: 81 \: + \: 3x^5}$

$\huge \boxed {\underline{Explanation:}}$

$\sf{3^4 \: + \: 3x^5}$

$\sf{3^4 \: = \: 81}$

$\sf{= \: 81 \: + \: 3x^5}$

Hope It Helps ! ~

#LearnWithBrainly

$\underline{Answer~:}$

Jace ~

Vesnalui [34]3 years ago

4 0

Answer:

Step-by-step explanation:

3^4 = 3*3*3*3 = 9*9= 81

3*5=15

81 + 15 = 96

Hope this helps!!

You might be interested in

Plz help thank you :) Karen spent $32 on fruit at the grocery store. She spent a total of $40 at the store. What percentage of t

Lostsunrise [7]

32 is 80% of 40. Karen spent 80% on fruit and 20% on misc items.
Answer: 80%

7 0

3 years ago

Read 2 more answers

7.) Given f(x) = 6(x - 1). what is the value of f(-8)? * A) 42 B) -42 C) 54 D) -54

natka813 [3]

Answer:

D) -54

Step-by-step explanation:

To solve this question, simply plug in -8 for x:

f(-8)=6(-8-1)

f(-8)=6(-9)

f(-8)=-54

Hope this helps!!

7 0

4 years ago

A giant tank in a shape of an inverted cone is filled with oil. the height of the tank is 1.5 metre and its radius is 1 metre. t

skad [1K]

The given height of the cylinder of 1.5 m, and radius of 1 m, and the rate

of dripping of 110 cm³/s gives the following values.

1) The rate of change of the oil's radius when the radius is 0.5 m is r' ≈ <u>9.34 × 10⁻⁵ m/s</u>

2) The rate of change of the oil's height when the height is 20 cm is h' ≈ <u>1.97 × 10⁻³ m/s</u>

3) The rate the oil radius is changing when the radius is 10 cm is approximately <u>0.175 m/s</u>

<h3>How can the rate of change of the radius & height be found?</h3>

The given parameters are;

Height of the tank, h = 1.5 m

Radius of the tank, r = 1 m

Rate at which the oil is dripping from the tank = 110 cm³/s = 0.00011 m³/s

$1) \hspace{0.15 cm}V = \frac{1}{3} \cdot \pi \cdot r^2 \cdot h$

From the shape of the tank, we have;

$\dfrac{h}{r} = \dfrac{1.5}{1}$

Which gives;

h = 1.5·r

$V = \mathbf{\frac{1}{3} \cdot \pi \cdot r^2 \cdot (1.5 \cdot r)}$

$\dfrac{d}{dr} V =\dfrac{d}{dr} \left( \dfrac{1}{3} \cdot \pi \cdot r^2 \cdot (1.5 \cdot r)\right) = \dfrac{3}{2} \cdot \pi \cdot r^2$

$\dfrac{dV}{dt} = \dfrac{dV}{dr} \times \dfrac{dr}{dt}$

$\dfrac{dr}{dt} = \mathbf{\dfrac{\dfrac{dV}{dt} }{\dfrac{dV}{dr} }}$

$\dfrac{dV}{dt} = 0.00011$

Which gives;

$\dfrac{dr}{dt} = \mathbf{ \dfrac{0.00011 }{\dfrac{3}{2} \cdot \pi \cdot r^2}}$

When r = 0.5 m, we have;

$\dfrac{dr}{dt} = \dfrac{0.00011 }{\dfrac{3}{2} \times\pi \times 0.5^2} \approx 9.34 \times 10^{-5}$

The rate of change of the oil's radius when the radius is 0.5 m is r' ≈ <u>9.34 × 10⁻⁵ m/s</u>

2) When the height is 20 cm, we have;

h = 1.5·r

$r = \dfrac{h}{1.5}$

$V = \mathbf{\frac{1}{3} \cdot \pi \cdot \left(\dfrac{h}{1.5} \right) ^2 \cdot h}$

r = 20 cm ÷ 1.5 = $13.\overline3$ cm = $0.1\overline3$ m

Which gives;

$\dfrac{dr}{dt} = \dfrac{0.00011 }{\dfrac{3}{2} \times\pi \times 0.1 \overline{3}^2} \approx \mathbf{1.313 \times 10^{-3}}$

$\dfrac{d}{dh} V = \dfrac{d}{dh} \left(\dfrac{4}{27} \cdot \pi \cdot h^3 \right) = \dfrac{4 \cdot \pi \cdot h^2}{9}$

$\dfrac{dV}{dt} = \dfrac{dV}{dh} \times \dfrac{dh}{dt}$

$\dfrac{dh}{dt} = \dfrac{\dfrac{dV}{dt} }{\dfrac{dV}{dh} }$ <em />

$\dfrac{dh}{dt} = \mathbf{\dfrac{0.00011}{\dfrac{4 \cdot \pi \cdot h^2}{9}}}$

When the height is 20 cm = 0.2 m, we have;

$\dfrac{dh}{dt} = \dfrac{0.00011}{\dfrac{4 \times \pi \times 0.2^2}{9}} \approx \mathbf{1.97 \times 10^{-3}}$

The rate of change of the oil's height when the height is 20 cm is h' ≈ <u>1.97 × 10⁻³ m/s</u>

3) The volume of the slick, V = π·r²·h

Where;

h = The height of the slick = 0.1 cm = 0.001 m

Therefore;

V = 0.001·π·r²

$\dfrac{dV}{dr} = \mathbf{ 0.002 \cdot \pi \cdot r}$

$\dfrac{dr}{dt} = \mathbf{\dfrac{0.00011 }{0.002 \cdot \pi \cdot r}}$

When the radius is 10 cm = 0.1 m, we have;

$\dfrac{dr}{dt} = \dfrac{0.00011 }{0.002 \times \pi \times 0.1} \approx \mathbf{0.175}$

The rate the oil radius is changing when the radius is 10 cm is approximately <u>0.175 m</u>

Learn more about the rules of differentiation here:

brainly.com/question/20433457

brainly.com/question/13502804

3 0

3 years ago

John can jog twice as fast as he can walk. He was able to jog the first 5 miles to his grandmothers house, but the he tired and

Arte-miy333 [17]

Formula for speed = distance/ time
S = 7/.9 = 7.77 miles per hour or 8 mph if you round up.

8 0

3 years ago

Helppppp BRAINLEST if right ASAP

icang [17]

Answer:

21.9

Step-by-step explanation:

3 0

3 years ago