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Anna35 [415]
3 years ago
12

What is the simplified form of the difference shown below? (3a^2-6a-1)-(5a^2-2a+3)

Mathematics
1 answer:
kodGreya [7K]3 years ago
8 0

Answer:

(3a^2-6a-1)-(5a^2-2a+3)

= 3a²-6a-1 -5a²+2a-3

=3a²-5a²-6a+2a-1-3

= -2a²-4a-4

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The height of a lamp is 100 centimeters. how tall is the lamp in inches?
lions [1.4K]

Answer:

39.37 in

Step-by-step explanation:

8 0
2 years ago
1260=(n-2)x180<br> find the value of n<br> can you please give me a explaination
mixas84 [53]
Answer:

Explanation:

1260 = (n - 2)180
1260/180 = n - 2
7 = n - 2
7 + 2 = n
9 = n

Therefore, n = 9
7 0
3 years ago
An Archer shoots an arrow horizontally at 250 feet per second. The bullseye on the target and the arrow are initially at the sam
Genrish500 [490]

Answer:

<h2>1.84feet</h2>

Step-by-step explanation:

Using the formula for finding range in projectile, Since range is the distance covered in the horizontal direction;

Range R = U\sqrt{\frac{H}{g} }

U is the velocity of the arrow

H is the maximum height reached = distance below the bullseye reached by the arrow.

R is the horizontal distance covered i.e the distance of the target from the archer.

g is the acceleration due to gravity.

Given R = 60ft, U = 250ft/s, g = 32ft/s H = ?

On substitution,

60 = 250\sqrt{\frac{H}{32}} \\\frac{60}{250} = \sqrt{\frac{H}{32}}\\\frac{6}{25} = \sqrt{\frac{H}{32}

Squaring both sides we have;

(\frac{6}{25} )^{2} = (\sqrt{\frac{H}{32} } )^{2} \\\frac{36}{625} =  \frac{H}{32} \\625H = 36*32\\H = \frac{36*32}{625} \\H = 1.84feet

The arrow will hit the target 1.84feet below the bullseye.

5 0
3 years ago
A tank in the form of a right-circular cylinder of radius 2 feet and height 10 feet is standing on end. If the tank is initially
k0ka [10]

Answer:

\frac{dh}{dt} = -3.48 \times 10^{-3}\sqrt{h} feet per second is the differential equation

Step-by-step explanation:

Given:

The radius of the cylindrical tank= 2 feet

The height of the cylindrical tank = 10 feet

The radius of the circular hole = 3/4 inches

To Find:

The differential equation for the height h of the water at time t.

Solution:

<u>Finding the surface area(A) of the tank</u>

Surface area  = \pi R^2

On substituting the values

Surface area =\pi(2)^2

= 4\pisquare feet

<u>Finding the surface area(a) of the hole</u>

The radius is given in  inches, so converting into feet we have

1 inch =  0.083 foot

similarly

\frac{3}{4} = 0.75 inches = 0.75 \times 0.083 =  0.0625 feet.

Now the surface area,

= \pi \times 0.0625

= 0.0625 \pi square feet

Now let the velocity of water through the hole is v

According law of conservation of energy, the penitential energy due to the height h of the  water gets converted into kinetic energy.

\frac{1}{2}mv^2 =mgh

v^2 = \frac{2mgh}{m}

v^2 = 2gh

v= \sqrt{2gh}

The rate of water flowing through the hole is  = a\times v

= >a \times \sqrt{2gh}

At any time t

V(t) = A \times h(t)

\frac{dV}{dt} = -a\sqrt{2gh}

\frac{d(Ah(t))}{dt} = -a\sqrt{2gh}

A \frac{d(h(t))}{d(t)} = -a\sqrt{2gh}

\frac{dh}{dt} = -\frac{a}{A} \sqrt{2gh}

On substituting the values, we get

\frac{dh}{dt} = -\frac{0.0625\pi}{4\pi}\sqrt{2\times 32 \times 10

\frac{dh}{dt} = -3.48 \times 10^{-3}\sqrt{h}  feet per second

3 0
3 years ago
# of yeast cells double every 4 hours how long will it take for 6billion cells
VLD [36.1K]

Answer:

60

Step-by-step explanation:

0.25

2×5

10

0.25×10

=2.5

;4×6

=24

:2.5×24

=60

6 0
3 years ago
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