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Nady [450]
3 years ago
14

What is the range of -5,1,4,6,9,0,-7,-1

Mathematics
2 answers:
tester [92]3 years ago
7 0

Answer: 16

Step-by-step explanation: 9 - -7 = 16

zloy xaker [14]3 years ago
4 0

Answer:

16

Step-by-step explanation:

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Which graph represents this exponential function? f(x)=3(2)^x-4
Rasek [7]

Answer:

A

Step-by-step explanation:

look up desmos graphing calculator and it graphs all of your equations for you. I used it all through high school.

4 0
3 years ago
What is the value of x?<br> 26°<br> オー<br> X=<br> [?]<br> Enter
Nat2105 [25]
273 and my proof is x is the value of 1 and the answer is 273 cause 26 degrees is the value of 273
3 0
2 years ago
Read 2 more answers
A world record was set for the men’s 100-m dash in the 2008 Olympic Games in Beijing by Usain Bolt of Jamaica. Bolt "coasted" ac
Andrei [34K]

Answer:

a) v = 12.21m/s

a = 4.07 m/s²

b)v = 11.24m/s

a =  3.75 m/s²

Step-by-step explanation:

a) Dividing the moviment into two parts:

I - With acceleration

v = v₀ + at

s = s₀ + v₀t + at²/2

  • v₀ = 0
  • s₀ = 0
  • a = ?
  • v = ?
  • t = 3s
  • s = x

v = v₀ + at

v = 3a ⇒ a = v/3

s = s₀ + v₀t + at²/2

x = v/3.3²/2

x = 3v/2

II - Uniform

s = s₀ + vt

s = 100

s₀ = x

v = v

t = 9.69 - 3 = 6.69s

s = s₀ + vt

100 = x + v*6.69

100 = x + 6.69v

As x = 3v/2

100 = 3v/2 + 6.69v

100 = 1.5v + 6.69v

100 = 8.19v

v = 12.21m/s

a = v/3 = 4.07 m/s²

b) Dividing the moviment into two parts:

I - With acceleration

v = v₀ + at

s = s₀ + v₀t + at²/2

  • v₀ = 0
  • s₀ = 0
  • a = ?
  • v = ?
  • t = 3s
  • s = x

v = v₀ + at

v = 3a ⇒ a = v/3

s = s₀ + v₀t + at²/2

x = v/3.3²/2

x = 3v/2

II - Uniform

s = s₀ + vt

s = 200

s₀ = x

v = v

t = 19.30 - 3 = 16.30s

s = s₀ + vt

200 = x + v*16.3

100 = x + 16.3v

As x = 3v/2

200 = 3v/2 + 16.3v

200 = 1.5v + 16.3v

200 = 17.8v

v = 11.24m/s

a = v/3 = 3.75 m/s²

8 0
2 years ago
Find the mass of the lamina that occupies the region D = {(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1} with the density function ρ(x, y) = xye
Alona [7]

Answer:

The mass of the lamina is 1

Step-by-step explanation:

Let \rho(x,y) be a continuous density function of a lamina in the plane region D,then the mass of the lamina is given by:

m=\int\limits \int\limits_D \rho(x,y) \, dA.

From the question, the given density function is \rho (x,y)=xye^{x+y}.

Again, the lamina occupies a rectangular region: D={(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1}.

The mass of the lamina can be found by evaluating the double integral:

I=\int\limits^1_0\int\limits^1_0xye^{x+y}dydx.

Since D is a rectangular region, we can apply Fubini's Theorem to get:

I=\int\limits^1_0(\int\limits^1_0xye^{x+y}dy)dx.

Let the inner integral be: I_0=\int\limits^1_0xye^{x+y}dy, then

I=\int\limits^1_0(I_0)dx.

The inner integral is evaluated using integration by parts.

Let u=xy, the partial derivative of u wrt y is

\implies du=xdy

and

dv=\int\limits e^{x+y} dy, integrating wrt y, we obtain

v=\int\limits e^{x+y}

Recall the integration by parts formula:\int\limits udv=uv- \int\limits vdu

This implies that:

\int\limits xye^{x+y}dy=xye^{x+y}-\int\limits e^{x+y}\cdot xdy

\int\limits xye^{x+y}dy=xye^{x+y}-xe^{x+y}

I_0=\int\limits^1_0 xye^{x+y}dy

We substitute the limits of integration and evaluate to get:

I_0=xe^x

This implies that:

I=\int\limits^1_0(xe^x)dx.

Or

I=\int\limits^1_0xe^xdx.

We again apply integration by parts formula to get:

\int\limits xe^xdx=e^x(x-1).

I=\int\limits^1_0xe^xdx=e^1(1-1)-e^0(0-1).

I=\int\limits^1_0xe^xdx=0-1(0-1).

I=\int\limits^1_0xe^xdx=0-1(-1)=1.

No unit is given, therefore the mass of the lamina is 1.

3 0
3 years ago
A sample of tritium-3 decayed to 94.5% of its original amount after a year.
Zolol [24]
(a) If y(t) is the mass after t days and y(0) = A then y(t) = Ae^{kt}.

y(1) = Ae^k = 0.945A \implies e^k = 0.945 \implies k = \ln 0.945

\text{Then } Ae^{(\ln 0.945)t} = \frac{1}{2}A\ \Leftrightarrow\ \ln e^{(\ln 0.945)t} = \ln \frac{1}{2}\ \Leftrightarrow\  ( \ln 0.945) t = \ln \frac{1}{2} \Leftrightarrow \\ \\&#10;t = - \frac{\ln 5}{\ln 0.945} \approx 12.25 \text{ years}

(b)
Ae^{(\ln 0.945) t} = 0.20 A \ \Leftrightarrow\ (\ln 0.945) t \ln \frac{1}{5}\ \implies \\ \\&#10;t = - \frac{\ln 5}{\ln 0.945} \approx 28.45\text{ years}
3 0
3 years ago
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