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defon
3 years ago
15

Data was collected on the average winter temperature and the number of days with

Mathematics
1 answer:
laila [671]3 years ago
6 0

Answer:

You should expect to see more days with snow when the average temperature is lower, and less days wth snow when the average winter temperature is higher.

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A school-bus is dropping students off after school. When the bus left the
Verdich [7]

Answer:

60%

Step-by-step explanation:

40-16=24

24/40=0.6

0.6*100=60

First, work out the difference (decrease) between the two numbers you are comparing.

Decrease = Original Number - New Number

Next, divide the decrease by the original number and multiply the answer by 100.

% Decrease = Decrease ÷ Original Number × 100

3 0
3 years ago
Read 2 more answers
Given f(x)= 2x^3+kx-1, and x+1 is a factor of f(x), then what is the value of k?
VMariaS [17]

Answer:

put x=-1, and the rest will be easy

5 0
3 years ago
Read 2 more answers
Solve the question: (-4)+(-1)+2+5+...+x=437
Alex Ar [27]
The sequence is an arithmetic sequence with
a₁ = -4

d = a₂ - a₁
d = -1 - (-4)
d = -1 + 4
d = 3

an = x

Sn = 437


General formula in arithmetic sequence
Formula to find nth term
an = a₁ + d(n - 1)
Formula to find sum of sequence (sn)
Sn = n/2 (a₁ + an)

We have to make an equation system based on the problem
plug the numbers into the formula
First equation
an = a₁ + d(n - 1)
x = -4 + 3(n - 1)
x = -4 + 3n - 3
x = 3n - 7

Second equation
Sn = n/2 (a₁ + an)
n/2 (a₁ + an) = 437
n/2 (-4 + x) = 437
n(x -  4) = 874
xn - 4n = 874

Solve the equation system by subtitution method
Subtitute x with 3n - 7 in the second equation
xn - 4n = 874
(3n - 7)n - 4n = 874
3n² - 7n - 4n = 874
3n² - 11n - 874 = 0
(3n + 46)(n - 19) = 0
n = -46/3 or n = 19
Because the number of terms shouldn't be negative, -46/3 isn't required, so the value of n is 19.

Solve for x, back to the first equatin
x = 3n - 7
x = 3(19) - 7
x = 57 - 7
x = 50

The solution is 50
6 0
3 years ago
Show that the line integral is independent of path by finding a function f such that ?f = f. c 2xe?ydx (2y ? x2e?ydy, c is any p
Juli2301 [7.4K]
I'm reading this as

\displaystyle\int_C2xe^{-y}\,\mathrm dx+(2y-x^2e^{-y})\,\mathrm dy

with \nabla f=(2xe^{-y},2y-x^2e^{-y}).

The value of the integral will be independent of the path if we can find a function f(x,y) that satisfies the gradient equation above.

You have

\begin{cases}\dfrac{\partial f}{\partial x}=2xe^{-y}\\\\\dfrac{\partial f}{\partial y}=2y-x^2e^{-y}\end{cases}

Integrate \dfrac{\partial f}{\partial x} with respect to x. You get

\displaystyle\int\dfrac{\partial f}{\partial x}\,\mathrm dx=\int2xe^{-y}\,\mathrm dx
f=x^2e^{-y}+g(y)

Differentiate with respect to y. You get

\dfrac{\partial f}{\partial y}=\dfrac{\partial}{\partial y}[x^2e^{-y}+g(y)]
2y-x^2e^{-y}=-x^2e^{-y}+g'(y)
2y=g'(y)

Integrate both sides with respect to y to arrive at

\displaystyle\int2y\,\mathrm dy=\int g'(y)\,\mathrm dy
y^2=g(y)+C
g(y)=y^2+C

So you have

f(x,y)=x^2e^{-y}+y^2+C

The gradient is continuous for all x,y, so the fundamental theorem of calculus applies, and so the value of the integral, regardless of the path taken, is

\displaystyle\int_C2xe^{-y}\,\mathrm dx+(2y-x^2e^{-y})\,\mathrm dy=f(4,1)-f(1,0)=\frac9e
8 0
3 years ago
A complete graph of the polynomial function f is provided. Use your knowledge of the domain, range, behavior at each x intercept
Papessa [141]

The graph represents a polynomial graph, and the equation of the polynomial graph is f(x) = -(x + 5)³(x - 2)²

<h3>How to determine the equation?</h3>

From the graph, we have the following observations:

  • The graph has a turning point at x = 2
  • The graph changes the factor of its direction at x = -5

This means that the graph has a multiplicity of 2 at x = 2 and a multiplicity of 3 at x = -5

So, we have:

f(x) = a(x + 5)³(x - 2)²

The curve is inverted,

This means that a < 0

Assume a = -1.

Then, we have:

f(x) = -(x + 5)³(x - 2)²

Hence, the equation of the polynomial graph is f(x) = -(x + 5)³(x - 2)²

Read more about polynomials at:

brainly.com/question/4142886

#SPJ1

3 0
2 years ago
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