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Sedbober [7]
3 years ago
9

The graphs below have the same shape. What is the equation of the red graph?

Mathematics
2 answers:
Sveta_85 [38]3 years ago
8 0

we have

g(x)=4-x^{2}

the vertex of the function g(x) is the point (0,4)

the vertex of the function f(x) is the point (0,0)

so

the rule of the translation of g(x) to f(x) is

(x,y)------> (x,y-4)

that means

the translation is 4 units down

therefore

the equation of the function f(x) is

f(x)=g(x)-4\\f(x)=(4- x^{2})-4 \\f(x)=- x^{2}

therefore

<u>the answer is</u>

f(x)=- x^{2}

Natasha2012 [34]3 years ago
4 0

g(x)=-x^2 is right for apex

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⇒x+48

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3 0
3 years ago
(a) Use the scale to change 40 kilometres to miles.
Nimfa-mama [501]

Answer:

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3 years ago
What four consecutive intergers have a sum of -34
Ivenika [448]

Answer:

-7, -8 , -9 and -10 are such four integers.

Step-by-step explanation:

Let the first integer = k

So, the next three consecutive integers are ( k+1), ((k + 1) + 1)  and (((k + 1) + 1) +1)

or, k , (k +1), (k +2) and (k +3) are four consecutive integers.

Now, their sum is -34.

⇒k   + (k +1) +  (k +2) +  (k +3)  = -34

or, 4k  + 6 = -34

or, 4k = - 34 - 6 = -40

⇒  k = -40/ 4 = -10

Hence, the first integer k = -10

The next integer = k + 1 = -10 + 1 = -9

Similarly next two integers are -8 and - 7.

So -7, -8 , -9 and -10 are such four integers.

7 0
3 years ago
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fredd [130]

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\dfrac{\partial\left(4x^2y^3+\frac{2y}{x+y^2}\right)}{\partial x}=8xy^3-\dfrac{2y}{(x+y^2)^2}

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\dfrac{\partial F}{\partial x}=2xy^4+\dfrac1{x+y^2}\implies F(x,y)=x^2y^4+\ln(x+y^2)+f(y)

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\boxed{x^2y^3+\ln(x+y^2)=8+\ln9}

8 0
3 years ago
Solve y'' + 10y' + 25y = 0, y(0) = -2, y'(0) = 11 y(t) = Preview
svetlana [45]

Answer:  The required solution is

y=(-2+t)e^{-5t}.

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y^{\prime\prime}+10y^\prime+25y=0,~~~~~~~y(0)=-2,~~y^\prime(0)=11~~~~~~~~~~~~~~~~~~~~~~~~(i)

Let us consider that

y=e^{mt} be an auxiliary solution of equation (i).

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y^prime=me^{mt},~~~~~y^{\prime\prime}=m^2e^{mt}.

Substituting these values in equation (i), we get

m^2e^{mt}+10me^{mt}+25e^{mt}=0\\\\\Rightarrow (m^2+10y+25)e^{mt}=0\\\\\Rightarrow m^2+10m+25=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow m^2+2\times m\times5+5^2=0\\\\\Rightarrow (m+5)^2=0\\\\\Rightarrow m=-5,-5.

So, the general solution of the given equation is

y(t)=(A+Bt)e^{-5t}.

Differentiating with respect to t, we get

y^\prime(t)=-5e^{-5t}(A+Bt)+Be^{-5t}.

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y(0)=-2\\\\\Rightarrow A=-2

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Thus, the required solution is

y(t)=(-2+1\times t)e^{-5t}\\\\\Rightarrow y(t)=(-2+t)e^{-5t}.

6 0
3 years ago
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