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3 years ago

What is the value of this expression? (37)0 0 37 1 307

Mathematics

2 answers:

Anni [7]3 years ago

5 0

37^0 = 1

Any number to the 0 exponent is equal 1, except 0

Answer

TiliK225 [7]3 years ago

4 0

Answer:

Step-by-step explanation:

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What is the equation for 5 units left and 3 units up from f(x)=x

erastovalidia [21]

The new equation after shifting will be:

$g(x) = (x+5)+3\\g(x) = x + 8$

Step-by-step explanation:

Function trnafomations upward and left are defined as:

Upward:

f(x) => f(x)+b where b is an integer

Left:

f(x) => f(x+b) where b is an integers

Given function is:

$f(x) = x$

Shifting the function 5 units left

$g(x) = f(x+5) => x+5\\g(x) = x+5$

Shifting the function upward 2 units

So,

$g(x) = (x+5)+3\\g(x) = x + 8$

The new equation after shifting will be:

$g(x) = (x+5)+3\\g(x) = x + 8$

Keywords: Functions, shifting

Learn more about functions at:

#LearnwithBrainly

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2 years ago

Is 8 yards graeter than 20 feet

Natasha_Volkova [10]

8 yards is 24 feet, so yes it’s greater

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3 years ago

Read 2 more answers

Solve y'' + 10y' + 25y = 0, y(0) = -2, y'(0) = 11 y(t) = Preview

svetlana [45]

Answer: The required solution is

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

Step-by-step explanation: We are given to solve the following differential equation :

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

Let us consider that

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

Then, we have

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

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

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

According to the given conditions, we have

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

and

$y^\prime(0)=11\\\\\Rightarrow -5(A+B\times0)+B=11\\\\\Rightarrow -5A+B=11\\\\\Rightarrow (-5)\times(-2)+B=11\\\\\Rightarrow 10+B=11\\\\\Rightarrow B=11-10\\\\\Rightarrow B=1.$