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12 times a number decreased by 5 is no more than 55

lions [1.4K]

Answer:

12n -5 ≤ 55

Step-by-step explanation:

12 times a number (represented by n) is 12n.

"Decreased by 5" means 5 is subtracted from the product, giving 12n-5.

"Is no more than 55" means "is less than or equal to 55".

Then the expression written using math symbols is ...

12n -5 ≤ 55

6 0

2 years ago

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If f(x)=3^x+10x and g(x)=4x-2, find (f+g)(x)

Vikentia [17]

Hi there!

First we need to remember the following.
$(f + g)(x) = f(x) + g(x)$

Now substitute both of the formulas.
$(f + g)(x) = ({3}^{x} + 10x)+ (4x - 2)$

Work out the parenthesis.
$(f + g)(x) = {3}^{x} + 10x + 4x - 2$

And finally collect terms.
$(f + g)(x) = {3}^{x} + 14x - 2$

~ Hope this helps you!

3 0

3 years ago

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Write an algebraic expression for " the minimum value of 2x + 1 is 13"

bezimeni [28]

2x+1=13 x=6 so an ok answer would be: 2x+10=40 x=15

6 0

3 years ago

Plz hellppppp me u won’t regret it

nexus9112 [7]

Answer:

Area of circle = 200.96 units²

Step-by-step explanation:

The formula to find the area of a circle is: πr², where r is the radius (which is half of the diameter)

So the diameter is 16 and of 16 is 8, which is the radius.

Sub it into the formula of a circle and you would get the area of a circle.

3.14 x 8²

Use calculator

Area of circle = 200.96 units²

8 0

2 years ago

Find the general solution to each of the following ODEs. Then, decide whether or not the set of solutions form a vector space. E

Ipatiy [6.2K]

Answer:

(A) $y=ke^{2t}$ with $k\in\mathbb{R}$ .

(B) $y=ke^{2t}/2-1/2$ with $k\in\mathbb{R}$

(C) $y=k_1e^{2t}+k_2e^{-2t}$ with $k_1,k_2\in\mathbb{R}$

(D) $y=k_1e^{2t}+k_2e^{-2t}+e^{3t}/5$ with $k_1,k_2\in\mathbb{R}$ ,

Step-by-step explanation

(A) We can see this as separation of variables or just a linear ODE of first grade, then $0=y'-2y=\frac{dy}{dt}-2y\Rightarrow \frac{dy}{dt}=2y \Rightarrow \frac{1}{2y}dy=dt \ \Rightarrow \int \frac{1}{2y}dy=\int dt \Rightarrow \ln |y|^{1/2}=t+C \Rightarrow |y|^{1/2}=e^{\ln |y|^{1/2}}=e^{t+C}=e^{C}e^t} \Rightarrow y=ke^{2t}$ . With this answer we see that the set of solutions of the ODE form a vector space over, where vectors are of the form $e^{2t}$ with $t$ real.

(B) Proceeding and the previous item, we obtain $1=y'-2y=\frac{dy}{dt}-2y\Rightarrow \frac{dy}{dt}=2y+1 \Rightarrow \frac{1}{2y+1}dy=dt \ \Rightarrow \int \frac{1}{2y+1}dy=\int dt \Rightarrow 1/2\ln |2y+1|=t+C \Rightarrow |2y+1|^{1/2}=e^{\ln |2y+1|^{1/2}}=e^{t+C}=e^{C}e^t \Rightarrow y=ke^{2t}/2-1/2$ . Which is not a vector space with the usual operations (this is because $-1/2$ ), in other words, if you sum two solutions you don't obtain a solution.

(C) This is a linear ODE of second grade, then if we set $y=e^{mt} \Rightarrow y''=m^2e^{mt}$ and we obtain the characteristic equation $0=y''-4y=m^2e^{mt}-4e^{mt}=(m^2-4)e^{mt}\Rightarrow m^{2}-4=0\Rightarrow m=\pm 2$ and then the general solution is $y=k_1e^{2t}+k_2e^{-2t}$ with $k_1,k_2\in\mathbb{R}$ , and as in the first items the set of solutions form a vector space.

(D) Using C, let be $y=me^{3t}$ we obtain that it must satisfies $3^2m-4m=1\Rightarrow m=1/5$ and then the general solution is $y=k_1e^{2t}+k_2e^{-2t}+e^{3t}/5$ with $k_1,k_2\in\mathbb{R}$ , and as in (B) the set of solutions does not form a vector space (same reason! as in (B)).

4 0

3 years ago