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ElenaW [278]
2 years ago
15

Pleaseeee helpppp!! thank you sm:(

Mathematics
2 answers:
Mariana [72]2 years ago
5 0

Answer:

A

Step-by-step explanation:

x-5 has to be less than -2 and the inequality sign goes the opposite direction than the arrow on the number line

sesenic [268]2 years ago
4 0

Answer:A

Step-by-step explanation:

You might be interested in
Please help on this one plzzzz
deff fn [24]

We know that Angle Bisector Divides an Angle into Two Equal Angles.

As UP is the Angle Bisector of Angle U, It Divides Angle U into two Equal Parts they are Angle(1) and Angle(2)

⇒ Angle(1) = Angle(2)

Given Angle(1) = 5x + 10 and Angle(2) = 3x + 14

⇒ 5x + 10 = 3x + 14

⇒ 2x = 4

⇒ x = 2

⇒ Angle(1) = 5x + 10 = 5(2) + 10 = 10 + 10 = 20

So : Measure of Angle(1) is 20

8 0
3 years ago
For f(x) = 3x - 1 and g(x) = x+2, find (f – g)(x).
kobusy [5.1K]
Well, I bet you want your answer right away! So here it is.


<span>Given <span>f (x) = 3x + 2</span> and <span>g(x) = 4 – 5x</span>, find <span>(f + g)(x), (f – g)(x), (f × g)(x)</span>, and <span>(f / g)(x)</span>.</span>

To find the answers, all I have to do is apply the operations (plus, minus, times, and divide) that they tell me to, in the order that they tell me to.

(f + g)(x) = f (x) + g(x)

= [3x + 2] + [4 – 5x]

= 3x + 2 + 4 – 5x

= 3x – 5x + 2 + 4

= –2x + 6

(f – g)(x) = f (x) – g(x)

= [3x + 2] – [4 – 5x]

= 3x + 2 – 4 + 5x

= 3x + 5x + 2 – 4

= 8x – 2

(f  × g)(x) = [f (x)][g(x)]

= (3x + 2)(4 – 5x)

= 12x + 8 – 15x2 – 10x

= –15x2 + 2x + 8

<span>\left(\small{\dfrac{f}{g}}\right)(x) = \small{\dfrac{f(x)}{g(x)}}<span><span>(<span><span>​g</span>​<span>​f</span><span>​​</span></span>)</span>(x)=<span><span>​<span>g(x)</span></span>​<span>​<span>f(x)</span></span><span>​​</span></span></span></span><span>= \small{\dfrac{3x+2}{4-5x}}<span>=<span><span>​<span>4−5x</span></span>​<span>​<span>3x+2</span></span><span>​​</span></span></span></span>

My answer is the neat listing of each of my results, clearly labelled as to which is which.

( f + g ) (x) = –2x + 6

( f – g ) (x) = 8x – 2

( f  × g ) (x) = –15x2 + 2x + 8

<span>\mathbf{\color{purple}{ \left(\small{\dfrac{\mathit{f}}{\mathit{g}}}\right)(\mathit{x}) = \small{\dfrac{3\mathit{x} + 2}{4 - 5\mathit{x}}} }}<span><span>(<span><span>​g</span>​<span>​f</span><span>​​</span></span>)</span>(x)=<span><span>​<span>4−5x</span></span>​<span>​<span>3x+2</span></span><span>​​


Hope I helped! :) If I did not help that's okay.


-Duolingo
</span></span></span></span>

7 0
3 years ago
5. After paying a tax of 8 palsa per rupee a man saves Rs. 18,400 per month Then his monthly income is 2 Points) 20,000​
Hunter-Best [27]

Answer:

92% of Income is  Rs 18400

Step-by-step explanation:

I = 18400/.92 = Rs 20000

8 0
2 years ago
Consider the following. (A computer algebra system is recommended.) y'' + 3y' = 2t4 + t2e−3t + sin 3t (a) Determine a suitable f
drek231 [11]

First look for the fundamental solutions by solving the homogeneous version of the ODE:

y''+3y'=0

The characteristic equation is

r^2+3r=r(r+3)=0

with roots r=0 and r=-3, giving the two solutions C_1 and C_2e^{-3t}.

For the non-homogeneous version, you can exploit the superposition principle and consider one term from the right side at a time.

y''+3y'=2t^4

Assume the ansatz solution,

{y_p}=at^5+bt^4+ct^3+dt^2+et

\implies {y_p}'=5at^4+4bt^3+3ct^2+2dt+e

\implies {y_p}''=20at^3+12bt^2+6ct+2d

(You could include a constant term <em>f</em> here, but it would get absorbed by the first solution C_1 anyway.)

Substitute these into the ODE:

(20at^3+12bt^2+6ct+2d)+3(5at^4+4bt^3+3ct^2+2dt+e)=2t^4

15at^4+(20a+12b)t^3+(12b+9c)t^2+(6c+6d)t+(2d+e)=2t^4

\implies\begin{cases}15a=2\\20a+12b=0\\12b+9c=0\\6c+6d=0\\2d+e=0\end{cases}\implies a=\dfrac2{15},b=-\dfrac29,c=\dfrac8{27},d=-\dfrac8{27},e=\dfrac{16}{81}

y''+3y'=t^2e^{-3t}

e^{-3t} is already accounted for, so assume an ansatz of the form

y_p=(at^3+bt^2+ct)e^{-3t}

\implies {y_p}'=(-3at^3+(3a-3b)t^2+(2b-3c)t+c)e^{-3t}

\implies {y_p}''=(9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c)e^{-3t}

Substitute into the ODE:

(9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c)e^{-3t}+3(-3at^3+(3a-3b)t^2+(2b-3c)t+c)e^{-3t}=t^2e^{-3t}

9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c-9at^3+(9a-9b)t^2+(6b-9c)t+3c=t^2

-9at^2+(6a-6b)t+2b-3c=t^2

\implies\begin{cases}-9a=1\\6a-6b=0\\2b-3c=0\end{cases}\implies a=-\dfrac19,b=-\dfrac19,c=-\dfrac2{27}

y''+3y'=\sin(3t)

Assume an ansatz solution

y_p=a\sin(3t)+b\cos(3t)

\implies {y_p}'=3a\cos(3t)-3b\sin(3t)

\implies {y_p}''=-9a\sin(3t)-9b\cos(3t)

Substitute into the ODE:

(-9a\sin(3t)-9b\cos(3t))+3(3a\cos(3t)-3b\sin(3t))=\sin(3t)

(-9a-9b)\sin(3t)+(9a-9b)\cos(3t)=\sin(3t)

\implies\begin{cases}-9a-9b=1\\9a-9b=0\end{cases}\implies a=-\dfrac1{18},b=-\dfrac1{18}

So, the general solution of the original ODE is

y(t)=\dfrac{54t^5 - 90t^4 + 120t^3 - 120t^2 + 80t}{405}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\dfrac{3t^3+3t^2+2t}{27}e^{-3t}-\dfrac{\sin(3t)+\cos(3t)}{18}

3 0
3 years ago
Tyler went to the supermarket to buy food for a food pantry. He has $36, and can carry up to 20 pounds of food in his backpack.
shutvik [7]
(12 8) is a solution
5 0
3 years ago
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