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Rina8888 [55]
3 years ago
7

Step my step I don’t understand(:Brainliest

Mathematics
1 answer:
PSYCHO15rus [73]3 years ago
3 0
Its the first one because if the circle is filled in than it is 2and more or the thing
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Their sum is −7 and their difference is 14 what are my numbers
Mashcka [7]

Answer:

Answer: 3 1/2 and -10 1/2 are the two numbers.

Step-by-step explanation:

Let x and y be the two unknown numbers.

x+y=-7 [Given]

x-y=14 [Given]

x-y=14 [Given]

x=y+14 [Add y to both sides]

x+y=-7 [Given]

(y+14)+y=-7 [Subtitution]

2y+14=-7 [Combine like terms]

2y=-21 [Subtract 14 from both sides]

y=-21/2 [Divide both sides by 2]

y=-10 1/2 [Division]

x-y=14 [Given]

x=y+14 [Add y to both sides]

x=-10 1/2 + 14 [Substitution]

x= 3 1/2 [Addition]

Check:

x+y=-7 [Given]

3 1/2 + -10 1/2?=-7 [Substition]

-7=-7 [Addition]

QED

x-y=14 [Given]

3 1/2 - -10 1/2?=14 [Substitution]

3 1/2 + 10 1/2?=14 [Change the sign of the subtrahend and add]

14=14 [Addition]

QED

Answer: 3 1/2 and -10 1/2 are the two numbers.

8 0
3 years ago
Solve for f.<br><br> f<br> 3<br> + -22<br> = -19<br><br><br> f =
IRISSAK [1]
F=1
Add 22 to both sides, then divide both sides by 3

3 0
3 years ago
Read 2 more answers
Simplify the following expression : (2x + 7y) (x – 3y) – (9x + 4) (3x - 2)
Dmitriy789 [7]
-25x^2+xy-21y^2+6x+8
8 0
3 years ago
Fill in the Table Below for the Corresponding x - valuesxy-2024
yaroslaw [1]

Here, to make your day a little better<3

7 0
3 years ago
3y''-6y'+6y=e*x sexcx
Simora [160]
From the homogeneous part of the ODE, we can get two fundamental solutions. The characteristic equation is

3r^2-6r+6=0\iff r^2-2r+2=0

which has roots at r=1\pm i. This admits the two fundamental solutions

y_1=e^x\cos x
y_2=e^x\sin x

The particular solution is easiest to obtain via variation of parameters. We're looking for a solution of the form

y_p=u_1y_1+u_2y_2

where

u_1=-\displaystyle\frac13\int\frac{y_2e^x\sec x}{W(y_1,y_2)}\,\mathrm dx
u_2=\displaystyle\frac13\int\frac{y_1e^x\sec x}{W(y_1,y_2)}\,\mathrm dx

and W(y_1,y_2) is the Wronskian of the fundamental solutions. We have

W(e^x\cos x,e^x\sin x)=\begin{vmatrix}e^x\cos x&e^x\sin x\\e^x(\cos x-\sin x)&e^x(\cos x+\sin x)\end{vmatrix}=e^{2x}

and so

u_1=-\displaystyle\frac13\int\frac{e^{2x}\sin x\sec x}{e^{2x}}\,\mathrm dx=-\int\tan x\,\mathrm dx
u_1=\dfrac13\ln|\cos x|

u_2=\displaystyle\frac13\int\frac{e^{2x}\cos x\sec x}{e^{2x}}\,\mathrm dx=\int\mathrm dx
u_2=\dfrac13x

Therefore the particular solution is

y_p=\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x

so that the general solution to the ODE is

y=C_1e^x\cos x+C_2e^x\sin x+\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x
7 0
3 years ago
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