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3 years ago

Find the value of the expression m − 4.6 for m = 7.

Mathematics

1 answer:

stich3 [128]3 years ago

3 0

Answer:

2.4

Step-by-step explanation:

If you replace m with 7, the expression is now 7-4.6.

7-4.6=2.4

The value of the expression is 2.4.

-hope it helps

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1. Over the past year, a university’s computer system has been struck by a virus at an average rate of 0.4 viruses per week. The

Marrrta [24]

Answer:

Step-by-step explanation:

Since 0.4virus occurs per week and it cost the university $1000 to remove one virus.

There next week we expect to 0.4 virus and cost required will be =0.4×$1000. this will amount to $400

5 0

3 years ago

Which value of x makes the equation -5x-(-7-4x) =-2(3x-4 true?

Ludmilka [50]

Answer:

x = 1/5

Step-by-step explanation:

Step 1: Write equation

-5x - (-7 - 4x) = -2(3x - 4)

Step 2: Solve for x

Distribute: -5x + 7 + 4x = -6x + 8
Combine like terms: -x + 7 = -6x + 8
Add 6x to both sides: 5x + 7 = 8
Subtract 7 on both sides: 5x = 1
Divide by 5 on both sides: x = 1/5

Step 3: Check

Plug in x to verify it's a solution.

-5(1/5) - (-7 - 4(1/5)) = -2(3(1/5) - 4)

-1 - (-7 - 4/5) = -2(3/5 - 4)

-1 - (-39/5) = -2(-17/5)

-1 + 39/5 = 34/5

34/5 = 34/5

5 0

3 years ago

Consider the following differential equation to be solved by undetermined coefficients. y(4) − 2y''' + y'' = ex + 1 Write the gi

kompoz [17]

Answer:

The general solution is

$y= (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x +\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))$

+ $\frac{x^2}{2}$

Step-by-step explanation:

Step :1:-

Given differential equation y(4) − 2y''' + y'' = e^x + 1

The differential operator form of the given differential equation

$(D^4 -2D^3+D^2)y = e^x+1$

comparing f(D)y = e^ x+1

The auxiliary equation (A.E) f(m) = 0

$m^4 -2m^3+m^2 = 0$

$m^2(m^2 -2m+1) = 0$

$(m^2 -2m+1) this is the expansion of (a-b)^2$

$m^2 =0 and (m-1)^2 =0$

The roots are m=0,0 and m =1,1

complementary function is $y_{c} = (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x$

Step 2:-

The particular equation is $\frac{1}{f(D)} Q$

P.I = $\frac{1}{D^2(D-1)^2} e^x+1$

P.I = $\frac{1}{D^2(D-1)^2} e^x+\frac{1}{D^2(D-1)^2}e^{0x}$

P.I = $I_{1} +I_{2}$

$\frac{1}{D^2} (\frac{x^2}{2!} )e^x + \frac{1}{D^{2} } e^{0x}$

$\frac{1}{D} means integration$

$\frac{1}{D^2} (\frac{x^2}{2!} )e^x = \frac{1}{2D} \int\limits {x^2e^x} \, dx$

applying in integration u v formula

$\int\limits {uv} \, dx = u\int\limits {v} \, dx - \int\limits ({u^{l}\int\limits{v} \, dx } )\, dx$

$I_{1}$ = $\frac{1}{D^2(D-1)^2} e^x$

$\frac{1}{2D} (e^x(x^2)-e^x(2x)+e^x(2))$

$\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))$

$I_{2}$ $= \frac{1}{D^2(D-1)^2}e^{0x}$

$\frac{1}{D} \int\limits {1} \, dx= \frac{1}{D} x$

again integration $\frac{1}{D} x = \frac{x^2}{2!}$

The general solution is $y = y_{C} +y_{P}$

$y= (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x +\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))$

+ $\frac{x^2}{2!}$

3 0

3 years ago

Can someone answer all 3 thanks.

MA_775_DIABLO [31]

Answer:

All of them are yes because you either subtract or add to put in the form of mx+b

6 0

3 years ago

Paige’s income statement for the month of December is shown. Paige monthly income statement for December. Total wages are 1,850

DedPeter [7]

Answer:

Paige’s net income for December is $950

Step-by-step explanation:

Here in this question, we are interested in calculating Paige’s net income for December

Mathematically;

Net income = Monthly wage - Total Expenses

From the question, monthly wage = $1,850 while the total expenses = $900

Thus, the net income will be ; $1,850 - $900 = $950

6 0

3 years ago