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2 years ago

When -3v = -18 is solved, the result is:

Mathematics

2 answers:

mel-nik [20]2 years ago

7 0

The result of this mathematical sentence is <u>v=6</u>

<u />

<h3>First degree equation</h3>

First degree equation is a mathematical sentence - having at least one unknown. Its representation is given by <u>ax + b = 0</u>. To calculate an equation, as it is in the expression in the statement, simply: isolate x and divide the second to the first member.

*Note: Equal signs, the result will be equated as positive.

-3v = -18

v = -18/-3

Therefore, the correct value of this equation is <u>v = 6</u>.

rjkz [21]2 years ago

4 0

Answer:

When -3v = -18 is solved, the result is:

21v

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At a greengrocer, two bananas and one apple cost ?1.16 and one banana and one apple cost ?0.71. How much does one apple cost?

Blababa [14]

Answer

Find out the how much does one apple cost .

To prove

Let us assume that the number of bananas be x .

Let us assume that the number of apples be y .

As given

At a greengrocer, two bananas and one apple cost $1.16 .

Than the equation becomes

2x + y = 1.16

one banana and one apple cost $0.71.

Than the equation becomes

x + y = 0.71

Subtracting x + y = 0.71 from 2x + y = 1.16

2x - x + y - y = 1.16 - 0.71

x = 0.45

Put in the equation x + y = 0.71

0.45 + y = 0.71

y = 0.71 - 0.45

y = $0.26

Therefore the cost of the one apple is $0.26 .

4 0

3 years ago

A) Find a particular solution to y" + 2y = e^3 + x^3. b) Find the general solution.

Reptile [31]

Answer:

a.P.I= $\frac{e^{3x}}{11}+\frac{1}{2}(x^3-3x)$

b.G.S= $C_1Cos \sqrt2 x+C_2 Sin\sqrt2 x+\frac{1}{11}e^{3x}+\frac{1}{2}(x^3-3x}$

Step-by-step explanation:

We are given that a linear differential equation

$y''+2y=e^{3x}+x^3$

We have to find the particular solution

P.I= $\frac{e^{3x}}{D^2+2}+\frac{x^3}{D^2+2}$

P.I= $\frac{e^{3x}}{3^2+2}+\frac{1}{2} x^3(1+\frac{D^2}{2})^{-2}$

P.I= $\frac{e^{3x}}{11}+\frac{1-2\frac{D^2}{4}+3\frac{D^4}{16}+...}{2}x^3$

P.I= $\frac{e^{3x}}{11}+\frac{x^3-2\frac{\cdot3\cdot 2x}{4}}{2}+0}$ (higher order terms can be neglected

P.I= $\frac{e^{3x}}{11}+\frac{1}{2}(x^3-3x)$

b.Characteristics equation

$D^2+2=0$

$D=\pm\sqrt2 i$

C.F= $C_1cos \sqrt2x+C_2 sin\sqrt2 x$

G.S=C.F+P.I

G.S= $C_1Cos \sqrt2 x+C_2 Sin\sqrt2 x+\frac{1}{11}e^{3x}+\frac{1}{2}(x^3-3x)$

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3 years ago

How did business leaders in the late nineteenth century utilize vertical integration? a. They maintained control of the raw mate

bulgar [2K]

Business leaders in the late nineteenth century utilized vertical integration by maintaining control of production and distribution of their products.

Answer: Option C

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Vertical integration is a competitive strategy that gives the company full control over one or more stages of product production or distribution. Rockefeller tirelessly tried to take full control of business 'oil refinery'. While other business people were flooding the area in search of quick fortune, Rockefeller was thinking of destroying his rivals and creating a real monopoly in the refining industry.

Looking for even more control, Rockefeller saw the benefits of organizing the transportation to his products. Then, he began to develop his business through vertical integration, in which the company analyses all aspects of the product life cycle, from raw material extraction, through the production process, to the final delivery of the product.

Other industrialists quickly followed, including Gustavus Swift, who at the end of the 19th century used vertical integration to dominate the American meat packaging industry.

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3 years ago

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AnnyKZ [126]

your answer

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