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3 years ago

Solving Equations ve x + 14 = 7x + 32 - 3x

Mathematics

1 answer:

Mkey [24]3 years ago

5 0

Answer:

-6

Step-by-step explanation:

Step 1: Simplify both sides of the equation.

x+14=7x+32−3x

x+14=7x+32+−3x

x+14=(7x+−3x)+(32)(Combine Like Terms)

x+14=4x+32

Step 2: Subtract 4x from both sides.

x+14−4x=4x+32−4x

−3x+14=32

Step 3: Subtract 14 from both sides.

−3x+14−14=32−14

−3x=18

Step 4: Divide both sides by -3.

−3x/−3=18/−3

x=−6

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aleksandrvk [35]

Answer:

The system cannot be solved by a unique cost for each soldier, because has infinite solutions

Step-by-step explanation:

Let

x-------> the cost of a legionary per day

y-------> the cost of an archer per day

we know that

$4x+4y=10$ ------> equation A

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Multiply equation B by $2$ both sides

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equation A and equation B is the same equation

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4 years ago

The monthly demand for a company’s sports caps varies directly as the amount spent on advertising and inversely as the square of

Amanda [17]

Answer: $\$16.432$

Step-by-step explanation:

Let be "y" the monthly demand for the company’s sports caps, "x" the amount in dollar spent on advertising and "z"the price ind dollars per cap.

The model for this situation is:

$y=\frac{kx}{z^2}$

Where "k" is the constant of proportionality.

Since $y=300$ when $z=15$ and $x=2,500$ , we can substitute values into $y=\frac{kx}{z^2}$ and solve for "k" to finds its value:

$300=\frac{k(2,500)}{15^2}\\\\\frac{(300)(15^2)}{2,500}=k\\\\k=27$

Then, in order to calculate what price yields a demand of 300 caps when advertising is increased to $3,000, we must substitute the following values into $y=\frac{kx}{z^2}$ :

$x=3,000\\\\y=300\\\\k=27$

Then:

$300=\frac{(27)(3,000)}{z^2}$

Solving for "z" we get:

$z^2=\frac{(27)(3,000)}{300}\\\\z=\sqrt{\frac{(27)(3,000)}{300}}\\\\z=16.432$

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4 years ago

A recursive rule for a geometric sequence is a1=2; an= 4/9 an-1

Oxana [17]

A recursive rule for a geometric sequence:

$a_1\\\\a_n=r\cdot a_{n-1}$

---------------------------------------------------

$a_1=2;\ a_n=\dfrac{4}{9}a_{n-1}\to r=\dfrac{4}{9}$

---------------------------------------------------

The exciplit rule:

$a_n=a_1r^{n-1}$

Substitute:

$a_n=2\left(\dfrac{4}{9}\right)^{n-1}=2\cdot\left(\dfrac{4}{9}\right)^n\cdot\left(\dfrac{4}{9}\right)^{-1}=2\cdot\left(\dfrac{4}{9}\right)^n\cdot\dfrac{9}{4}\\\\\boxed{a_n=\dfrac{9}{2}\cdot\left(\dfrac{4}{9}\right)^n}$

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3 years ago

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Mars2501 [29]

Answer:

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6 0

2 years ago

Read 2 more answers

I need an answer NOT A LINK!!! ASAP

pantera1 [17]

Answer:

Why are you saying l dont want link

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3 0

3 years ago