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2 years ago

PLS WILL MARK BRAINLIEST

Mathematics

2 answers:

muminat2 years ago

7 0

Answer:

x=-9 and 3

Step-by-step explanation:

FOIL method

x^2+6x-27=0

Find the common numbers that multiply to 27 that will also equal 6.

(27x6) or (9x3)

the 6 is positive while 27 is negative so you need to use a + and - in the parenthesis.

then FOIL them

(x+9)(x-3)

FIRST: x*x=x^2

OUTSIDE: x*-3=-3x

INNER: 9*x=9x

LAST: +9*-3=-27

x^2(-3x9x)-27

x^2+6x-27

now that the equation matches you solve for x.

(x+9)=0 x=-9

(x-3)=0 x=3

loris [4]2 years ago

5 0

Answer:

Solution

=−9

x=3

x=−9

x=3x=−9

Step-by-step explanation:

Boom

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Step-by-step explanation:

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3 years ago

Evaluate 8a - 1 + 0.5b when a= 1/4 and b=10.

nikitadnepr [17]

Answer:

Step-by-step explanation:

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6 0

3 years ago

The voltage in an electrical circuit is multiplied by itself each time it is reduced. The voltage is 27/125 of a volt and it has

ANTONII [103]

Answer:

The voltage in exponential form would be $(\frac{3}{5})^3$ .

The original voltage in the circuit is $\frac{3}{5}$ of a volt.

Step-by-step explanation:

Let x represent the original voltage.

We have been given that the voltage in an electrical circuit is multiplied by itself each time it is reduced. The voltage is 27/125 of a volt and it has been reduced 3 times.

The original voltage multiplied by itself for 3 times would be $x\cdot x\cdot x=x^3$ .

Now, we will equate $x^3$ with 27/125 as:

$x^3=\frac{27}{125}$

We know that 27 is cube of 3 and 125 is cube of 5, so we can represent our equation as:

$x^3=\frac{3^3}{5^3}$

$x^3=(\frac{3}{5})^3$

Therefore, the voltage in exponential form would be $(\frac{3}{5})^3$

Now, we will take cube root of both sides to find original voltage in the circuit as:

$\sqrt[3]{x^3} =\sqrt[3]{(\frac{3}{5})^3}$

Using property $\sqrt[n]{a^n} =a$ , we will get:

$x =\frac{3}{5}$

Therefore, the original voltage in the circuit is $\frac{3}{5}$ of a volt.

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4 years ago

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m_a_m_a [10]

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