Ask question

Ask question

All categories

3 years ago

7

1. Write as a base with a positive exponent, then evaluate. Write your final answer as a

1 answer:

vodomira [7]3 years ago

5 0

Answer:

Well a base with a positive exponent would be 5^3. All this means is 5x5x5 which is 125.

Step-by-step explanation:

Hope this helps!

You might be interested in

Use completing the square to solve for x in the equation (x+7)(x-9) = 25.

balandron [24]

Answer:

x = 1 ±√89

Step-by-step explanation:

We have the equation:

(x+7)(x-9) = 25

Using distributive property:

x(x-9) + 7(x-9) = 25

x²- 9x + 7x - 63 -25 = 0

x²- 2x - 88 = 0

To complete squares we need to add and subtract 1, as follows:

x²- 2x - 88 +1 -1 = 0

x²- 2x +1 -88 -1 = 0 (this is a perfect square)

(x - 1)² - 89 = 0

Solving for x:

(x - 1)² = 89

x - 1 = ±√89

x = 1 ±√89

6 0

3 years ago

Can anyone help me solve this

pshichka [43]

Answer:

well i think the answer is 80

Step-by-step explanation:

180=130 + 137+x

130-50+x

130-50=80

3 0

3 years ago

Read 2 more answers

An electronics company produces transistors, resistors, and computer chips. Each transistor requires 3 units of copper, 1 unit

padilas [110]

Answer:

An electronics company can be produce 350 transistors and 340 computer chips, they can´t produce resistors.

Step-by-step explanation:

1. We will name the variables for transistors, resistors and the computer chips.

a = Transistors

b= Resistors

c = Computer chips

2. We propose three linear equations, one for the copper, one for the zinc and one for the glass.

$\left \{ {{3a+3b+2c=1730} \atop {a+2b+c=690}}\atop {2a+b+2c=1380}} \right.$

3. We write the matrix form as Ax=d

$A=\left(\begin{array}{ccc}3&3&2\\1&2&1\\2&1&2\end{array}\right)$

$x=\left(\begin{array}{ccc}a\\b\\c\end{array}\right)$

$A=\left(\begin{array}{ccc}1730\\690\\1380\end{array}\right)$

With this formula the solution of x is $x=\frac{d}{A}$ or $x=A^{-1}d$

4. We will find the inverse matrix $A^{-1}$ using the formula:

$A^{-1} = \frac{1}{detA} (C_{A})^{T}$

a. det A

$det A=\left[\begin{array}{ccc}3&3&2\\1&2&1\\2&1&2\end{array}\right] =3*(4-1)-3*(2-2)+2*(1-4)=9-0-6=3$

b. $(C_{A})^{T}$

$C_{A}=\left(\begin{array}{ccc}4-1&.(2-2)&1-4\\-(6-2)&6-4&-(3-6)\\3-4&-(3-2)&6-3\end{array}\right)$

$C_{A}=\left(\begin{array}{ccc}3&.0&-3\\-4&2&3\\-1&-1&3\end{array}\right)$

$(C_{A}) ^T=\left(\begin{array}{ccc}3&0&-3\\-4&2&3\\-1&-1&3\end{array}\right)^T$

$(C_{A}) ^T=\left(\begin{array}{ccc}3&-4&-1\\0&2&-1\\-3&3&3\end{array}\right)$

c. $A^{-1}$

$A^{-1}=\frac{1}{3} \left(\begin{array}{ccc}3&-4&-1\\0&2&-1\\-3&3&3\end{array}\right)$

5. As $x=\frac{d}{A}$ or $x=A^{-1}d$ , the solution of x is:

$x=\frac{1}{3}\left(\begin{array}{ccc}3&-4&-1\\0&2&-1\\-3&3&3\end{array}\right)\left(\begin{array}{ccc}1730\\690\\1380\end{array}\right)$

$x=\frac{1}{3}\left(\begin{array}{ccc}(3*1730)+(-4*690)+(-1*1380)\\(0*1730)+(2*690)+(-1*1380)\\(-3*1730)+(3*690)+(3*1380)\end{array}\right)$

$x=\frac{1}{3}\left(\begin{array}{ccc}1050\\0\\1020)\end{array}\right)$

$X=\left[\begin{array}{ccc}350\\0\\340\end{array}\right]$

<u><em>Therefore:</em></u>

<u><em>a= 350 Transistors</em></u>

<u><em>b=0 Resistors</em></u>

<u><em>c=340 Computer chips</em></u>

4 0

3 years ago

Simplify -12x^4/x^4+8x^5

alexgriva [62]

Answer:

OPTION A: $\frac{12}{1 + 8x}$ , where $x \ne - \frac{1}{8}$ .

Step-by-step explanation:

Given: $\frac{- 12x^4}{x^4 + 8x^5}$

Taking $x^4$ common outside in the denominator, we get:

$= \frac{-12x^4}{(x^4)(1 + 8x)}$

$x^4$ will get cancelled on the numerator and denominator, we get:

$= \frac{-12}{1 + 8x}$

we know that the denominator can not be zero.

That means, 1 + 8x $\ne$ 0.

$\implies 8x \ne -1$

$\implies x \ne \frac{-1}{8}$

So, the answer is: $\frac{-12}{1 + 8x}$ , where $x \ne \frac{-1}{8}$ .

6 0

3 years ago

How do I tell if a fraction is bigger or smaller than another fraction?

rodikova [14]

Answer:

compare them

Step-by-step explanation:

first make the denominators the same, then compare the numerators

6 0

3 years ago