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3 years ago

Simplify: 3^2 · 3^4 A) 3^6 B) 3^8 C) 3^2 D) 9^6

2 answers:

5 0

It is A Bc if they have the same base (here, they have the same base as 3) THEN AND ONLY THEN can you add the exponents together

Greeley [361]3 years ago

3 0

Answer:

3^6

Step-by-step explanation:

3^2 * 3^4

You add the exponents together.

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Sam needs a web designer. Designer A is offering her services for an initial $550 in addition to $105 per hour. Designer B is of

Wewaii [24]

After two hours the prices will meet a 760

$550+105=655 |655+105= 760

660+50= 710 |710+50= 760

They meet in the second hour so in the second hour they will charge the same.

5 0

3 years ago

ITS EASY JUST PLEASE HELP

svet-max [94.6K]

Letter B is the answer to your question

8 0

3 years ago

The length of a regulation soccer field is 110 meters. The diagonal length of the same soccer field is about 133.14 meters. Abou

dexar [7]

Answer:

Step-by-step explanation:

Imagine this as a right angle triangle, where the diagonal length is the hypotenuse, the length is one side, and the width is the other.

We can therefore use Pythagoras' Theorem (or Pythagorean Theorem) to solve. The formula for this is a²+b²=c², where c is the hypotenuse, and a and b are the sides.

We can input the values we know to this formula to get the width. This gives 110²+b²=133.14² or 12100+b²=17 726.2596.

From there subtracting 12100 from both sides gives b²=5626.2596.

Square rooting b isolates it, leaving b=75.0083969.

Since the value of the diagonal was approximate, this can be assumed the b is 75m.

**This content involves Pythagoras' Theorem/Pythagorean Theorem, which you may wish to revise. I'm always happy to help!

8 0

3 years ago

Use the Newton-Raphson method to find the root of the equation f(x) = In(3x) + 5x2, using an initial guess of x = 0.5 and a stop

xxMikexx [17]

Answer with explanation:

The equation which we have to solve by Newton-Raphson Method is,

f(x)=log (3 x) +5 x²

$f'(x)=\frac{1}{3x}+10 x$

Initial Guess =0.5

Formula to find Iteration by Newton-Raphson method

$x_{n+1}=x_{n}-\frac{f(x_{n})}{f'(x_{n})}\\\\x_{1}=x_{0}-\frac{f(x_{0})}{f'(x_{0})}\\\\ x_{1}=0.5-\frac{\log(1.5)+1.25}{\frac{1}{1.5}+10 \times 0.5}\\\\x_{1}=0.5- \frac{0.1760+1.25}{0.67+5}\\\\x_{1}=0.5-\frac{1.426}{5.67}\\\\x_{1}=0.5-0.25149\\\\x_{1}=0.248$

$x_{2}=0.248-\frac{\log(0.744)+0.30752}{\frac{1}{0.744}+10 \times 0.248}\\\\x_{2}=0.248- \frac{-0.128+0.30752}{1.35+2.48}\\\\x_{2}=0.248-\frac{0.17952}{3.83}\\\\x_{2}=0.248-0.0468\\\\x_{2}=0.2012$

$x_{3}=0.2012-\frac{\log(0.6036)+0.2024072}{\frac{1}{0.6036}+10 \times 0.2012}\\\\x_{3}=0.2012- \frac{-0.2192+0.2025}{1.6567+2.012}\\\\x_{3}=0.2012-\frac{-0.0167}{3.6687}\\\\x_{3}=0.2012+0.0045\\\\x_{3}=0.2057$

$x_{4}=0.2057-\frac{\log(0.6171)+0.21156}{\frac{1}{0.6171}+10 \times 0.2057}\\\\x_{4}=0.2057- \frac{-0.2096+0.21156}{1.6204+2.057}\\\\x_{4}=0.2057-\frac{0.0019}{3.6774}\\\\x_{4}=0.2057-0.0005\\\\x_{4}=0.2052$

So, root of the equation =0.205 (Approx)

Approximate relative error

$=\frac{\text{Actual value}}{\text{Given Value}}\\\\=\frac{0.205}{0.5}\\\\=0.41$

Approximate relative error in terms of Percentage

=0.41 × 100

= 41 %

7 0

2 years ago

12−(−2) over a line 12−6÷2

joja [24]

Given 12−(−2) over a line 12−6÷2

We can use PEMDAS rule and solve the expression in the order as given below:-

1. Parentheses

2. Exponents

3. Multiplication

4. Division

5. Addition

6. Subtraction

Now solving for given expression:-

$\frac{12-(-2)}{12-(\frac{6}{2})}\\\\=\frac{12+2}{12-3}\\\\=\frac{14}{9}$

So final answer is 14/9 i.e. 14 over 9.

7 0

3 years ago