All categories

1 year ago

Select all radical expressions that are equivalent to 3 4/7 Sorry for the bad quality!!

Mathematics

2 answers:

tatuchka [14]1 year ago

7 0

Answer:

Correct \/

Step-by-step explanation:

I just took the test

AnnZ [28]1 year ago

5 0

Answer:

$(\sqrt[7]{3} )^{4}$

Step-by-step explanation:

The only radical that matches the equivalent answer is the fifth one down. We can easily eliminate the radicals without exponents on the outside, since we know they won't create leftover fractions. So that leaves us with the second, fourth and fifth answers to contemplate.

Let's look at $(\sqrt[4]{3} )^{7}$ and $\sqrt[4]{3^{7} }$ first. It's good to know that these are equivalent radicals. The numbers are the same, and they will produce the same answers.

When you do the math, the exponent rule gives us fractions of $\frac{7}{4}$ for exponents, and eventually, a $3 * 3^{\frac{3}{4} }$ for both answers. So these are eliminated.

Now, for $(\sqrt[4]{3}^{7}$ , we can easiy simplify by changing the 7th root to a fraction in our exponent. Use the rule: $\sqrt[n]{x} = x^{\frac{1}{n} }$

$(3^{\frac{1}{7} } )^{4}$
Multiply the exponents: $\frac{1}{7} * 4 = \frac{4}{7}$
Insert the product into the exponent: $3^{\frac{4}{7} }$

And we can see the answer we're looking for! If you use this method to look at the other problems, you'll see that this is the only radical that simplifies to the required answer.

You might be interested in

Show that the Fibonacci numbers satisfy the recurrence relation fn = 5fn−4 + 3fn−5 for n = 5, 6, 7, . . . , together with the in

Sonja [21]

Answer with step-by-step explanation:

We are given that the recurrence relation

$f_n=5f_{n-4}+3f_{n-5}$

for n=5,6,7,..

Initial condition

$f_0=0,f_1=1,f_2=1,f_3=2,f_4=3$

We have to show that Fibonacci numbers satisfies the recurrence relation.

The recurrence relation of Fibonacci numbers

$f_n=f_{n-1}+f_{n-2}$ , $f_0=0,f_1=1$

Apply this

$f_n=(f_{n-2}+f_{n-3})+f_{n-2}=2f_{n-2}+f_{n-3}$

$f_n=2(f_{n-3}+f_{n-4})+f_{n-3}=3f_{n-3}+2f_{n-4}$

$f_n=3(f_{n-4}+f_{n-5})+2f_{n-4}=5f_{n-4}+3f_{n-5}$

Substitute n=2

$f_2=f_1+f_0=1+0=1$

$f_3=f_2+f_1=1+1=2$

$f_4=f_3+f_2=2+1=3$

Hence, the Fibonacci numbers satisfied the given recurrence relation .

Now, we have to show that $f_{5n}$ is divisible by 5 for n=1,2,3,..

Now replace n by 5n

$f_{5n}=5f_{5n-4}+3f_{5n-5}$

Apply induction

Substitute n=1

$f_5=5f_1+3f_0=5+0=5$

It is true for n=1

Suppose it is true for n=k

$f_{5k}=5f_{5k-4}+3f_{5k-5}$ is divisible 5

Let $f_{5k}=5q$

Now, we shall prove that for n=k+1 is true

$f_{5k+5}=5f_{5k+5-4}+3f_{5k+5-5}=5f_{5k+1}+3f_{5k}=5f_{5k+1}+3(5q)$

$f_{5k+5}=5(f_{5k+1}+3q)$

It is multiple of 5 .Therefore, it is divisible by 5.

It is true for n=k+1

Hence, the $f_{5n}$ is divisible by 5 for n=1,2,3,..

8 0

1 year ago

Tasha invests in an account that pays 1.5% compound interest annually. She uses the expression P(1+r)t to find the total value o

zhenek [66]

$\bf ~~~~~~ \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$3000\\
r=rate\to 1.5\%\to \frac{1.5}{100}\to &0.015\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annually, thus once}
\end{array}\to &1\\
t=years\to &5
\end{cases}
\\\\\\
A=3000\left(1+\frac{0.015}{1}\right)^{1\cdot 5}\implies A=3000(1.015)^5\implies A\approx 3231.852$

5 0

1 year ago

Read 2 more answers

The figure below shows a triangular wooden frame ABC. The side AD of the frame has rotted and needs to be replaced: What is the

Svet_ta [14]

The piece of wood is 5.1 inches

4 0

2 years ago

Read 2 more answers

Joey is building a frame for a sandbox. The sandbox is going to be a quadrilateral that has the lengths shown. A rectangle is sh

adelina 88 [10]

Answer- it’s A

Step-by-step explanation:

8 0

1 year ago

Read 2 more answers

Fredrick and Kate have 1713 cups of flour. They want to make loaves of banana bread for a bake sale. Fredrick has a recipe for b

Mariana [72]

Answer:

Fredrick's recipe will make 10 whole loaves.

There will be less flour left over if Frederick's recipe is used.

Step-by-step explanation:

Fredrick and Kate have 17 1/3 cups of flour. They want to make loaves of banana bread for a bake sale. Fredrick has a recipe for banana bread that calls for 1 5/8 cups of flour. Kate has a recipe that calls for 1 3/4 cups of flour. Select all the statements that are true about the recipes.

1. Fredrick's recipe will make 10 whole loaves.

2. Kate's recipe will make 11 whole loaves.

3. There will be less flour left over if Frederick's recipe is used.

4. There will be less flour left over if Kate's recipe is used.

5. More loaves of bread can be made using Frederick's recipe than Kate's recipe.

Solution:

Total cups of flour = 17 1/3

Fredrick recipe = 1 5/8 cups of flour

Kate = 1 3/4 cups of flour

Number of banana bread made with Fredrick's recipe = Total flour / flour per Fredrick's recipe

= 17 1/3 ÷ 1 5/8

= 52/3 ÷ 13/8

=52/3 × 8/13

= 416 / 39

= 10.67 loaves of banana bread

Number of banana bread made with Kate's recipe = Total flour / flour per Kate's recipe

= 17 1/3 ÷ 1 3/4

= 52/3 ÷ 7/4

= 52/3 × 4/7

= 216 / 21

= 10.29 loaves of banana bread

The true statements are:

Fredrick's recipe will make 10 whole loaves.

There will be less flour left over if Frederick's recipe is used

5 0

1 year ago