7. simplify:
√144 ______ √361 ______
−√144 ______ −√49 ______
−√1 ______ −√225 ______

8. simplify each expression:
3√49 = ______
√576 / 8 = ______
√121 − 4 = ______

9. which best describes √35 ?
a. exactly 5 c. between 5 and 6
b. exactly 17.5 d. between 31 and 37

10. circle all numbers that lie between
2.8×10⁰ and √12.25
3% (\frac{11}{3}) 3.6 2.8
√13 294% 3.5×10¹

11. selecting a queen from a deck of cards,
keeping it and then selecting a heart.
independent events or dependent events

12. a chorus class has 50 8th graders and 35
7th graders. two students are selected at
random to sing a duet. what is the probability
that both are 7th graders (without
replacement)?
( p(7^ ext{th} ext{ grader}, 7^ ext{th} ext{ grader}) )

13. when ordering from least to greatest, which
number would come first?
( 7.5 imes 10^{-1} ), ( 70% ), ( \frac{7}{100} ), ( 0.0075 )

14. plot the ( sqrt{113} ) on the number line.
number line with two ticks
what whole number is ( sqrt{113} ) closest to on the
number line?

Question

7. simplify:
√144 ______ √361 ______
−√144 ______ −√49 ______
−√1 ______ −√225 ______

8. simplify each expression:
3√49 = ______
√576 / 8 = ______
√121 − 4 = ______

9. which best describes √35 ?
  a. exactly 5  c. between 5 and 6
  b. exactly 17.5  d. between 31 and 37

10. circle all numbers that lie between
2.8×10⁰ and √12.25
3%  (\frac{11}{3})  3.6  2.8
√13  294%  3.5×10¹

11. selecting a queen from a deck of cards,
keeping it and then selecting a heart.
independent events or dependent events

12. a chorus class has 50 8th graders and 35
7th graders. two students are selected at
random to sing a duet. what is the probability
that both are 7th graders (without
replacement)?
( p(7^	ext{th} 	ext{ grader}, 7^	ext{th} 	ext{ grader}) )

13. when ordering from least to greatest, which
number would come first?
( 7.5 	imes 10^{-1} ), ( 70% ), ( \frac{7}{100} ), ( 0.0075 )

14. plot the ( sqrt{113} ) on the number line.
number line with two ticks
what whole number is ( sqrt{113} ) closest to on the
number line?

Accepted Answer

### Question 7
# Explanation:
## Step1: Simplify $\sqrt{144}$
We know that $12	imes12 = 144$, so $\sqrt{144}=12$.
## Step2: Simplify $\sqrt{361}$
We know that $19	imes19 = 361$, so $\sqrt{361}=19$.
## Step3: Simplify $-\sqrt{144}$
From Step1, $\sqrt{144}=12$, so $-\sqrt{144}=- 12$.
## Step4: Simplify $-\sqrt{49}$
We know that $7	imes7 = 49$, so $-\sqrt{49}=-7$.
## Step5: Simplify $-\sqrt{1}$
We know that $1	imes1 = 1$, so $-\sqrt{1}=-1$.
## Step6: Simplify $-\sqrt{225}$
We know that $15	imes15 = 225$, so $-\sqrt{225}=-15$.
# Answer:
$\sqrt{144}=\boldsymbol{12}$; $\sqrt{361}=\boldsymbol{19}$; $-\sqrt{144}=\boldsymbol{-12}$; $-\sqrt{49}=\boldsymbol{-7}$; $-\sqrt{1}=\boldsymbol{-1}$; $-\sqrt{225}=\boldsymbol{-15}$

### Question 8
# Explanation:
## Step1: Simplify $3\sqrt{49}$
We know that $\sqrt{49} = 7$, so $3\sqrt{49}=3	imes7 = 21$.
## Step2: Simplify $\frac{\sqrt{576}}{8}$
We know that $\sqrt{576}=24$, so $\frac{\sqrt{576}}{8}=\frac{24}{8}=3$.
## Step3: Simplify $\sqrt{121}-4$
We know that $\sqrt{121} = 11$, so $\sqrt{121}-4=11 - 4=7$.
# Answer:
$3\sqrt{49}=\boldsymbol{21}$; $\frac{\sqrt{576}}{8}=\boldsymbol{3}$; $\sqrt{121}-4=\boldsymbol{7}$

### Question 9
# Explanation:
We know that $5^{2}=25$ and $6^{2}=36$. Since $25<35<36$, taking square roots, we get $5 < \sqrt{35}<6$. So $\sqrt{35}$ is between 5 and 6.
# Answer:
c. between 5 and 6

### Question 10
First, simplify the bounds:
- $2.8	imes10^{0}=2.8$
- $\sqrt{12.25} = 3.5$

Now, simplify each number:
- $3\%=0.03$
- $\frac{11}{3}\approx3.67$
- $3.628$
- $\sqrt{13}\approx3.606$
- $294\% = 2.94$
- $3.5	imes10^{1}=35$

Now, find numbers between $2.8$ and $3.5$:
- $294\%=2.94$ (between 2.8 and 3.5)
- $\frac{11}{3}\approx3.67$ (greater than 3.5)
- $3.628$ (greater than 3.5)
- $\sqrt{13}\approx3.606$ (greater than 3.5)
- $3\% = 0.03$ (less than 2.8)
- $3.5	imes10^{1}=35$ (greater than 3.5)

So the number between $2.8	imes10^{0}$ and $\sqrt{12.25}$ is $294\%$.
# Answer:
Circle $\boldsymbol{294\%}$

### Question 11
When we select a queen from a deck of cards and keep it, the total number of cards left in the deck changes, and also the number of hearts (if the queen was a heart) or the number of non - heart cards changes. The outcome of the first event (selecting a queen) affects the probability of the second event (selecting a heart). So these are dependent events.
# Answer:
Dependent Events

### Question 12
# Explanation:
## Step1: Find the total number of students
The total number of students is $50 + 35=85$.
## Step2: Probability of first 7th grader
The probability of selecting a 7th grader first is $P(	ext{7th grader})=\frac{35}{85}=\frac{7}{17}$.
## Step3: Probability of second 7th grader (without replacement)
After selecting one 7th grader, there are $35 - 1 = 34$ 7th graders left and a total of $85 - 1=84$ students left. So the probability of selecting a 7th grader second is $P(	ext{7th grader after 7th grader})=\frac{34}{84}=\frac{17}{42}$.
## Step4: Probability of both events
Using the multiplication rule for dependent events, $P(	ext{7th grader, 7th grader})=\frac{35}{85}	imes\frac{34}{84}$.
Simplify $\frac{35}{85}	imes\frac{34}{84}=\frac{35	imes34}{85	imes84}=\frac{35	imes34}{85	imes84}=\frac{35	imes34}{(17	imes5)	imes(12	imes7)}=\frac{34}{(17	imes12)}=\frac{2}{12}=\frac{1}{6}\approx0.1667$ (or we can simplify as $\frac{35}{85}	imes\frac{34}{84}=\frac{7}{17}	imes\frac{17}{42}=\frac{7	imes17}{17	imes42}=\frac{1}{6}$)
# Answer:
$\boldsymbol{\frac{1}{6}}$ (or approximately $0.167$)

### Question 13
# Explanation:
First, convert all numbers to decimal form:
- $7.5	imes10^{-1}=0.75$
- $70\% = 0.7$
- $\frac{7}{100}=0.07$
- $0.0075$

Now, compare the decimals: $0.0075<0.07<0.7<0.75$. So the smallest number is $0.0075$.
# Answer:
$\boldsymbol{0.0075}$

### Question 14
To plot $\sqrt{113}$ on the number line:
We know that $10^{2}=100$ and $11^{2}=121$. So $\sqrt{100}=10$ and $\sqrt{121}=11$. Since $100<113<121$, $\sqrt{113}$ is between 10 and 11 on the number line.

To find the whole number closest to $\sqrt{113}$:
Calculate the distance between $\sqrt{113}$ and 10: $d_1=\sqrt{113}-10$
Calculate the distance between 11 and $\sqrt{113}$: $d_2 = 11-\sqrt{113}$

We know that $10.6^{2}=112.36$ and $10.7^{2}=114.49$. So $\sqrt{113}\approx10.63$.

$d_1=\sqrt{113}-10\approx10.63 - 10 = 0.63$

$d_2=11-\sqrt{113}\approx11 - 10.63 = 0.37$

Since $d_2<d_1$, $\sqrt{113}$ is closer to 11.

# Answer:
When plotting $\sqrt{113}$ on the number line, it is between 10 and 11. The whole number closest to $\sqrt{113}$ is $\boldsymbol{11}$.

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Question 7

Step1: Simplify $\sqrt{144}$

Step2: Simplify $\sqrt{361}$

Step3: Simplify $-\sqrt{144}$

Step4: Simplify $-\sqrt{49}$

Step5: Simplify $-\sqrt{1}$

Step6: Simplify $-\sqrt{225}$

Step1: Simplify $3\sqrt{49}$

Step2: Simplify $\frac{\sqrt{576}}{8}$

Step3: Simplify $\sqrt{121}-4$

Answer:

Question 8