Which of the following statements is correct?

alexdok [17]

The expression obtained after solving the given expression is, $\rm \frac{(15x+3)}{y}$

<h3>What is a BODMAS rule?</h3>

For the Bracket, Order, Division, Multiplication, Addition, and Subtraction rules, BODMAS stands for Bracket, Order, Division, Multiplication, Addition, and Subtraction.

Given expression ;

9(5x + 1) ÷ 3y

Follow the BODMAS rule;

Step 1; Divide

$\rm \frac{9(5x+1)}{3y} \\\\ \frac{3(5x+1)}{y} \\\\$

Step 2; Multiply

$\rm \frac{9(5x+1)}{3y} \\\\ \frac{(15x+3)}{y} \\\\$

Hence the expression obtained after solving the given expression is, $\rm \frac{(15x+3)}{y}$

To learn more about the BODMAS rule, refer to brainly.com/question/23827397.

#SPJ1

4 0

2 years ago

Solve 2csc^2x-2cscx-1=0 <br> for 0° ≤ x ≤ 360°

umka21 [38]

Answer:

Step-by-step explanation:

2 csc²x-2 csc x-1=0

or

$\frac{2}{sin^2x} -\frac{2}{sin x} -1=0\\multiply~by~sin^2x\\2-2sin~x-sin^2x=0\\sin^2x+2sinx-2=0\\sin ~x=\frac{-2 \pm\sqrt{2^2-4*1*(-2)} }{2*1} \\=\frac{-2 \pm\sqrt{4+8} }{2} \\=\frac{-2 \pm\sqrt{4*3} }{2} \\=\frac{-2 \pm2\sqrt{3} }{2} \\=-1\pm\sqrt{3} \\|sin~x| \le ~1\\so~sin~x=-1+\sqrt{3} \\sin~x=\sqrt{3} -1\\x=sin^{-1}(\sqrt{3} -1) \approx 47.06^\circ,180-47.06\\or~x=47.06^\circ,132.94^\circ$

6 0

3 years ago

What equation would help determine the cost of 4 markers if five markers are $6.55 and remember I am asking for the equation not

harkovskaia [24]

Answer:

4($6.55)/5

Step-by-step explanation:

If five markers cost $6.55

then equate or cross multiply inorder to get the cost of 4 markers.

4($6.55)/5

7 0

3 years ago

Please help asap there both together

lana66690 [7]

Step 2 , supposed to be 4^-15 instead of 4^-2. Always multiply the exponents together.

Answer: Chris made a mistake in step 2 -C

4 0

3 years ago

Read 2 more answers

Let Z={a,c,{a,b}}. What is |Z|?

Lina20 [59]

$Z=\{a,c,\{a,b\}\}$

$\boxed{|Z|=3}$ (treat $\{a,b\}$ as one element of $Z$ )

The power set of $Z$ is

$\boxed{2^Z=\bigg\{\{\},\{a\},\{c\},\big\{\{a,b\}\big\},\{a,c\},\big\{a,\{a,b\}\big\},\big\{c,\{a,b\}\big\},\big\{a,c,\{a,b\}\big\}\bigg\}}$

1. $\{a,c\}\subseteq Z$ is true because both $a\in Z$ and $c\in Z$ .

2. $a\in Z$ is true.

3. $\{c\}\subseteq Z$ is true (same reason as part 1).

4. $\{c\}\in Z$ is false because $Z$ does not contain the set $\{c\}$ , rather just the element $c$ itself.

5. $b\in Z$ is false because the element $b$ on its own simply is not in $Z$ . That $b\in\{a,b\}$ does not mean $b\in Z$ , but that $b$ belongs to a subset of $Z$ .

6. $\{a,b\}\in Z$ is true.

6 0

4 years ago