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gizmo_the_mogwai [7]

2 years ago

9

4.) Determine the missing side of a rectangle with an area of 480 cm² and a length of 32 cm. Please show your work.

1 answer:

asambeis [7]2 years ago

8 0

Answer:

w = 15 cm

Step-by-step explanation:

We want to find the width, as the area of a rectangle is length (l) * width (w).

Since we know the area and the length, we have 480 = 32w

Isolating w by dividing both sides by 32 gives us w = 15 cm

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Mrs. Kim is teaching a 5th grade class. She is standing 48 feet in front of Lester. Carey is sitting to Lester's right. If Carey

victus00 [196]

Lester and Carey are 14 feet apart

Step-by-step explanation:

Mrs. Kim is teaching a 5th grade class

She is standing 48 feet in front of Lester
Carey is sitting to Lester's right
Carey and Mrs. Kim are 50 feet apart

We need to find how far apart Lester and Carey

Look to the attached figure

The positions of Mrs. Kim, Lester, and Carey formed a right triangle

∵ The distance between Mrs. Kim and Carey is 50 feet

∴ The length of the hypotenuse of the Δ = 50 feet

∵ The distance between Mrs. Kim and Lester is 48 feet

∴ The length of one leg of the Δ = 48 feet

The formula of length of the sides of the right Δ is:

$(hypotenuse)^{2}=(leg_{1})^{2}+(leg_{2})^{2}$

Substitute the length of the hypotenuse and the length of the given leg in the formula above

∵ (50)² = (48)² + (leg)²

∴ 2500 = 2304 + (leg)²

- Subtract 2304 from both sides

∴ 196 = (leg)²

- Take square for both sides

∴ 14 = leg

∵ This leg represents the distance between Lester and Carey

∴ The distance between Lester and Carey = 14 feet

Lester and Carey are 14 feet apart

Learn more:

You can learn more about the right triangle in brainly.com/question/1238144

#LearnwithBrainly

8 0

3 years ago

Write a solution in Interval Notation - (you don't have to help me on all, 1 or 2 is fine c: )

Gemiola [76]

QUESTION 1

The given inequality is

$|m|-2>0$

We group like terms to get,

$|m|>2$

This implies that,

$-m>2$ or $m>2$ .

We simplify the inequality to get,

$m$ or $m>2$ .

We can write this interval notation to get,

$(-\infty,-2)\cup (2,+\infty)$ .

QUESTION 2

$|x-4|-3\:>\:5$ .

We group like terms to get,

$|x-4|\:>\:5+3$ .

$|x-4|\:>\:8$

We split the absolute value sign to get,

$-(x-4)\:>\:8$ or $x-4\:>\:8$

This implies that,

$x-4\:$ or $x-4\:>\:8$

$x\:$ or $x\:>\:8+4$

$x\:$ or $x\:>\:12$

We can write this interval notation to get,

$(-\infty,-4)\cup (12,+\infty)$ .

QUESTION 3

The given inequality is

$|6+9x|\leq 24$

We split the absolute value sign to obtain,

$-(6+9x)\leq 24$ or $(6+9x)\leq 24$

This simplifies to

$6+9x\ge -24$ and $6+9x\leq 24$

$9x\ge -24-6$ and $9x\leq 24-6$

$9x\ge -30$ and $9x\leq 18$

$x\ge -\frac{10}{3}$ and $x\leq 2$

$-\frac{10}{3}\leq x\leq2$

We write this in interval form to get,

$[-\frac{10}{3},2]$

QUESTION 4

The given inequality is

$|1-5a|>29$

We split the absolute value sign to get,

$-(1-5a)>29$ or $1-5a>29$

This simplifies to,

$1-5a\:$ or $1-5a\:>\:29$

This implies that,

$-5a\:$ or $-5a\:>\:29-1$

$-5a\:$ or $-5a\:>\:28$

$a\:>\:6$ or $a\:$

We write this in interval notation to get,

$(-\infty,-\frac{28}{5})\cup (6,+\infty)$

7 0

3 years ago

What is the solution set for 5x>15?

denis23 [38]

Answer= X>3

5x>15

Divide both sides by 5x

$\frac{5 x}{5x} > \frac{15}{5x}$

5x will cancel with 15. You get 3

So, x>3

6 0

3 years ago

Solve for the indicated variable r-6m=n for "r"

olchik [2.2K]

R - 6m = n.....add 6m to both sides
r = n + 6m <==

7 0

3 years ago

How to do?? <br> find for x<br> 10 points

laila [671]

Check the picture below.

5 0

3 years ago