In inches, Micah is 5 inches shorter than twice the height of Jasmine, who is 32 inches tall. How tall is Micah in inches?

2 answers:

5 0

Micah is 5 inches shorter than twice the height of Jasmine

Micah = 5 inches shorter than 2 x Jasmine

Micah = 2 x Jasmine - 5 <em>(jasmine is 32 inches tall. Replace jasmine with 32)
</em>
Micah = 2 x 32 - 5

Micah = 64 - 5

Micah = 59

Micah is 59 inches tall

stealth61 [152]3 years ago

4 0

Answer: Micah is 59 inches tall.

Step-by-step explanation:

Given : The height of Jasmine = 32 inches

Also, Micah is 5 inches shorter than twice the height of Jasmine.

Therefore, the height of Micah = $2(32)-5$ inches

i.e. the height of Micah $=64-5=59$ inches

i.e. the height of Micah $=59$

Theerfore , the Micah is 59 inches tall.

You might be interested in

The tree in milan's backyard is 8.4m high. how high is it in centimeters? be sure to include the correct unit in your answer.

Alex787 [66]

840 centimeters ...there 100 centimeters in one meter

5 0

2 years ago

a customer went to a garden shop and bought some potting soil for $11.50 and 9 shrubs. the total bill was $94.75. write and solv

Mrac [35]

To answer this problem, let p be the price of each scrub. The total bill (t) is the sum of the price of the potting soil (s) and of the 9 scrubs (n). This is depicted in the equation,
$11.50 + 9p = $94.75
The value of p from the equation is 9.25. Thus, each scrub costs $9.25. The answer is letter C.

8 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 $Z$ .