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Irina-Kira [14]

3 years ago

5

A cedar chest is a shaped like a rectangular prism. The cedar chest is 2 feet tall, 4 feet wide, and 2 feet deep What is the sur

face area of the cedar chest? Enter your answer in the box

1 answer:

Nastasia [14]3 years ago

5 0

Answer:

40 ft^2

Step-by-step explanation:

lateral surface area -Slat = 24 ft^2

top surface area-Stop = 8 ft^2

bottom surface area-Sbot = ft m^2

24+8+8= 40

total surface area-Stot = 40 ft^2

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Pani-rosa [81]

Answer:

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Step-by-step explanation:

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3 years ago

A glass marble has a radius of 1 cm.How much glass is in the marble?

Alika [10]

I think if you just ask google he will give you the answer

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AT&T charges $76.00 for 500 minutes or Sprint charges $54.00 or 450 minutes

zlopas [31]

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3 years ago

A) how many men have a waist measurement of more than 85 cm

KATRIN_1 [288]

Answer:

(a)29

(b)91cm

(c)13cm

Step-by-step explanation:

(a)From the graph, 11 men have a waist measurement of 85cm and below.

Since the cumulative frequency is 40

Number of men who have a waist measurement of more than 85 cm

$=40-11\\=29$

(b)Median

$\text{Median}=\dfrac{N}{2}\\N=40\\Median=\dfrac{40}{2}=20th$ item$

Tracing 20 from the y-axis to the x-axis, the median waist measurement is 91cm.

(c)Interquartile Range

Interquartile Range = $Q_3-Q_1$

$Q_1=\dfrac{N}{4}=\dfrac{40}{4}=10th$ item$

From the graph, at y=10, x=84 cm

$Q_3=\dfrac{3N}{4}=\dfrac{3*40}{4}=30th$ item$

From the graph, at y=30, x=97 cm

Therefore:

Interquartile Range=97-84=13cm

5 0

3 years ago

The greatest common factor (GCF) of x^3, x^7, x^9 is ?<br> Please help

solong [7]

Answer:

$x^3$

Step-by-step explanation:

Let the greatest common factor of $a,b,c$ such that $a,b,c \in\mathbb{Z}$ and they are not all equal to zero, $d$ is the common divisor of $a$ and $b$ . Therefore, $d \mid a$ and $d \mid b$

You can write

$D(a) = \{d \in\mathbb{Z} : d \mid a\}$

$D(b) = \{d \in\mathbb{Z} : d \mid b\}$

The greatest common factor of $a,b$ is given as

$D(a,b) = \{d \in\mathbb{Z} : d \mid a \text{ and } d \mid b\}$

and

$D(a, b) = D(a) \cap D(b)$

This happens because $D(a,b)$ is upper bounded because if $a\neq 0$ then $d \leq |a|$

for all $d\in D(a, b)$ . Therefore, the set $D(a, b)$ has the greatest elements.

Taking $x^3, x^7, x^9$ such that $x>0$

You can note that $x^7 = x^3 \cdot x^4$ and $x^9 = x^3 \cdot x^6 = x^3 \cdot x^3 \cdot x^3$

Therefore, the greatest common factor is $x^3$

Note: $x^3 \cap x^7 \cap x^9 = x^3$

6 0

3 years ago