All categories

2 years ago

the diagram shows a 5cm x 5cm x 5cm cube calculate the length of the diagonal AB give your answer correct to 1 decimal place

Mathematics

1 answer:

Lynna [10]2 years ago

4 0

Answer:

√3 * 5 = 5√3 cm

Step-by-step explanation:

→ ABCDEFGH is a cube.

→ CF = Diagonal of cube.

→ CH = Diagonal of Base Face BCDH.

→ Let the side of Each cube = a.

Than,

in Right ∆CFH, By Pythagoras Theoram, we have,

→ CH² + FH² = CF² --------- Equation (1)

and, Similarly, in Right ∆CDH ,

→ CD² + DH² = CH² ------- Equation (2).

Putting Value of Equation (2) in Equation (1), we get,

→ (CD² + DH²) + FH² = CF²

→ a² + a² + a² = CF²

→ CF² = 3a²

→ CF = √3a .

Hence, we can say That Diagonal of a cube is √3 times of its sides.

__________________

Given:-

Side of cube = 5cm.

So,

→ Diagonal of cube = √3 * 5 = 5√3 cm. (Ans.)

You might be interested in

vegetable ghee is stored in the cylindrical vessel of internal radius 1.4m height 1.5m. if it is transferred into the rectangula

irina1246 [14]

Answer:

<em>5,598 cans are required to empty the vessel</em>

Step-by-step explanation:

The volume of a cylinder of radius r and height h is:

$V = \pi r^2h$

The volume of a box of dimensions X, Y, and Z is:

V=X.Y.Z

A cylinder of r=1.4 m and height h=1.5 is used to store vegetable ghee. It contains a volume of:

$V_1 = \pi 1.4^2*1.5$

$V_1=9.236\ m^3$

Converting to cubic cm:

$V_1=9.236*1,000,000\ cm^3$

$V_1=9,236,282\ cm^3$

The volume of each rectangular tin can is:

$V_2=33*10*5=1,650\ cm^3$

$V_2=1,650\ cm^3$

The number of cans required to empty the vessel is:

$V_1/V_2=9,236,282/1,650 = 5,598$

5,598 cans are required to empty the vessel

5 0

2 years ago

Suppose babies born in a large hospital have a mean weight of 4095 grams, and a standard deviation of 569 grams. If 130 babies a

Butoxors [25]

Answer:

$P =0.3998$

Step-by-step explanation:

Let ${\overline{x}}$ be the average of the sample, and the population mean will be $\mu$

We know that:

$\mu = 4095$ gr

Let $\sigma$ be the standard deviation and n the sample size, then we know that the standard error of the sample is:

$E=\frac{\sigma}{\sqrt{n}}$

Where

$\sigma=569$

$n=130$

In this case we are looking for:

$P(|{\displaystyle{\overline{x}}}- \mu|>42)$

This is:

${\displaystyle{\overline{x}}}- \mu>42$ or ${\displaystyle{\overline{x}}}- \mu$

$P=P({\displaystyle{\overline{x}}}- \mu>42)+ P({\displaystyle{\overline{x}}}- \mu$

Now we get the z score

$Z=\frac{{\displaystyle{\overline{x}}}-\mu}{\frac{\sigma}{\sqrt{n}}}$

$P=P(z>\frac{42}{\frac{569}{\sqrt{130}}}) + P(z$

$P=P(z>0.8416) + P(z$

Looking at the tables for the standard nominal distribution we get

$P =0.1999+0.1999$

$P =0.3998$

6 0

2 years ago

How many extraneous solutions does the equation below have? StartFraction 2 m Over 2 m + 3 EndFraction minus StartFraction 2 m O

xeze [42]

Answer:

Step-by-step explanation:

We have the fraction $\frac{2m}{2m+3} -\frac{2m}{2m-3}$

Step 1. Use LCM of the fraction, (2m+3)(2m-3), to simplify the fraction:

$\frac{(2m-3)(2m)-(2m+3)(2m)}{(2m+3)(2m-3)} =\frac{2m^2-6m-[2m^2+6m]}{(2m+3)(2m-3)} =\frac{2m^2-6m-2m^2-6m}{(2m+3)(2m-3)} =\frac{-12m}{(2m+3)(2m-3)}$

Step 2. Equate the resulting fraction to zero and solve for $m$ :

$\frac{-12m}{(2m+3)(2m-3)} =0$

$-12m=0[(2m+3)(2m-3)]$

$-12m=0$

$m=\frac{0}{-12}$

$m=0$

Step 3. Replace the value in the original equation and check if it holds:

$\frac{-12m}{(2m+3)(2m-3)} =0$

$-12m=0$

Since $m=0$ ,

$-12(0)=0$

$0=0$

Since the only solution of the equation holds, the equation bellow doesn't have any extraneous solution

5 0

2 years ago

Read 2 more answers

PLEASE HELP!

Bond [772]

Let x be the number of Standard Specials sold last Monday and y be the number of Deluxe specials sold last Monday.

The Standard special sells for $2, then x Standard Specials cost $2x.

The Deluxe special sells for $4.50, then y Deluxe specials cost $4.50y.

1. Last Monday, PB&P sold at least $200 worth of Standard and Deluxe peanut butter and pickle sandwich specials. This means that

2x+4.50y≥200.

2. When all related business expenses are included, the Standard special costs $0.50 to prepare, then x Standard Specials cost $0.50x.

When all related business expenses are included, the Deluxe special costs $1.25 to prepare, then y Deluxe specials cost $1.25y.

Expenses were no more than $100, then

0.50x+1.25y≤100.

3. At least 30 Standard special were sold, then x≥30.

4. Graph all these inequalities (see attached diagram). As you can see only point (60,70) doesn't belong to needed region.

Answer: all points except point (60,70)

8 0

2 years ago

Read 2 more answers

Mrs. Melanie Lazo pays Php 15.75 for each hair clip to its supplier. She decided to adds markup of Php 3.15. What is the markup

malfutka [58]

Answer:

the answer to the problem is 20.8%

8 0

2 years ago