All categories

3 years ago

I WILL MARK BRAINLIEST :) Determine which segment is shortest in the diagram.

Mathematics

2 answers:

erastovalidia [21]3 years ago

8 0

Answer:

Step-by-step explanation:

Nana76 [90]3 years ago

7 0

Answer:

C. Segment AC

Step-by-step explanation:

I am going to figure all the angles first. Triangles always add up to 180 degrees so I am going to use that.

50 + 49 = 99

180 - 99 = 81

<C = 81°

99 + 37 = 136

180 - 136 = 44

<A = 44°

The shortest segment is always going to be opposite the smallest angle. The smallest angle is 37°. Therefore, Segment AC is the shortest.

You might be interested in

Chan jogs 10 miles a week. How many miles will she jog in 52 weeks?

igor_vitrenko [27]

520 miles because 52x10=520

4 0

3 years ago

Read 2 more answers

Someone help please

ratelena [41]

Answer:

Step-by-step explanation:

HAV A GREAT DAY hope this helps

4 0

3 years ago

Help fast!!! Give me the real solution pls. First correct awnser get brainliest!

Over [174]

Answer:

She subtracted the two 10x's where instead you're supposed to add them together.

Step-by-step explanation:

3 0

3 years ago

Have a wonderful day lvu<3 free pts...... also if you go for biden plz dont come in saying trump sucks or sum like that ive b

Morgarella [4.7K]

✽ - - - - - - - - - - - - - - - ~Hello There!~ - - - - - - - - - - - - - - - ✽

➷ $455+446= 901$

✽

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

↬ May ♡

4 0

3 years ago

A closed, rectangular-faced box with a square base is to be constructed using only 36 m2 of material. What should the height h a

kkurt [141]

Answer:

$b=h=\sqrt{6}$ m

Step-by-step explanation:

Let

Bas length of box=b

Height of box=h

Material used in constructing of box=36 square m

We have to find the height h and base length b of the box to maximize the volume of box.

Surface area of box= $2b^2+4bh$

$2b^2+4bh=36$

$b^2+2bh=18$

$2bh=18-b^2$

$h=\frac{18-b^2}{2b}$

Volume of box, V= $b^2h$

Substitute the values

$V=b^2\times \frac{18-b^2}{2b}$

$V=\frac{1}{2}(18b-b^3)$

Differentiate w. r.t b

$\frac{dV}{db}=\frac{1}{2}(18-3b^2)$

$\frac{dV}{db}=0$

$\frac{1}{2}(18-3b^2)=0$

$\implies 18-3b^2=0$

$\implies 3b^2=18$

$b^2=6$

$b=\pm \sqrt{6}$

$b=\sqrt{6}$

The negative value of b is not possible because length cannot be negative.

Again differentiate w.r.t b

$\frac{d^2V}{db^2}=-3b$

At $b=\sqrt{6}$

$\frac{d^2V}{db^2}=-3\sqrt{6}$

Hence, the volume of box is maximum at $b=\sqrt{6}$ .

$h=\frac{18-(\sqrt{6})^2}{2\sqrt{6}}$

$h=\frac{18-6}{2\sqrt{6}}$

$h=\frac{12}{2\sqrt{6}}$

$h=\sqrt{6}$

$b=h=\sqrt{6}$ m

7 0

3 years ago