All categories

3 years ago

Look at the picture below // 99 points // PLEASEE HELPP

Mathematics

2 answers:

Katena32 [7]3 years ago

7 0

It would be D right? Because they are obviously congruent, parallel and the same length. Just an educated guess..

steposvetlana [31]3 years ago

7 0

Is given that the figure on the problem is a square, and we know that the square's sides are congruent or have the same length.

•The correct answer is letter D, Yes, the opposite sides are parallel, and all sides are the same length.

You might be interested in

Please answer these<br> i need this

Len [333]

Answer:

1. no

2. 8

Step-by-step explanation:

1. 20 does not equal 4/5

2. 6x=48 divide 6 on both sides

x=8

5 0

2 years ago

Read 2 more answers

Help i will give brainlist

Arisa [49]

Answer:

3. 2/10 4/20 3/15

4. 12/1 48/4 36/3

Step-by-step explanation:

3 0

3 years ago

WILL GIVE BRAINLIEST GIVE EXPLANATION

alex41 [277]

Answer:

5z+30

Step-by-step explanation:

5(z+6)

^ multiply each term in the parentheses by 5

5z+5x6

then,

5z+5x6 (multiply the numbers)

the answer will be 5z+30

5 0

2 years ago

if on a scale drawing 35 feet are represented by 25 inches, then a scale of 1/8 inch represents how many feet?

kirill115 [55]

Answer:

7/40 feet

Step-by-step explanation:

Solve using proportions:

$\frac{25in}{35ft} =\frac{1/8 in}{?ft}$

Find the scale factor by dividing the numerators 25 by 1/8.

25 ÷ 1/8 = 200 This is the common ratio

To get from the top left number to the top right number, you divide 25 by 200.

To get from the bottom left number to the bottom right number, you divide 35 by 200.

35 ÷ 200 = 0.175

1/8 of an inch represents 0.175 feet.

To find 0.175 in fraction form, reduce 35/200 to its lowest terms.

35/200 = 7/40

1/8 of an inch represents 7/40 feet.

8 0

3 years ago

Derive this with respect to x<br><img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B3%7D%7B%281%20%2B%20%20secx%29%20%7D%20" id="Tex

Elenna [48]

$~~~~\dfrac{d}{dx} \left(\dfrac{3}{1 + \sec x} \right)\\\\\\=3 \dfrac{d}{dx} \left( \dfrac 1{ 1+ \sec x} \right)\\\\\\=3 \dfrac{d}{dx} (1+ \sec x)^{-1}\\\\\\=3 (-1) (1 + \sec x )^{-1 -1} \dfrac{d}{dx}( 1 + \sec x)\\\\\\=-3(1 + \sec x)^{-2} ( 0 + \sec x \tan x)\\\\\\=-\dfrac{3\sec x \tan x}{(1 + \sec x)^2}$

8 0

1 year ago