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3 years ago

14

HELP PLEASE! it would help a ton!

1 answer:

Aneli [31]3 years ago

6 0

Answer:

so basically their is a x axis and a y axis go up on the rectangle by rotating the paper so the answer is D

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If you make me laugh you get points free points best joker will be marked brainliest

Mars2501 [29]

Answer:

Step-by-step explanation:

Show that x+x=1

5 0

3 years ago

Read 2 more answers

Help please......!! :(

Blababa [14]

6. X= -32 and 32
7. Y= -3
im not sure on the other questions and i LOVE doing math and im a sophmore in highschool so at least i could help get a couple of them

7 0

3 years ago

Read 2 more answers

lys-0071 [83]

Answer: x < 1

Step-by-step explanation:

Step 1: Add 4 to both sides.

-11x - 4 + 4 > -15 + 4

-11x > - 11

Step 2: Divide both sides by -11.

-11x/11 > -11/-11

x < 1

6 0

3 years ago

A wheel has 16 spokes. How many spaces are there between the spokes?

leva [86]

The wheel has 16 spokes.

There will be 16 spaces between the spokes.

See attachment. There are 8 spokes and 8 spaces between them.

Similarly, in the other, there are 16 spokes and 16 spaces.

you can see that the number of spaces between the spokes is equal to the number of spaces.

4 0

3 years ago

A design for party favors are in the shape of a regular hexagonal pyramid. The hexagon has a side length of 6 inches. The latera

Nookie1986 [14]

Answer:

Part 1) $LA=108\sqrt{2}\ in^2$

Part 2) $B=54\sqrt{3}\ in^2$

Step-by-step explanation:

Part 1) Find the lateral area of the pyramid

we know that

The lateral area of the pyramid, is equal to the area of six congruent isosceles triangles

so

$LA=6[\frac{1}{2}(b)(h)]$

we have

$b=6\ in$

Applying Pythagorean Theorem calculate the height of triangle

$9^2=(6/2)^2+h^2$

$h^2=81-9\\h^2=72\\h=\sqrt{72}\ in$

simplify

$h=6\sqrt{2}\ in$

Find the lateral area

$LA=6[\frac{1}{2}(6)(6\sqrt{2})]\\\\LA=108\sqrt{2}\ in^2$

Part 2) Find the base area

we know that

The area of a regular hexagon is equal to the area of six equilateral triangles

so

Applying the law of sines to calculate the area of triangle

$B=6[\frac{1}{2}b^2sin(60^o)]$

we have

$b=6\ in$

$sin(60^o)=\frac{\sqrt{3}}{2}$

substitute

$B=6[\frac{1}{2}6^2(\frac{\sqrt{3}}{2})]$

$B=54\sqrt{3}\ in^2$

3 0

4 years ago