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koban [17]
3 years ago
12

Movement of our arms and legs is possible when muscles pull on our _________________________________.

Mathematics
1 answer:
cricket20 [7]3 years ago
5 0

Answer:

Isn't this supposed to be a science question

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Can u pls help me with this question ​
Tresset [83]
The answer is a b c e
3 0
3 years ago
What is the area of the largest rectangle that can be inscribed in an isosceles triangle with side lengths $8$, $\sqrt{80}$, and
e-lub [12.9K]
First let's solve the general case for an isosceles triangle of base 2 and height k. If the triangle is drawn so the base is on the x-axis and the apex is on the y-axis, the equation for the line containing the right side is
  y = -k(x-1)
Then a rectangle with x-dimension w will have an area that is the product of this width and the height y = -k((w/2)-1).
  area = w(-k(w/2 -1)) = (-k/2)w² +kw
The derivative of area with respect to w will be zero where the area is a maximum:
  d(area)/dw = 0 = -kw +k
  w = 1 . . . . . . add kw and divide by k
Using the formula for area, we find
  area = 1(-k(1/2-1)) = k/2
This value is 1/2 the total area of the triangle we started with.

If the side lengths of your triangle are 8, √80, and √80, the height will be given by the Pythagorean theorem as
  h = √((√80)² - (1/2·8)²) = √(80-16) = 8
Your triangle's area will be
  triangle area = (1/2)(8)(8) = 32

The maximum area of the inscribed rectangle will be half this value,
  16 square units

4 0
3 years ago
How do i solve this problem? how did they get 2.5?
WARRIOR [948]
Take the GCF (-8t) of the given expression.

You'll get -8t(2t-5).

Now, set each equal to zero and solve for x.

-8t = 0 2t-5=0

t=0 t = 5/2 (Or 2.5 seconds)
6 0
3 years ago
If a pyramid had a base of area 4 and height of 6, what is the volume
lilavasa [31]
24 because 4x6 is 24
4 0
3 years ago
Read 2 more answers
I need help please with this 1 problem
Lyrx [107]

Answer:

See below, please

Step-by-step explanation:

We know

f(x) =  {x}^{2}  + 3x - 7

g(x) = 3x + 5

h(x) = 2 {x}^{2}  - 4

Hence

f(g(x))

=  {(3x + 5)}^{2}  + 3 \times (3x + 5) - 7

= (9 {x}^{2}  + 30x + 25) + (9x + 15) - 7

= 9 {x}^{2}  + 39x + 33

h(g(x))

= 2 \times  {(3x + 5)}^{2}  - 4

2 \times (9 {x}^{2}  + 30x + 25) - 4

= 18 {x}^{2}  + 60x + 50 - 4

= 18 {x}^{2}  + 60x + 46

(h - f)(x)

= h(x) - f(x)

= (2 {x}^{2}  - 4) - ( {x}^{2}  + 3x - 7)

=  {x}^{2}  - 3x + 3

(f + g)(x)

= f(x) + g(x)

= ( {x}^{2}  + 3x - 7) + (3x + 5)

=  {x}^{2}  + 6x - 2

5 0
2 years ago
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