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3 years ago

I need the first and second one, please anyone

Mathematics

1 answer:

tia_tia [17]3 years ago

5 0

Answer:

f(-1) = 0

f(2) = 16

Step-by-step explanation:

f(-1) = 4(-1) + 4 = 0

f(2) = 4(2) + 8 = 16

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la pyramide du Louvre cest une pyramide a base carrée de 35,4m de large et de 21,6m de haut calculer l'inclinaison dune face par

MrRissso [65]

Answer:

c'est une pyramide régulière à base carrée de35.4 m de large et de 21.6 m de haut. calculer l'inclinaison d'une face par rapport à l'horizontale. ... dans le triangle rectangle SHA: tg SAH = 21,6/17,7 donc l'angle SAH = 50,66° ...

Step-by-step explanation:

7 0

3 years ago

Can someone help please 3x-y=13 A. (6,5) B.(3,-4) C. (6,5) and (3,-4) D. neither

gavmur [86]

Answer:

C. (6, 5) and (3, -4)

Step-by-step explanation:

Given the equation 3x - y = 13, we need to figure out which points satisfy it. In order for an ordered pair to satisfy an equation, when we plug the x-coordinate in for x and the y-coordinate in for y, the equation should hold true.

Let's try with (6, 5):

3x - y = 13

3 * 6 - 5 =? 13

18 - 5 =? 13

13 = 13

Since this is true, we know that (6, 5) is indeed a solution.

Now let's try with (3, -4):

3x - y = 13

3 * (3) - (-4) =? 13

9 + 4 =?13

13 = 13

Again, since this is true, then (3, -4) must be a solution.

Thus, the answer is C.

<em>~ an aesthetics lover</em>

6 0

3 years ago

Read 2 more answers

Number 2 rewrite as a single power

Paul [167]

Answer:

-625

Step-by-step explanation:

I think so..

4 0

3 years ago

meriva

Answer:

67.38

Step-by-step explanation:

SinX=opp/hyp

Angles then arcsin

7 0

3 years ago

A wire b units long is cut into two pieces. One piece is bent into an equilateral triangle and the other is bent into a circle.

mezya [45]

1. Divide wire b in parts x and b-x.

2. Bend the b-x piece to form a triangle with side (b-x)/3

There are many ways to find the area of the equilateral triangle. One is by the formula A= $\frac{1}{2}sin60^{o}side*side= \frac{1}{2} \frac{ \sqrt{3} }{2} (\frac{b-x}{3}) ^{2}= \frac{ \sqrt{3} }{36}(b-x)^{2}$
A= $\frac{ \sqrt{3} }{36}(b-x)^{2}=\frac{ \sqrt{3} }{36}( b^{2}-2bx+ x^{2} )=\frac{ \sqrt{3} }{36}b^{2}-\frac{ \sqrt{3} }{18}bx+ \frac{ \sqrt{3} }{36}x^{2}$

Another way is apply the formula A=1/2*base*altitude,
where the altitude can be found by applying the pythagorean theorem on the triangle with hypothenuse (b-x)/3 and side (b-x)/6

3. Let x be the circumference of the circle.

$2 \pi r=x$

so $r= \frac{x}{2 \pi }$

Area of circle = $\pi r^{2}= \pi ( \frac{x}{2 \pi } )^{2} = \frac{ \pi }{ 4 \pi ^{2} }* x^{2} = \frac{1}{4 \pi } x^{2}$

4. Let f(x)= $\frac{ \sqrt{3} }{36}b^{2}-\frac{ \sqrt{3} }{18}bx+ \frac{ \sqrt{3} }{36}x^{2}+\frac{1}{4 \pi } x^{2}$

be the function of the sum of the areas of the triangle and circle.

5. f(x) is a minimum means f'(x)=0

f'(x)= $\frac{ -\sqrt{3} }{18}b+ \frac{ \sqrt{3} }{18}x+\frac{1}{2 \pi } x$ =0

$\frac{ -\sqrt{3} }{18}b+ \frac{ \sqrt{3} }{18}x+\frac{1}{2 \pi } x=0$

$(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) x=\frac{ \sqrt{3} }{18}b$

$x= \frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$

6. So one part is $\frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$ and the other part is b- $\frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$

4 0

4 years ago

Read 2 more answers