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3 years ago

Yeah, uhm can someone help me with this?

Mathematics

1 answer:

hram777 [196]3 years ago

8 0

Answer:

-24

Step-by-step explanation:

$x/4 + 8 = 2\\x/4 = 2 - 8\\x/4 = -6\\x = -6 *4\\x = -24$

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urgent! i need an explanation please *picture included* also can someone make sure the second one is b or not?

Schach [20]

Answer:

For the first one it is D.

Step-by-step explanation:

I added 84.6 and 30.4 which gives me 115. I then added 66 to get 181, triangles angles only add up to 180 degrees, therefore they are not similar.

6 0

2 years ago

A fitness club charges $25 per month as well as a one-time $80 fee of all new members which equation represents the cost in doll

madam [21]

Answer: So if you want to find the total cost (c) for a certain number of months (m) and it costs $35 per month, you just need to multiply the number of months by the cost per month. So

35m=c

If you want to see the units cancel

($35/month)(months)=$

The months will cancel out, and you will be left with total dollars.

Not sure about the second part of your question, but if you wanted to figure out how much it would cost to be a member for 6 months, you would just plug in 6 for m.

35(6)=c

c=210

So $210

Step-by-step explanation:

4 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

3 years ago

Read 2 more answers

Find the product, using suitable properties:

Ray Of Light [21]

Answer: a) 6700 b) 2233

Step-by-step explanation:

$(-67)*(-99)+(-67)*(-1)=\\(-67)*((-99)+(-1))=\\(-67)*(-100)=\\6700$