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2 years ago

I NEED HELP PLEASE! If you can help me with this you amazing! Thank you

Mathematics

1 answer:

krek1111 [17]2 years ago

6 0

Answer:

$a=-3$

Step-by-step explanation:

we have

$a-2=\frac{3+6a}{3}$

Solve for a

That means ----> Isolate the variable a

Factor 3 in the numerator right side

$a-2=\frac{3(1+2a)}{3}$

Simplify right side

$a-2=1+2a$

subtract a both sides

$a-2-a=1+2a-a$

$-2=1+a$

subtract 1 both sides

$-2-1=1+a-1$

$-3=a$

Rewrite

$a=-3$

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Please help :) Is the relationship shown by the data linear? If so model the data with an equation.

MrRissso [65]

Answer:

y-5=5/4(x-1)

Step-by-step explanation:

the slope is 1.25

m=y2-y1/x2-x1=10-5/5-1=5/4=1.25

y-5=(5)/(4)(x-1)=

y=(5)/(4)x+(15)/(4) this one is linear

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2 years ago

Solve this system of equations.<br> y = 2x – 12<br> y = -x + 3

3241004551 [841]

X = 5, y = -2
-x +3 = 2x - 12
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2 years ago

The zoo needed to raise 25000.zoo officials planned a dinner/dance event and charged $300 per couple. What is the fewest number

garri49 [273]

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3 years ago

Read 2 more answers

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

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3 years ago

Read 2 more answers

The diameter of a circle is 7 cm. Find the circumference to the nearest tenth

Anton [14]

Answer:38.5 cm

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1 year ago