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3 years ago

Solve for c a=(b - c)/c

1 answer:

6 0

(b-c)/c can be separated into b/c-c/c, which is the same as b/c - 1.
so
a= b/c -1
add one to both sides of equations
a+1 = b/c -1+1
1 cancels out on the right side leaving
a+1 = b/c
multiply both sides by c
c ×(a+1)= (b/c)×c
c cancels out on the right side making
c×(a+1) = b
divide both sides by a+1
c =b/(a+1)

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Billy likes to go cycling.

Agata [3.3K]

Answer:

560 metres

Step-by-step explanation:

80 cm in metres is 0.8 metres.

After 1 revolution any point on the wheel will be back to the original position before moving. 0.8 times 2, which gives us 1.6 metres. In 1 revolution the bike travels 1.6 metres.

Now 1.6 metres times 350 gives us 560.

So the answer is 560 metres.

4 0

2 years ago

For the function f(x)=8(x+10), find f^-1(x)

NeTakaya

In other words, find the inverse of the given function f(x).

1. Replace f(x) with "y." y = 8(x+10)
2. Interchange x and y: x = 8(y+10)
3. Solve this for y: y+10 = x/8, or y = x/8 - 10

-1
4. Replace "y" with "f (x) "

-1 x-80
f (x) = ---------- (Answer)
8

6 0

3 years ago

The coordinates of the endpoints of AB and CD are A(2,

Ratling [72]

Answer:

Option 1: CD is a perpendicular bisector of AB

Step-by-step explanation:

Let us find out the slopes of various line segments and the Distances and then we will draw the conclusions accordingly.

Formula to find slope

$m= \frac{y_2-y_1}{x_2-x_1}$

Formula to Find Distance between two points

$D=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}$

mAB ( represents , Slope of AB )

1. $mAC= \frac{3-2}{2-5}=\frac{1}{-3}=-\frac{1}{3}$

2. $mBC=\frac{2-1}{5-8}=\frac{1}{-3}=-\frac{1}{3}$

3. $mCD=\frac{5-2}{6-5}=\frac{3}{1}=3$

4. $AC=\sqrt{(3-2)^2+(2-5)^2} =\sqrt{(1)^2+(-3)^2}=\sqrt{1+9}=\sqrt{10}$

5. $BC=\sqrt{(2-1)^2+(5-8)^2} =\sqrt{(1)^2+(-3)^2}=\sqrt{1+9}=\sqrt{10}$

mAC = mBC , and C is common point , hence these three are collinear points making a straight line whole slope is $-\frac{1}{3}$

$mAB=-\frac{1}{3}$

$mCD=3$

$mAB \times mCD = -\frac{1}{3} \times 3 = -1$

Hence CD ⊥ AB

Also

From Point 4 and point 5 above , we see that

AC = CB

Hence CD bisect AB at C, also CD ⊥ AB

There fore

CD is a perpendicular bisector of AB

Therefor option 1 is true

4 0

3 years ago

At what point does a line with a slope of 34 and a y-intercept of -5 intersect a line with a slope of −14 and a y-intercept of 3

maw [93]

Set them into slope-intercept forms of equations then set those equations equal to each other and solve for x and y. y = 34x - 5 and y = -14x+3. Set those equal to each other, solve for x to get x = 1/6. Sub in that x value to get y = 2/3. The point where they intersect is the solution of the system; in other words, where on both lines we have the exact same (x, y) value.

4 0

3 years ago

Right triangles can be proven to be congruent by knowing that the hypotenuse and leg of one are congruent to the hypotenuse and

Slav-nsk [51]

The answer to the question is true

8 0

3 years ago

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