T/s - b/s =1 solve for s

The owner of a bike shop sells unicycles and bicycles and keeps inventory by counting seats and wheels. On day, she counts 15 se

ycow [4]

Hey there Wcrystasia!!!!! The first thing that you need to realize is what slope-intercept form is. Remember that it is y=mx+b. I always say one variable on the left, while everything else is on the right. After you do that, you need to set up some equations. Seats Each cycle has 1 seat. You know there are 21 seats. So

u+b=21

Manipulate that equation into something that looks like slope-intercept form.

u+b=21

u+b−b=21−b

u=21−b

Now just rearrange the terms on the right side. You have an answer. If you needed the other equation, it would be set up like

u+2b=30

SO the answer is 30 !!!! HOPE I HELPED!!!!!!!!!!!

6 0

3 years ago

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Please help!! Find the center, vertices, and foci of the ellipse with equation

elena55 [62]

Answer:

center (0;0), vertices (0;10) and (0;-10), foci (0;6) and (0;-6)

Step-by-step explanation:

3 0

2 years ago

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F(x)=bx^2+32 For the function f defined above, b is a constant and f(2)=40. What is the value of f(-2)?

zhuklara [117]

Answer:

f(-2) = 0

Step-by-step explanation:

Given that:

f(x) = bx^2 + 32

f(2) = b(2)^2 + 32

f(2) = 4b + 32

f(2) = 4b = -32

f(2) = b = -32/4

f(2) = b = -8

Thus;

f(-2) = -8(-2)^2 + 32

f(-2) = -8(4) + 32

f(-2) = -32 + 32

f(-2) = 0

6 0

3 years ago

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Suppose y varies jointly as x and z. If y=15 when x=1/2 and z=6, what is the constant of variation?

Mashcka [7]

Answer:

y= 15

Step-by-step explanation:

Y= Kxz

y=6 x=1/2 z=6

15= K (1/2) (6)

15= K (3)

K= 5

SO

y= (5) (1/2) (6)

Y= 15

4 0

2 years ago

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Maggie graphed the image of a 90 counterclockwise rotation about vertex A of . Coordinates B and C of are (2, 6) and (4, 3) and

Lemur [1.5K]

Answer:

A(2,2)

Step-by-step explanation:

Let the vertex A has coordinates $(x_A,y_A)$

Vectors AB and AB' are perpendicular, then

$\overrightarrow {AB}=(2-x_A,6-y_A)\\ \\\overrightarrow {AB'}=(-2-x_A,2-y_A)\\ \\\overrightarrow {AB}\perp\overrightarrow {AB'}\Rightarrow \overrightarrow {AB}\cdot \overrightarrow {AB'}=0\Rightarrow (2-x_A)(-2-x_A)+(6-y_A)(2-y_A)=0$

Vectors AC and AC' are perpendicular, then

$\overrightarrow {AC}=(4-x_A,3-y_A)\\ \\\overrightarrow {AC'}=(1-x_A,4-y_A)\\ \\\overrightarrow {AC}\perp\overrightarrow {AC'}\Rightarrow \overrightarrow {AC}\cdot \overrightarrow {AC'}=0\Rightarrow (4-x_A)(1-x_A)+(3-y_A)(4-y_A)=0$

Now, solve the system of two equations:

$\left\{\begin{array}{l}(2-x_A)(-2-x_A)+(6-y_A)(2-y_A)=0\\ \\(4-x_A)(1-x_A)+(3-y_A)(4-y_A)=0\end{array}\right.\\ \\\left\{\begin{array}{l}-4-2x_A+2x_A+x_A^2+12-6y_A-2y_A+y^2_A=0\\ \\4-4x_A-x_A+x_A^2+12-3y_A-4y_A+y_A^2=0\end{array}\right.\\ \\\left\{\begin{array}{l}x_A^2+y_A^2-8y_A+8=0\\ \\x_A^2+y_A^2-5x_A-7y_A+16=0\end{array}\right.$

Subtract these two equations:

$5x_A-y_A-8=0\Rightarrow y_A=5x_A-8$

Substitute it into the first equation:

$x_A^2+(5x_A-8)^2-8(5x_A-8)+8=0\\ \\x_A^2+25x_A^2-80x_A+64-40x_A+64+8=0\\ \\26x_A^2-120x_A+136=0\\ \\13x_A^2-60x_A+68=0\\ \\D=(-60)^2-4\cdot 13\cdot 68=3600-3536=64\\ \\x_{A_{1,2}}=\dfrac{60\pm8}{2\cdot 13}=\dfrac{34}{13},2$

Then

$y_{A_{1,2}}=5\cdot \dfrac{34}{13}-8 \text{ or } 5\cdot 2-8\\ \\=\dfrac{66}{13}\text{ or } 2$

Rotation by 90° counterclockwise about A(2,2) gives image points B' and C' (see attached diagram)

8 0

3 years ago