Question

Mathematics

2 answers:

icang [17]1 year ago

6 0

Answer:

$18.00

Step-by-step explanation:

1.5 times 10 + 3

15 + 3

Mariana [72]1 year ago

4 0

Answer:

It would cost $18 to ride the taxi 10 miles.

You might be interested in

There are an infinite number of vector and parametric equations for a given plane. Why is the scalar equation of a given plane u

Sonbull [250]

Answer:

The answer is below

Step-by-step explanation:

When you refer to a normal vector you mean the form a*x + b*y + c*z = d, if that's the case then it's not unique in the nose because it gives you its normal vector. Taking into account that uniqueness only supports multiplicative constants, which means that you can multiply the equation with whatever you want, that is, it remains the same

5 0

2 years ago

Match the parabolas represented by the equations with their vertices. y = x2 + 6x + 8 y = 2x2 + 16x + 28 y = -x2 + 5x + 14 y = -

GaryK [48]

Consider all parabolas:

$y = x^2 + 6x + 8,\\y=x^2+6x+9-9+8,\\y=(x^2+6x+9)-1,\\y=(x+3)^2-1.$

When x=-3, y=-1, then the point (-3,-1) is vertex of this first parabola.

$y = 2x^2 + 16x + 28=2(x^2+8x+14),\\y=2(x^2+8x+16-16+14),\\y=2((x^2+8x+16)-16+14),\\y=2((x+4)^2-2)=2(x+4)^2-4.$

When x=-4, y=-4, then the point (-4,-4) is vertex of this second parabola.

$y =-x^2 + 5x + 14=-(x^2-5x-14),\\y=-(x^2-5x+\dfrac{25}{4}-\dfrac{25}{4}-14),\\y=-((x^2-5x+\dfrac{25}{4})-\dfrac{25}{4}-14),\\y=-((x-\dfrac{5}{2})^2-\dfrac{81}{4})=-(x-\dfrac{5}{2})^2+\dfrac{81}{4}.$

When x=2.5, y=20.25, then the point (2.5,20.25) is vertex of this third parabola.

$y =-x^2 + 7x + 7=-(x^2-7x-7),\\y=-(x^2-7x+\dfrac{49}{4}-\dfrac{49}{4}-7),\\y=-((x^2-7x+\dfrac{49}{4})-\dfrac{49}{4}-7),\\y=-((x-\dfrac{7}{2})^2-\dfrac{77}{4})=-(x-\dfrac{7}{2})^2+\dfrac{77}{4}.$

When x=3.5, y=19.25, then the point (3.5,19.25) is vertex of this fourth parabola.

$y =2x^2 + 7x +5=2(x^2+\dfrac{7}{2}x+\dfrac{5}{2}),\\y=2(x^2+\dfrac{7}{2}x+\dfrac{49}{16}-\dfrac{49}{16}+\dfrac{5}{2}),\\y=2((x^2+\dfrac{7}{2}x+\dfrac{49}{16})-\dfrac{49}{16}+\dfrac{5}{2}),\\y=2((x+\dfrac{7}{4})^2-\dfrac{9}{16})=2(x+\dfrac{7}{4})^2-\dfrac{9}{8}.$

When x=-1.75, y=-1.125, then the point (-1.75,-1.125) is vertex of this fifth parabola.

$y =-2x^2 + 8x +5=-2(x^2-4x-\dfrac{5}{2}),\\y=-2(x^2-4x+4-4-\dfrac{5}{2}),\\y=-2((x^2-4x+4)-4-\dfrac{5}{2}),\\y=-2((x-2)^2-\dfrac{13}{2})=-2(x-2)^2+13.$