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2 years ago

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Fill in the blanks with the answer to each question based on information provided.Two toy robots are having a race. Robot A can

go 14 meter in 15 seconds. Robot B can go 1 meter in 50 seconds.1) If Robot A continues at that same speed, how long will it take to go 1 meter? Remember units! __________2) How long for Robot A to go 4 meters at that speed? Remember units! _________

1 answer:

GalinKa [24]2 years ago

5 0

Answer:

1 1/14 seconds
4 2/7 seconds

Step-by-step explanation:

We are being asked for times, so we can express the rate in terms of seconds per meter.

(15 seconds)/(14 meters) = (15/14) seconds per meter = 1 1/14 seconds/meter

__

1) It will take Robot A 1 1/14 seconds to go 1 meter.

2) it will take Robot A 4 2/7 seconds to go 4 meters. (4 times as long)

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Sholpan [36]

P=12T

Q=66T

-----------

Toys added to each bag:

P=12T+nT

Q=66T+nT

Therefore, 66T+nT=3(12T+nT)

------------

66T+nT=36T+3nT

T(66+n)=T(36+3n)

66+n=36+3n

3n-n=66-36

2n=30

Therefore, n=30/2=15

-------------

*She added 15 toys to each bag.

6 0

3 years ago

Which function has a vertex on the y-axis? f(x) = (x – 2)2 f(x) = x(x + 2) f(x) = (x – 2)(x + 2) f(x) = (x + 1)(x – 2)

strojnjashka [21]

we know that

If the vertex is on the y-axis, then the x-coordinate of the vertex is equal to zero

we are going to verify the vertex of each one of the functions to determine the solution

Remember that

The equation in vertex form of a vertical parabola is equal to

$y=a(x-h)^{2} +k$

where

(h,k) is the vertex

if $a>0$ -------> the parabola open upward (vertex is a minimum)

if $a$ -------> the parabola open downward (vertex is a maximun)

<u>case A)</u> $f(x)=(x-2)^{2}$

This is a vertical parabola open upward

the vertex is the point $(2,0)$

therefore

The function $f(x)=(x-2)^{2}$ does not have a vertex on the y-axis

<u>case B)</u> $f(x)=x(x+2)$

$f(x)=x(x+2)=x^{2}+2x$

convert to vertex form

Complete the square. Remember to balance the equation by adding the same constants to each side.

$f(x)+1=x^{2}+2x+1$

Rewrite as perfect squares

$f(x)+1=(x+1)^{2}$

$f(x)=(x+1)^{2}-1$

the vertex is the point $(-1,-1)$

therefore

The function $f(x)=x(x+2)$ does not have a vertex on the y-axis

<u>case C)</u> $f(x)=(x-2)(x+2)$

$f(x)=(x-2)(x+2)=x^{2}-2^{2}$

$f(x)=x^{2}-4$

the vertex is the point $(0,-4)$

The x-coordinate of the vertex is equal to zero

therefore

The function $f(x)=(x-2)(x+2)$ has a vertex on the y-axis

<u>case D)</u> $f(x)=(x+1)(x-2)$

$f(x)=(x+1)(x-2)\\ \\f(x)= x^{2}-2x+x-2 \\ \\f(x)= x^{2} -x-2$

convert to vertex form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$f(x)+2= x^{2} -x$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$f(x)+2+0.25= x^{2} -x+0.25$

$f(x)+2.25= x^{2} -x+0.25$

Rewrite as perfect squares

$f(x)+2.25= (x-0.50)^{2}$

$f(x)=(x-0.50)^{2}-2.25$

the vertex is the point $(0.5,-2.25)$

therefore

The function $f(x)=(x+1)(x-2)$ does not have a vertex on the y-axis

<u>the answer is</u>

$f(x)=(x-2)(x+2)$

6 0

3 years ago

Read 2 more answers

98 POINTSSSSSS!!!!!!!!!

ddd [48]

Answer:

the answer is b -2u +w

or -2w - u

8 0

3 years ago

What are two subsets of real numbers that form the set of real numbers

Ulleksa [173]

Answer:

Two subsets that form set of real numbers

You can take positive numbers and negative numbers subset

One subset become (- infinity,0]

other subset (0, infinity)

4 0

3 years ago

What is the area of the partial circle with a radius of 8 cm? Use 3. 14 for pi.

Olegator [25]

Radius=r=8cm

$\\ \tt\hookrightarrow Area=\pi r^2$

$\\ \tt\hookrightarrow Area=\pi 8^2$

$\\ \tt\hookrightarrow Area=64\pi$

$\\ \tt\hookrightarrow Area=64(3.14)$

$\\ \tt\hookrightarrow Area=200.96cm^2$

8 0

3 years ago

Read 2 more answers