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2 years ago

Find the rate of change for the situation.You run 7 miles in one hour and 21 miles in three hours.

Mathematics

1 answer:

likoan [24]2 years ago

5 0

Answer: 0

Step-by-step explanation: 21/7=3, so there is no change.

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Please help me thanks :)

Lina20 [59]

Answer:

Step-by-step explanation:

To sovle this we must combine the numbers togther and the variables togther.

We have:

8c+6-3c-2

Lets take out the c's and combine them:

8c -3c

Now lets take out the numbers:

6-2

Now lets put these back together and we get the expression:

5c+4

This looks like A.

Hope this helps! :D

8 0

2 years ago

784,072 rounded to the nearest hundred thousand

Fittoniya [83]

When you round; you round down if the number in the selected column is below 5 and you round up if the number is 5 and above.

Thus; if you apply this to your question:

784,072 then round up and you get 790,000

3 0

3 years ago

Wegnerkolmp or someone please help me with this question about slope....

NNADVOKAT [17]

Answer:

The slope is -3/4

Step-by-step explanation:

We need two points to find the slope

We have one point at (0,5) and we have one point at (4,2)

We can use the slope formula

m = (y2-y2)/(x2-x1)

= (2-5)/( 4-0)

= -3/4

The slope is -3/4

3 0

3 years ago

Read 2 more answers

In Exercises 5-7, find all the exact t-values for which the given statement is true,

bearhunter [10]

Answer: See Below

Step-by-step explanation:

NOTE: You need the Unit Circle to answer these (attached)

5) cos (t) = 1

Where on the Unit Circle does cos = 1?

Answer: at 0π (0°) and all rotations of 2π (360°)

In radians: t = 0π + 2πn

In degrees: t = 0° + 360n

******************************************************************************

$6)\quad sin (t) = \dfrac{1}{2}$

Where on the Unit Circle does $sin = \dfrac{1}{2}$

Hint: sin is only positive in Quadrants I and II

$\text{Answer: at}\ \dfrac{\pi}{6}\ (30^o)\ \text{and at}\ \dfrac{5\pi}{6}\ (150^o)\ \text{and all rotations of}\ 2\pi \ (360^o)$

$\text{In radians:}\ t = \dfrac{\pi}{6} + 2\pi n \quad \text{and}\quad \dfrac{5\pi}{6} + 2\pi n$

In degrees: t = 30° + 360n and 150° + 360n

******************************************************************************

$7)\quad tan (t) = -\sqrt3$

Where on the Unit Circle does $\dfrac{sin}{cos} = \dfrac{-\sqrt3}{1}\ or\ \dfrac{\sqrt3}{-1}\quad \rightarrow \quad (1,-\sqrt3)\ or\ (-1, \sqrt3)$

Hint: sin and cos are only opposite signs in Quadrants II and IV

$\text{Answer: at}\ \dfrac{2\pi}{3}\ (120^o)\ \text{and at}\ \dfrac{5\pi}{3}\ (300^o)\ \text{and all rotations of}\ 2\pi \ (360^o)$

$\text{In radians:}\ t = \dfrac{2\pi}{3} + 2\pi n \quad \text{and}\quad \dfrac{5\pi}{3} + 2\pi n$

In degrees: t = 120° + 360n and 300° + 360n

4 0

2 years ago

he triangles below are similar. Triangle Y X Z. Side Y X is 6 centimeters, X Z is 7 centimeters, and Z Y is 3 centimeters. Trian

MissTica

YX=6
XZ=7
ZY=3

And

AB=12
BC=6
AC=14

Check

AB/YX=BC/ZY=AC/XZ
12/6=6/3=14/7
2=2=2

Hence

ABC=XYZ

ZYX≈CBA

Option C

6 0

1 year ago