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musickatia [10]

4 years ago

7

You 495 miles from home and you are driving toward home at an average of 45mph. How long will it take to drive home if you maint

ain the same speed? (Hint: When you reach home, the x-value is zero.)

1 answer:

iogann1982 [59]4 years ago

3 0

<span>I am not sure about that x value stuff....but time = distance / speed......time = 495/45 = 11 hrs to reach home </span>

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Can somebody plz help answer these questions correctly thanks! (Grade7math) btw :)

Monica [59]

Answer:

1. both boxes 3

2. both boxes 7

3. both boxes 5

4. both boxes 2

5. top box 12 bottom 18

6. top box 9 bottom box 11

7. top box 42 bottom box 60

8. top box 4 bottom box 5

7 0

3 years ago

A trapezoid has base lengths of 12 centimeters and 13 centimeters. The other sides have lengths of 5 centimeters and 10 centimet

r-ruslan [8.4K]

Answer:

The area of the composite figure is 392.12 $cm^{2}$ .

Step-by-step explanation:

The area of the composite figure = area of trapezoid + area of rectangle

Area of trapezium = $\frac{1}{2}$ ( a +b)h

Where: a is the length of the first base, b the length of the second base and h is the height of the trapzium.

Applying Pythagoras theorem, the height, h, is;

h = $\sqrt{5^{2} - 1^{2} }$

= $\sqrt{24}$

h = 2 $\sqrt{6}$

Area of trapezium = $\frac{1}{2}$ ( a +b)h

= $\frac{1}{2}$ (13 + 12) × 2 $\sqrt{6}$

= 156 $\sqrt{6}$

= 382.12 $cm^{2}$

Area of trapezium is 382.12 $cm^{2}$

Area of rectangle = length × width

= 5 × 2

= 10 $cm^{2}$

Area of rectangle = 10 $cm^{2}$

Therefore,

area of the composite figure = 382.12 + 10

= 392.12 $cm^{2}$

8 0

3 years ago

Plz help I need this by tomorrow

Rufina [12.5K]

Ik this doesn't help but maybe look the questions up online bc those are even confusing for me to understand. Really sorry I couldnt help :(

6 0

3 years ago

Read 2 more answers

Simply the expression <br> (15a^2y^-3)^4(5ay^-2)^-5

ankoles [38]

Answer:

81a3/5y2

Step-by-step explanation:

7 0

4 years ago

I need help finding the inverse of this problem

fomenos

$y=\sqrt{x-1}+6$

(Notice that we always have $y\ge6[tex].)Swap [tex]x$ and $y$ , then solve for $y$ :

$x=\sqrt{y-1}+6$

$x-6=\sqrt{y-1}$

$(x-6)^2=(\sqrt{y-1})^2$

$(x-6)^2=y-1$

$y=(x-6)^2+1$ (this is the inverse)

###

This inverse is valid only for $x\ge6$ . Why?

Suppose we take $x=0$ . Then

$y=(0-6)^2+1=37$

This would suggest that in the original equation, we should get $x=37$ when $y=0$ . But when we check this, we end up with

$0=\sqrt{37-1}+6=\sqrt{36}+6=6+6=12$

which is clearly not true.

6 0

3 years ago