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3 years ago

100 points brainliest for correct answer please help me with this question and please show all work thanks

Mathematics

2 answers:

pychu [463]3 years ago

8 0

Step-by-step explanation:

14: The print and the canvas should have similar dimensions to one another.

Let x be the height of the canvas.

The print is 24 by 36, and the canvas is 12 by x.

24/36 = 12/x, x = 18.

Hence the canvas is 18 inches tall.

15: Ann's room is 10 by 12.5 while her sketch of the room is 8 by 10.

Since 10/12.5 = 8/10, it is a scale drawing. The scale factor is 1:0.8.

16: Both rectangles are similar to each other.

14/9 = x/4.5, therefore x = 7.

17: There is a similar relationship between Bill's height + shadow, and the tree's height + shadow.

Bill's height : Bill's shadow length

= Tree's height : Tree's shadow length

Let y be the height of the tree.

This means that 6:4 = y:16, so y = 24.

Hence the tree is 24 feet tall.

jarptica [38.1K]3 years ago

6 0

Answer:

14. 18 inches tall

Step-by-step explanation:

14. 24in wide = 12in wide

36in tall = 18in tall

So technically the relationship of 24 to 12 is dividing it by 2 so you would do the same for 36in tall wall.

Sorry I could only do one I’m lazy and in a rush but hope this helps!

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3 years ago

Simplify the expression 2√6 -5√54

Art [367]

Answer:

- 13 $\sqrt{6}$

Step-by-step explanation:

Using the rule of radicals

$\sqrt{a}$ × $\sqrt{b}$ ⇔ $\sqrt{ab}$

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$\sqrt{54}$

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3 years ago

Help!!!<br><br> Simplify (3x^2y^5)^3/(2xy^3)^4 show your work

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9/16 x^2y^3

Step-by-step explanation:

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3 years ago

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3 years ago

Suppose r(t) = (et cos t)i + (et sin t)j. Show that the angle between r and a never changes. What is the angle?

tatyana61 [14]

Answer:

Angle = Ф = $cos^{-1}$ (0) = 0

Hence, it is proved that angle between position vector r and acceleration vector a = 0 and is it never changes.

Step-by-step explanation:

Given vector r(t) = $e^{t}cost i + e^{t}sint j$

As we know that,

velocity vector = v = $\frac{dr}{dt}$

Implies that

velocity vector = $(e^{t} cost - e^{t} sint)i + (e^{t} sint - e^{t}cost )j$

As acceleration is velocity over time so:

acceleration vector = a = $\frac{dv}{dt}$

Implies that

vector a =

$(e^{t}cost - e^{t}sint - e^{t}sint - e^{t}cost )i + ( e^{t}sint + e^{t}cost + e^{t}cost - e^{t}sint )j$

vector a = $(-2e^{t}sint) i + ( 2e^{t}cost)j$

Now scalar product of position vector r and acceleration vector a:

r. a = $. \\$

r.a = $-2e^{2t}sintcost + 2e^{2t}sintcost$

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Now, for angle between position vector r and acceleration vector a is given by:

cosФ = $\frac{r.a}{|r|.|a|}$ = $\frac{0}{|r|.|a|} = 0$

Ф = $cos^{-1}$ (0) = 0

Hence, it is proved that angle between position vector r and acceleration vector a = 0 and is it never changes.

4 0

3 years ago