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valkas [14]
3 years ago
15

There are 2.54 centimeters in 1 inch. There are 100 centimeters in 1 meter. To the nearest inch, how many inches are in 2 meters

?​
Mathematics
1 answer:
kiruha [24]3 years ago
5 0

Answer:

1400 cm/ 2.54 cm per inch = 551 inches. Step-by-step explanation: There are 39.3701 inches in 1 meter, so all you have to do is multiply. :) 551 inches.Step-by-step explanation:

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2ln(4x –7) –1=11 Solve for x to the nearest 100th
ArbitrLikvidat [17]
2(4x - 7) -1 = 11
8x - 14 -1 = 11
8x - 15 = 11
     + 15     + 15
8x = 26
/8      /8
x = 3. 25 ------> Rounded: 3.30
3 0
4 years ago
I need blank 1 and blank 2
lord [1]

Answer:

Blank 1: 40

Blank 2: 10

Step-by-step explanation:

You can start by representing the speed of the water as x and the speed of the dolphin as y, and writing an equation.

y+x=50

y-x=30

Adding these two equations together, you get:

2y=80

y=40 for the speed of the dolphin in still water. Now, you an use one of the previous equations to find the speed of the current.

40-x=30

x=10

Hope this helps!

8 0
4 years ago
Hello there!
jok3333 [9.3K]

Answer:

4. dy/dx = -2

8. dy/dx = 1/2 x^(-3/2)

10/ dy/dr = 4 pi r^2

Step-by-step explanation:

4.  y = -2x+7

dy/dx = -2(1)

dy/dx = -2

8.  y = 4 - x^-1/2

 dy/dx =  - (-1/2x^ (-1/2-1)

 dy/dx = 1/2 x^(-3/2)

10.  y = 4/3 pi r^3

dy/dr = 4/3 pi  (3r^2)

dy/dr = 4 pi r^2

7 0
3 years ago
A cyclist travels 17 km from home to the park.
kupik [55]

Answer:

34 km/h

Step-by-step explanation:

speed = distance/time

Distance = 17km

Time = 0.5 or 1/2 hours because it's 30 minutes between 17:15 to 17:45, and 30 minutes is a half of an hour and you want km over hours

So

speed = 17/0.5

speed = 34 km/h

7 0
3 years ago
The point (1, −1) is on the terminal side of angle θ, in standard position. What are the values of sine, cosine, and tangent of
Lilit [14]

Check the picture below.

\bf (\stackrel{a}{1}~,~\stackrel{b}{-1})\qquad \impliedby \textit{let's find the hypotenuse} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies c = \sqrt{a^2+b^2} \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ c=\sqrt{1^2+(-1)^2}\implies c=\sqrt{2} \\\\[-0.35em] ~\dotfill

\bf sin(\theta ) \implies \cfrac{\stackrel{opposite}{-1}}{\stackrel{hypotenuse}{\sqrt{2}}}\implies \cfrac{-1}{\sqrt{2}}\cdot \cfrac{\sqrt{2}}{\sqrt{2}}\implies -\cfrac{\sqrt{2}}{(\sqrt{2})^2}\implies -\cfrac{\sqrt{2}}{2}

\bf cos(\theta ) \implies \cfrac{\stackrel{adjacent}{1}}{\stackrel{hypotenuse}{\sqrt{2}}}\implies \cfrac{1}{\sqrt{2}}\cdot \cfrac{\sqrt{2}}{\sqrt{2}}\implies \cfrac{\sqrt{2}}{(\sqrt{2})^2}\implies \cfrac{\sqrt{2}}{2} \\\\\\ tan(\theta ) = \cfrac{\stackrel{opposite}{-1}}{\stackrel{adjacent}{1}}\implies tan(\theta ) = -1

7 0
3 years ago
Read 2 more answers
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