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3 years ago

I Need more help plz I need help help help

Mathematics

2 answers:

sergij07 [2.7K]3 years ago

6 0

Answer:

1.715 kilometers

Step-by-step explanation:

If you know it travels 343 meters per second for 5 seconds, you simply multiply 343 by 5 to get 1,715. There is 1000 meters in a kilometer, so you divide by 1000 to get 1.715 kilometers.

Elden [556K]3 years ago

5 0

Answer:

1.715 kilometers

Step-by-step explanation:

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Leona [35]

To find the answer you have to find the area of the fountain using A= pi r^2.
=(3.14)(64)
=200.96

Then you have to find the area of the big circle.
=(3.14)(11^2)
=379.94

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3 years ago

Zipora wants to construct a triangle that has the following angle measures: 50°, 75°, and 55°. How many triangles can she make?

mr Goodwill [35]

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3 years ago

Find the area of the figure in cm

vazorg [7]

Answer:

260.55cm

Step-by-step explanation:

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3 years ago

Find the equation of the line that pass through the points (4,8) and (6,2

jeyben [28]

Answer:

3x + y = 20

Step-by-step explanation:

The equation of line passing through two points is determined by formula:

$y-y_{1}=\frac{y_{2}-y_{1}}{x_{2} - x_{1}}(x - x_{1})$

Here, (x₁ , y₁) = (4, 8)

and (x₂, y₂) = (6, 2)

Putting these value in above formula. We get,

$y-8=\frac{2-8}{6 - 4}(x - 4)$

⇒ $y-8=\frac{-6}{2}(x - 4)$

⇒ ( y - 8) = -3 (x - 4)

⇒ y - 8 = -3x + 12

⇒ 3x + y = 20

which is required equation.

6 0

4 years ago

Identify the "inside function" u = f(x) and the "outside function" y = g(u). Then find dy/dx using the Chain Rule.

skad [1K]

DfLet $f(x)=\sec x$ and $g(x)=\sqrt x$ . Then

$y=\sec\sqrt x=\sec(g(x))=f(g(x))=f\circ g(x)$

By the chain rule,

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dy}{\mathrm du}\dfrac{\mathrm du}{\mathrm dx}$

where $u=g(x)=\sqrt x$ , so that $y=f(g(x))=f(u)=\sec u$ . We have

$\dfrac{\mathrm du}{\mathrm dx}=\dfrac1{2\sqrt x}$
$\dfrac{\mathrm d\sec u}{\mathrm du}=\sec u\tan u$

and so

$\dfrac{\mathrm dy}{\mathrm dx}=\sec u\tan u\dfrac{\mathrm dy}{\mathrm du}=\dfrac{\sec\sqrt x\,\tan\sqrt x}{2\sqrt x}$

5 0

3 years ago