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3 years ago

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Mathematics

1 answer:

Dafna1 [17]3 years ago

3 0

The picture below shows the box plot.

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The two largest lizards in the United States are the Gila monster and the chuckwalla. The average Gila monster is 0.608 meter lo

HACTEHA [7]

Rounded to the nearest hundredth: Gila Monster: 600; Chuckwalla: 400. This should help you.

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3 years ago

Let be an exponential function such that (3)/(2) = 4. Determine the value of the following ratios.

statuscvo [17]

This is hard to answer

6 0

2 years ago

What is the derivative of y^3+xe^4=3x^2-10?

Pani-rosa [81]

Answer:

${y}^{3} + x {e}^{4} = 3 {x}^{2} - 10$

introduce dy/dx :

$\frac{d( {y}^{3} + x {e}^{4} ) }{dx} = \frac{d( {3x}^{2} - 10) }{dx} \\ \\ 3 {y}^{2} \frac{dy}{dx} + {e}^{4} = 6x - 0 \\ \\ {3y}^{2} \frac{dy}{dx} = 6x - {e}^{4} \\ \\ { \boxed{ \boxed{\frac{dy}{dx} = \frac{6x - {e}^{4} }{ {3y}^{2} } }}}$

7 0

2 years ago

Claire and her partner, Grace, are throwing for the javelin event as a team. Claire threw the javelin 42 5/8 feet and Grace thre

Gennadij [26K]

The answer is 121.225.

Step-by-step explanation:

Because you would start with turning 42 and 5/8 into 341/8 then do 341 divided by 8 to get 42.625 as the decimal, then start turning 393/5 into a decimal by 393 divided by 5 to get 78.6 as your decimal then add them together like this: 42.625 + 78.6 = 121.225 as your answer.

Claire threw the javelin 42.625 feet but grace threw the javelin 78.6 feet.

Your welcome!.

3 0

2 years ago

Read 2 more answers

A circle has a diameter 9 cm.

vredina [299]

Answer:

Step-by-step explanation:

Since we are to show by Pythagoras that the square will fit inside the circle without touching the edge of the circle , this means that we must find the value of the length of the diagonal of the square using Pythagoras.

For the square to fit in without touching the edge of the circle , this means that the diagonal of the square must be less than the diameter of the circle.

Length of the square is 5cm.

Recall Pythagoras theorem: it states that

the square on the hypotenuse of a right-angled triangle is equal in area to the sum of the squares on the other two sides

$D^{2}$ = $5^{2}$ + $5^{2}$

$D^{2}$ = 25 + 25

$D^{2}$ = 50

Take the square root of both sides , that is

D = $\sqrt{50}$

D = $\sqrt{25}$ X $\sqrt{2}$ , since 50 = 25 x 2

D = 5 $\sqrt{2}$

D ≈ 7.07

Note that the diameter of the circle given is 9 and D < 9 , we can therefore conclude that the square will fit inside the circle without touching the edge of the circle since the diagonal of the square is less than the diameter of the circle

8 0

2 years ago