Answer:
The player throws 127.3 ft from second base to home plate.
Step-by-step explanation:
Given:
Distance from home to first base = 90 ft
Distance from first base to second base = 90 ft
We need to find the distance from second base to home.
Solution:
Now we can assume the complete scenario to be formed as a right angled triangle with two sides given and to find the third side.
Now by using Pythagoras theorem which states that;
Square of the hypotenuse side is equal to sum of squares of other two sides.
framing in equation form we get;
distance from second base to home = 
Rounding to nearest tent we get;
distance from second base to home = 127.3 ft
Hence The player throws 127.3 ft from second base to home plate.
Answer:
Odd
Step-by-step explanation:
You are given that PS ≅ RS as indicated by the single has mark.
You are given that ∠PST ≅ ∠RST as indicated by the single arc.
You know the segment ST is congruent to itself by the reflexive property.
That is, you have two congruent sides and the angle between them, so you can claim the triangles are congruent by the SAS postulate.
Selection C is appropriate.
THIS IS THE COMPLETE QUESTION BELOW
The demand equation for a product is p=90000/400+3x where p is the price (in dollars) and x is the number of units (in thousands). Find the average price p on the interval 40 ≤ x ≤ 50.
Answer
$168.27
Step by step Explanation
Given p=90000/400+3x
With the limits of 40 to 50
Then we need the integral in the form below to find the average price
1/(g-d)∫ⁿₐf(x)dx
Where n= 40 and a= 50, then if we substitute p and the limits then we integrate
1/(50-40)∫⁵⁰₄₀(90000/400+3x)
1/10∫⁵⁰₄₀(90000/400+3x)
If we perform some factorization we have
90000/(10)(3)∫3dx/(400+3x)
3000[ln400+3x]₄₀⁵⁰
Then let substitute the upper and lower limits we have
3000[ln400+3(50)]-ln[400+3(40]
30000[ln550-ln520]
3000[6.3099×6.254]
3000[0.056]
=168.27
the average price p on the interval 40 ≤ x ≤ 50 is
=$168.27
Answer:
Step-by-step explanation:
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