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Gnom [1K]
4 years ago
13

Which symbol makes -121. That make -21 a true statement

Mathematics
1 answer:
baherus [9]4 years ago
8 0
This question doesn’t make sense. Please make sure you write the question properly without any grammatical mistakes.
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What is the percent that is equal to 0.02
alekssr [168]
2%. You must move the decimal two places to the right.
3 0
3 years ago
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On a baseball field, the pitcher’s mound is 60.5 feet from home plate. During practice, a batter hits a ball 195 feet at an angl
worty [1.4K]

In this problem, we can imagine that all the points connect to form a triangle. The three point or vertices are located on the pitcher mount, the home plate and where the outfielder catches the ball. So in this case we are given two sides of the triangle and the angle in between the two sides.

<span>With the following conditions, we can use the cosine law to solve for the unknown 3rd side. The formula is:</span>

c^2 = a^2 + b^2 – 2 a b cos θ

Where,

a = 60.5 ft

b = 195 ft

θ = 32°

Substituting the given values:

c^2 = (60.5)^2 + (195)^2 – 2 (60.5) (195) cos 32

c^2 = 3660.25 + 38025 – 20009.7

c^2 = 21,675.56

c = 147.23 ft

<span>Therefore the outfielder throws the ball at a distance of 147.23 ft towards the home plate.</span>

8 0
3 years ago
June served 4/9 of 18 liters of fruit punch. How many liters does she have left?
victus00 [196]
She has 10 liters left.  Find 4/9 of 18 by multiplying 4/9 and 18.  4/9 of 18 is 8, so she served 8 of the 18, leaving 10 liters.
3 0
3 years ago
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Question 7 of 10
klemol [59]
I think the answer is C. 45 units
4 0
3 years ago
Money Flow  The rate of a continuous money flow starts at $1000 and increases exponentially at 5% per year for 4 years. Find the
77julia77 [94]

Answer:

Present value =  $4,122.4

Accumulated amount = $4,742

Step-by-step explanation:

Data provided in the question:

Amount at the Start of money flow = $1,000

Increase in amount is exponentially at the rate of 5% per year

Time = 4 years

Interest rate = 3.5%  compounded continuously

Now,

Accumulated Value of the money flow = 1000e^{0.05t}

The present value of the money flow = \int\limits^4_0 {1000e^{0.05t}(e^{-0.035t})} \, dt

= 1000\int\limits^4_0 {e^{0.015t}} \, dt

= 1000\left [\frac{e^{0.015t}}{0.015} \right ]_0^4

= 1000\times\left [\frac{e^{0.015(4)}}{0.015} -\frac{e^{0.015(0)}}{0.015} \right]

= 1000 × [70.7891 - 66.6667]

= $4,122.4

Accumulated interest = e^{rt}\int\limits^4_0 {1000e^{0.05t}(e^{-0.035t}} \, dt

= e^{0.035\times4}\times4,122.4

= $4,742

8 0
3 years ago
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