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iren [92.7K]
2 years ago
7

Does 28.40 equal with 28.400?

Mathematics
1 answer:
Nastasia [14]2 years ago
4 0

Question

Does 28.40 equal with 28.400?

Answer:

The Answer Is  yes 28.40 Does Equal 28.400

Hope this helps!!

xXxAnimexXx

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find the average rate of change from t=2 to t=5 for the velocity function v(t)=t^2-t+10. WILL GIVE THE BRAINLIEST ANSWER--EXPLAI
Elina [12.6K]
Answer: The average rate of change is 6.
First, plug in each value of <em>t</em> into the function, v(t) to find there coordinate pairs.

v(2) = (2)^2 - (2) + 10
v(2) = 4 + 8
v(2) = 12

v(5) = (5)^2 - (5) + 10
v(5) = 25 + 5
v(5) = 30

You can write these values as coordinate pairs, like so: (2, 12) and (5, 30).
The formula for the average rate of change is A(x) =  \frac{f(x)-(f(a)}{x-a}. When you plug in the values from this particular case, the average rate of change formula becomes A(t) = \frac{30-12}{5-2}, or A(t)= \frac{18}{3}.

Looking at the equation, you can solve for the average rate of change between t = 2 and t = 5, which equals 6.
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3 years ago
Need asap final exam help fast
antiseptic1488 [7]

I uploaded the answer to a file hosting. Here's link:

tinyurl.com/wpazsebu

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HELP IT"S ALGEBRA CAN SOMEONE DO THIS QUICK
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You purchase a car in 2010 for $25,000. The value of the car decreases by 14% annually. What would the value of the car be in 20
Fed [463]

Answer: -$6,500

Step-by-step explanation:

Here we could , use the arithmetic progression where

T(2020 - 2010) = a + ( n - 1 )d

T10 = a + ( 10 - 1 )d --------------- 1

a = $25,000, n = 10 and d = 14% of $25,000 = $3,500 the common difference.

Note since it decreases the common difference d = -$3,500.

Now substitute for the values in the equation above.

T10 = 25,000 + 9 x -3,500

= $25,000 - $31,500

= -$6,500 (deficit )

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3 years ago
CAN ANYBODY HELP ME OUT
bija089 [108]

Answer:

Correct option is

b. If two sides and one included angle are equal in triangles PQS and PRS, then their corresponding sides are also equal.

Step-by-step explanation:

Here, we are given the line RQ, which is divided in two equal parts by a line PS which is perpendicular to RQ.

The foot S of PS is on the line RQ.

First of all, let us do a construction here.

Join the point R with P and P with Q.

Please refer to the attached image.

Now, let us consider the triangles  PQS and PRS:

  • Side QS = RS (as given)
  • \angle PSR = \angle PSQ = 90^\circ
  • Side PS = PS (Common side in both the triangles)

Now, Two sides and the angle included between the two triangles are equal.

So by SAS congruence we can say that \triangle PRS \cong \triangle PQS

Therefore, the corresponding sides will also be equal.

RP = QP

RP is the distance between R and P.

QP is the distance between Q and P.

Hence, to prove that P is equidistant from R and Q, we have proved that:

b. If two sides and one included angle are equal in triangles PQS and PRS, then their corresponding sides are also equal.

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