Normal body temperature is 98.6°F. Ellen's temperature is 100.8°F. How many degrees abovenormal body temperature is this?

Answers

Answer 1
Answer:

Answer:2.2

Step-by-step explanation:100.8-98.6=2.2


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Towns A and B are 400 miles apart. If a train leaves A in the direction of B at 50 miles per hour, how long will it take before that train meets another train, going from B to A,
at a speed of 30 miles per hour?

Answers

from A :50/hour. *x(hour)
From B : 30/hour *x(hour)
50 *x +30 *x =400
80 *x=400
x=5(hour)
Then the train from A will make 50 *5=250 miles
(These is fisik)

How do you work out percentages

Answers

if you are converting, make the denomenator 100 for fractions or look at 2 decimal places for decimals. if you want to find a certain percentage for a total, make the total of the item 100% and from there find 1% and multiply to the wanted percentage

1.43 x (-19/37) please answer
.....

Answers

1.43 X (-19/37)=1.43 X -0.51=-0.73
1.43 × (-19/37) = -0.734324

Subtract.(x + 1) - (-2x - 5)
A) 6 –x
B) -x - 4
C) 3x - 4
D) 3x + 6

Answers

(x + 1) - (-2x - 5)

Distribute the negative into the parenthesis

x + 1 + 2x + 5

Rearrange the terms

x + 2x + 1 + 5

Add like terms

3x + 6

Your answer is D.

H(t) = -5t^2+20t+1 what time does the ball reach the same height it was kicked at again? When does the ball reach its max height? What is the max height?

Answers

h(t) = -5t² + 20t + 1
-5t² + 20t + 1 = 0
t = -(20) +/- √((20)² - 4(-5)(1))
                      2(-5)
t = -20 +/- √(400 + 20)
                  -10
t = -20 +/- √(420)
              -10
t = -20 +/- 2√(105)
              -10
t = -20 + 2√(105)      t = -20 - 2√(105)
             -10                           -10
t = 2 - 0.2√(105)      t = 2 + 0.2√(105)
h(t) = -5t² + 20t + 1
h(2 - 0.2√(105)) = -5(2 - 0.2√(105))² + 20(2 - 0.2√(105)) + 1
h(2 - 0.2√(105)) = -5(2 - 0.2√(105))(2 - 0.2√(105)) + 20(2) - 20(0.2√(105)) + 1
h(2 - 0.2√(105)) = -5(4 - 0.4√(105) - 0.4√(105) + 0.04√(11025)) + 40 - 4√(105) + 1
h(2 - 0.2√(105)) = -5(4 - 0.8√(105) + 0.04(105)) + 40 + 1 - 4√(105)
h(2 - 0.2√(105)) = -5(4 - 0.8√(105) + 4.2) + 41 - 4√(105)
h(2 - 0.2√(105)) = -5(4 + 4.2 - 0.8√(105)) + 41 - 4√(105)
h(2 - 0.2√(105)) = -5(8.2 - 0.8√(105)) + 41 - 4√(105)
h(2 - 0.2√(105)) = -5(8.2) - 5(-0.8√(105)) + 41 - 4√(105)
h(2 - 0.2√(105)) = -41 + 4√(105) + 41 - 4√(105)
h(2 - 0.2√(105)) = -41 + 41 + 4√(105) - 4√(105)
h(2 - 0.2√(105)) = 0 + 0
h(2 - 0.2√(105)) = 0
(t, h(t)) = (2 - 0.2√(105), 0)
or
h(t) = -5t² + 20t + 1
h(2 + 0.2√(105)) = -5(2 + 0.2√(105))² + 20(2 + 0.2√(105)) + 1
h(2 + 0.2√(105)) = -5(2 + 0.2√(105))(2 + 0.2√(105)) + 20(2) + 20(0.2√(105)) + 1
h(2 + 0.2√(105)) = -5(4 + 0.4√(105) + 0.4√(105) + 0.04√(11025)) + 40 + 4√(105) + 1
h(2 + 0.2√(105)) = -5(4 + 0.8√(105) + 0.04(105)) + 40 + 4√(105) + 1
h(2 + 0.2√(105)) = -5(4 + 0.8√(105) + 4.2) + 40 + 1 + 4√(105)
h(2 + 0.2√(105)) = -5(4 + 4.2 + 0.8√(105)) + 41 + 4√(105)
h(2 + 0.2√(105)) = -5(8.2 + 0.8√(105)) + 41 + 4√(105)
h(2 + 0.2√(105)) = -5(8.2) - 5(0.8√(105)) + 41 + 4√(105)
h(2 + 0.2√(105)) = -41 - 4√(105) + 41 + 4√(105)
h(2 + 0.2√(105)) = -41 + 41 - 4√(105) + 4√(105)
h(2 + 0.2√(105)) = 0 + 0
h(2 + 0.2√(105)) = 0
(t, h(t)) = (2 + 0.2√(105), 0)

A line passes through the point (-2, 7) and has a slope of -5. What is the value of a if the point (a, 2) is also on the line?
A. -7
B. -1
C. 1
D. 7

Answers

Answer:

B. -1

Step-by-step explanation: