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Serjik [45]
3 years ago
13

. For the function given, state the starting point for a sample period:

Mathematics
1 answer:
Aneli [31]3 years ago
8 0
I think the answer is π?
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The temperature at 10 AM is 12°F. The temperature at 6 AM was -7°F. How many degrees did the temperature rise?
KatRina [158]

Answer:

19 degrees more

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Hey i really need help with this. thank you:))
balandron [24]

Given:

L=2\dfrac{1}{2} ft and W=3\dfrac{2}{5} ft.

P=2L+2W

To find:

The value of P.

Solution:

We have,

P=2L+2W

Substituting L=2\dfrac{1}{2} and W=3\dfrac{2}{5}, we get

P=2\times 2\dfrac{1}{2}+2\times 3\dfrac{2}{5}

P=2\times \dfrac{2(2)+1}{2}+2\times \dfrac{3(5)+2}{5}

P=2\times \dfrac{5}{2}+2\times \dfrac{17}{5}

P=5+\dfrac{34}{5}

Taking LCM, we get

P=\dfrac{5(5)+34}{5}

P=\dfrac{25+34}{5}

P=\dfrac{59}{5}

P=11\dfrac{4}{5}

Therefore, the value of P is 11\dfrac{4}{5} ft.

7 0
3 years ago
Given the function h(x) =1/3 |x-6| +4, evaluate the function when x = - 3, - 2, and 0
Masja [62]

|x| = x for x ≥ 0

examples:

|3| = 3; |0.56| = 0.56; |102| = 102

|x| = -x for x < 0

examples:

|-3| = -(-3) = 3; |-0.56| = -(-0.56) = 0.56; |-102| = 102

--------------------------------------------------------------------------------

Use PEMDAS:

P Parentheses first

E Exponents (ie Powers and Square Roots, etc.)

MD Multiplication and Division (left-to-right)

AS Addition and Subtraction (left-to-right)

--------------------------------------------------------------------------------

h(x)=\dfrac{1}{3}|x-6|+4

Put the values of x to the equation of the function h(x):

x=-3\to h(-3)=\dfrac{1}{3}|-3-6|+4=\dfrac{1}{3}|-9|+4=\dfrac{1}{3}(9)+4=3+4=7\\\\x=-2\to h(-2)=\dfrac{1}{3}|-2-6|+4=\dfrac{1}{3}|-8|+4=\dfrac{1}{3}(8)+4=\dfrac{8}{3}+\dfrac{12}{3}=\dfrac{20}{3}\\\\x=0\to h(0)=\dfrac{1}{3}|0-6|+4=\dfrac{1}{3}|-6|+4=\dfrac{1}{3}(6)+4=2+4=6


6 0
3 years ago
An equilateral ∆ has sides of length 16 cm. Find the length of an altitude.
inn [45]

The length of the altitude is 8\sqrt{3}

Explanation:

Let ABC be an equilateral triangle.

It has sides of length 16 cm

Let AD be the altitude of the triangle.

We need to determine the length of an altitude.

Let AC = 16 cm and CD = 8 cm

Let us consider the right angled triangle ADC

Using the Pythagorean theorem, we have,

AC^2=AD^2+DC^2

Substituting the values, we get,

 16^2=AD^2+8^2

 256=AD^2+64

 192=AD^2

8\sqrt{3}=AD

The length of the altitude is 8\sqrt{3}

5 0
3 years ago
PLEASE HELP ME GEOMETRY
Aloiza [94]
Answer is B. second choice

Segments DE and AC are parallel by construction


4 0
3 years ago
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