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Vilka [71]
3 years ago
14

F(t) = t - 6 f(u + 6) =

Mathematics
1 answer:
vovikov84 [41]3 years ago
4 0

Answer:

  f(u +6) = u

Step-by-step explanation:

Put (u+6) in place of t and simplify:

  f(u+6) = (u+6) -6

  f(u+6) = u

You might be interested in
How many oz of nuts are there in a box that holds 3.2 pounds
melisa1 [442]

Answer:

51.2 oz

Step-by-step explanation:

Stoichiometry

3.2 lb x 16 oz

          ÷ 1 lb      =51.2 oz

3 0
3 years ago
A radio station is going to construct a 5-foot tower on top of a building. the tower will be supported by three cables, each att
KonstantinChe [14]
You need to use Pythagoras’ Theorem. 
A² + B² = c² and c is the hypotenuse of the right angled triangle.

The height of the tower (5 feet) is A and the distance from the end of the cable and the base of the tower (12 feet) is B. The length of ONE cable is c. So:

A² + B² = c²
5² + 12² = c²
25 + 144 = c²
169 = c²
13 = c. This is the length of one cable.

3 x 13 = 39 therefore the total length of the cables is 39 feet.



25 
8 0
3 years ago
What value, written as a decimal, should Lena use as the common ratio?
inn [45]

Answer:

an=1*2.5^(n-1)

=2.5^(n-1)

Step-by-step explanation:

Complete question below:

What value, written as a decimal, should Lena use as the common ratio? Lena is asked to write an explicit formula for the graphed geometric sequence. On a coordinate plane, 3 points are plotted. The points are (1, 1), (2, 2.5), (3, 6.25).

Solution

Point (1, 1), (2, 2.5), (3, 6.25).

a=1

ar=2.5

ar^2=6.25

From ar and ar^2

r=6.25/2.5

=2.5

r=2.5

an=ar^(n-1)

Therefore, the explicit formula is

an=1*2.5^(n-1)

=2.5^(n-1)

7 0
3 years ago
The sum of the diagonals of a rhombus is 5√2.
Alex
Greetings!
Let ABCD be a rhombus
AC + BD = 5√2 cm

Area of ABCD = Ar△ABD + Ar △BCD
= \frac{1}{2} \times BD \times AO + \frac{1}{2} \times BD \times OC
= \frac{1}{2} BD (AO + OC)
\frac{1}{2} BD \times AC

So, \frac{ BD \times AC}{2} = 4 \: cm {}^{2}
BD \times AC = 8
Now, AC + \: BD = 5 \sqrt{2}

Squaring both sides, we get
AC {}^{2} + BD {}^{2} + 2 AC.BD =50
AC {}^{2} + BD {}^{2} + 2 \times 8 = 50
AC {}^{2} + BD {}^{2} = 50 - 16
AC {}^{2} + BD {}^{2} = 34

In △AOB, we have
OA {}^{2} + OB {}^{2} =AB {}^{2}
( \frac{ AC }{2} ) {}^{2} + ( \frac{ BD}{2} ) {}^{2} = AB {}^{2}
\frac{ AC {}^{2} } {4} + \frac{ BD {}^{2} }{4} = AB {}^{2}
AC {}^{2} + BD {}^{2} = 4 AB {}^{2}
34 = 4 AB {}^{2}

Square rooting both sides
\sqrt{34} = 2 AB
Perimeter = 4 \: AB \\ = 2 \times 2 \: AB \\ = 2 \: \times \sqrt{34 } \\ = 2 \sqrt{34 \: } units.

Hope it helps!

7 0
3 years ago
Subtract what she has now from the total she needs. She needs ten. She has one 2 1/3 and 3 1/3 added is 5 2/3
tatiyna

Answer:

4 ⅓

Step-by-step explanation:

10 - 5 ⅔

10 - 5 × 3 + 2/3

10 - 17/3

3 × 10 - 17/3

30 - 17/3

13/3 = 4 ⅓

<u>-TheUnknownScientist</u>

6 0
3 years ago
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