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Nadya [2.5K]
3 years ago
9

Need help with circle measurements in Geometry.

Mathematics
1 answer:
Ymorist [56]3 years ago
5 0
I got 0.999 which is 999/1000
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The table shows jennifer spent £119 on shopping
IrinaK [193]
£53 would be the correct answer man
6 0
3 years ago
Two equations are given below:
Evgesh-ka [11]

The solution to the set of equations is (0,2).

Reasoning:

m + 4n = 8 -> 0 + 4(2) = 8

m = n - 2 -> 0 = 2 - 2

6 0
3 years ago
I need to verify identity functions for a one to one function.
Eddi Din [679]

We have the function

f(x)=(x+6)^3

1. For f^-1:

Let y = f(x) = (x+6)^3

Switch x and y to get:

x=(y+6)^3

And solve for y

\begin{gathered} x^{\frac{1}{3}}=y+6 \\ x^{\frac{1}{3}}-6=y+6-6 \\ x^{\frac{1}{3}}-6=y \end{gathered}

And we have y = f^-1(x)

Answer blank 1:

f^{-1}(x)=x^{\frac{1}{3}}-6

2. For f o f^-1 (x):

(f\circ f^{-1})(x)=f(f^{-1}(x))

And solve

\begin{gathered} =f(x^{\frac{1}{3}}-6) \\ =(x^{\frac{1}{3}}-6+6)^3 \\ =(x^{\frac{1}{3}})^3 \\ =x \end{gathered}

answer blank 2

x^{\frac{1}{3}}-6

answer blank 3

x^{\frac{1}{3}}-6

answer blank 4

x^{\frac{1}{3}}

3. For f^-1 o f:

(f^{-1}\circ f)(x)=f^{-1}(f(x))

Solve

\begin{gathered} =f^{-1}((x+6)^3) \\ =\sqrt[3]{(x+6)^3}-6 \\ =x+6-6 \\ =x \end{gathered}

answer blank 5

(x+6)^3

answer blank 6

(x+6)^3

answer blank 7

x+6

4 0
1 year ago
How many quarts are in 1200 fl oz?
Semenov [28]

Answer:

37 1/2

Step-by-step explanation:

hope this helps:)

5 0
3 years ago
Read 2 more answers
The sample loses 1/3of it’s mass every ___ days
eimsori [14]

Answer:

The sample loses \frac{1}{3} of its mass every \frac{1}{3} days.

Step-by-step explanation:

Expression that shows the relation between final mass of the element after 't' days is,

M(t) = 900(\frac{8}{27})^t

If M(t) = two third of the initial mass [After the loss of one third of the initial mass remaining mass of the element = 1 - \frac{1}{3} = \frac{2}{3}rd of the initial mass]

900\times \frac{2}{3}=900(\frac{8}{27})^t

\frac{2}{3}=(\frac{8}{27})^t

\frac{2}{3}=[(\frac{2}{3})^3]^t

(\frac{2}{3})^1=(\frac{2}{3})^{3t}

By comparing powers on both the sides

3t = 1

t = \frac{1}{3} day

Therefore, The sample loses \frac{1}{3} of its mass every \frac{1}{3} days.

5 0
3 years ago
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