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3 years ago

Plz help 50 points

Mathematics

2 answers:

cluponka [151]3 years ago

8 0

9/2 that was so easy tho

geniusboy [140]3 years ago

6 0

Answer: So the answer to number 2 is D! And the answer to number 3 is C I think I am not pretty sure about #3 but I am sure its D for 2 and pretty sure for #3 is C!

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Which expression represents the absolute value of - 10

mote1985 [20]

Answer:

|-10|=10

Step-by-step explanation:

Absolute value is 10

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How many ways can five people line up at a checkout counter in a supermarket?

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120
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3 years ago

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412.92meter _______m______cm

makvit [3.9K]

Answer:

412.92meter = 412m and 92 cm

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3 years ago

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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

Arvin has $10000 to invest. He invests part in a term deposit paying 5%/year, and the remainder in Canada savings bonds paying 3

Delicious77 [7]

Answer:

$4,562.5

Step-by-step explanation:

The amount Arvin has to invest, P = $10,000

The interest paid on the investment in the term deposit = 5%/year

The interest paid om the investment in Canada savings bonds = 3.4%/year

The amount Arvin earned at the of the year as simple interest, A = $413

Let, x, represent the amount Arvin invested in the term deposit and let, y, represent the amount he invested in Canada savings bonds, we can get the following system of equations

x + y = 10,000...(1)

0.05·x + 0.034·y = 413...(2)

Making y the subject of equation (1) and substituting the value in equation (2), we get;

From equation (1), we get, y = 10,000 - x

Plugging the above value of y in equation (2) gives;

0.05·x + 0.034 × (10,000 - x) = 413

∴ 0.05·x - 0.034·x + 340 = 413

x = (413 - 340)/(0.05 - 0.034) = 4,562.5

Therefore, the amount Arvin invested in the term deposit at 5%, x = $4,562.5

(y = 10,000 - x

∴ y = 10,000 - 4,562.5 = 5,437.5

The amount Arvin invested in Canada savings bonds, y = $5,437.5.)

8 0

3 years ago