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2 years ago

Need precalc help. i’ll give brainliest

Mathematics

2 answers:

Ilya [14]2 years ago

8 0

Answer in picture below

Leviafan [203]2 years ago

5 0

Answer:

F= 2-(x+5)^2/gx

Step-by-step explanation:

Hope this helps

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Sue chooses a dress and necklace to wear when she performs at a violin recital. The probability that Sue selects a black dress i

Verizon [17]

Answer:

Sue has a 12% chance of choosing a black dress and a gold necklace (as a fraction: 3/25, and as a decimal: 0.12).

Step-by-step explanation:

To find the probability of two separate events both occurring (in this case, selecting a black dress and a gold necklace), the probabilities are multiplied usually as fractions or decimals.

As fractions:

40% = 4/10

30% = 3/10

4/10 * 3/10 = 12/100 = 3/25, 0.12, or 12%

As decimals:

40% = 0.4

30% = 0.3

0.4 * 0.3 = 0.12 = 3/25, 0.12, or 12%

8 0

3 years ago

Read 2 more answers

Q20 y varies directly as the square of (x – 3). y= 16 when x=1. Find y when x= 10.

enot [183]

-56

Mark brainliest please

Hope this helps you

3 0

3 years ago

hus, D satisfying (ABC)DequalsI exists. Why does the expression for D found above also satisfy D(ABC)equalsI, thereby showing

Assoli18 [71]

Complete Question

The complete question is shown on the first uploaded image

Answer:

First Question

Option A is correct

Second Question

Option C is correct

Third Question

$D = A^{-1} * B^{-1} * C^{-1}$

Fourth Question

So substituting for D in (ABC) D = I

$(ABC) * A^{-1} * B^{-1} * C^{-1} = I$

$I = I$

This proof that ABC is invertible

Step-by-step explanation:

From the question we are told that

A , B and C are invertible which means that $A^{-1} , B^{-1}, C^{-1}$ exist

Now

From the question

(ABC) D = I

Where I is an identity matrix

Now when we multiply both sides by $A^{-1}$ we have

$A^{-1} A BCD = A^{-1} * I$

$IBCD = A^{-1}$

Now when we multiply both sides by $B^{-1}$ we have

$B^{-1 } *I BCD = A^{-1} * B^{-1}$

$I CD = A^{-1} * B^{-1}$

Now when we multiply both sides by $C^{-1}$ we have

$C^{-1} * I CD = A^{-1} * B^{-1} * C^{-1}$

$I D = A^{-1} * B^{-1} * C^{-1}$

$D = A^{-1} * B^{-1} * C^{-1}$

So substituting for D in the above equation

$(ABC) * A^{-1} * B^{-1} * C^{-1} = I$

$I = I$

This proof that ABC is invertible

4 0

3 years ago

An arithmetic sequence is shown below.

Elena L [17]

Answer:

B. a (n) = -3n + 13

Step-by-step explanation:

Given arithmetic sequence is: 10, 7, 4, 1, -2, ...

Here a = 10, d = 7 - 10 = - 3

$a_n = a + (n - 1)d \\ \\ a_n = 10+ (n - 1)( - 3) \\ \\ a_n = 10 - 3n + 3 \\ \\ \purple{ \bold{a_n = - 3n +1 3 }}\\ \\$

7 0

3 years ago

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How to rename the number 3

Norma-Jean [14]

Three is another name for 3

4 0

3 years ago

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