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3 years ago

Need help ASAP due in 30 minutes!

Mathematics

2 answers:

Whitepunk [10]3 years ago

7 0

Y=3x-4

Hope you get it turned in in time!!!

Serjik [45]3 years ago

4 0

Answer:y=3x-4 is the answer

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I please need help solving for x .

Alekssandra [29.7K]

Answer:

16=x

Step-by-step explanation:

So, The angle is 180 degrees

180-87=93

93=(6x-3)

Add 3 to both sides

96=6x

Divide 6 on both sides

16=x

7 0

3 years ago

Read 2 more answers

The edges of a box are 4 cm, 6 cm and 8 cm. Find the length

qaws [65]

Answer:

Largest Face: 90 degree angle

Step-by-step explanation:

7 0

2 years ago

Here is the graph of the sequence M define M recursively using function notation<br>

aev [14]

The graph of the sequence M , we define M recursively using function notation

$f(n+1)=f(n)-2$

Given :

the graph of the sequence M

Lets write the recursive function using the given points

First point is (1,7) and second point is (2,5) then (3,3) and so on

For recursive function , we use the initial point (1,7)

when n=1, f(n)=7

the y values of the point are decreasing by 2. Common difference is -2

Recursive function is

$f(n+1)=f(n)-diffeference \\f(n+1)=f(n)-2$

where n=2,3,4...

Learn more : brainly.com/question/10676131

8 0

2 years ago

Tom wants to place a photograph that is 8.5 inches wide on his science fair poster. His poster board is 21.25 inches wide. How f

SOVA2 [1]

Answer:

6 3/8 inches (6.375 inches)

Step-by-step explanation:

For simplicity we'll assume that both the photo and the poster board are square. To determine the width of the border, subtract 8.5 inches from the poster board width 21.25 inches, obtaining 12.75 inches, and then divide that 12.75 inches by 2: 6.375 inches (6 3/8 inches).

Set the photo 6 3/8 inches from each edge of the poster board.

6 0

3 years ago

Prove the function f: R- {1} to R- {1} defined by f(x) = ((x+1)/(x-1))^3 is bijective.

Eduardwww [97]

Answer:

See explaination

Step-by-step explanation:

given f:R-\left \{ 1 \right \}\rightarrow R-\left \{ 1 \right \} defined by f(x)=\left ( \frac{x+1}{x-1} \right )^{3}

let f(x)=f(y)

\left ( \frac{x+1}{x-1} \right )^{3}=\left ( \frac{y+1}{y-1} \right )^{3}

taking cube roots on both sides , we get

\frac{x+1}{x-1} = \frac{y+1}{y-1}

\Rightarrow (x+1)(y-1)=(x-1)(y+1)

\Rightarrow xy-x+y-1=xy+x-y-1

\Rightarrow -x+y=x-y

\Rightarrow x+x=y+y

\Rightarrow 2x=2y

\Rightarrow x=y

Hence f is one - one

let y\in R, such that f(x)=\left ( \frac{x+1}{x-1} \right )^{3}=y

\Rightarrow \frac{x+1}{x-1} =\sqrt[3]{y}

\Rightarrow x+1=\sqrt[3]{y}\left ( x-1 \right )

\Rightarrow x+1=\sqrt[3]{y} x- \sqrt[3]{y}

\Rightarrow \sqrt[3]{y} x-x=1+ \sqrt[3]{y}

\Rightarrow x\left ( \sqrt[3]{y} -1 \right ) =1+ \sqrt[3]{y}

\Rightarrow x=\frac{\sqrt[3]{y}+1}{\sqrt[3]{y}-1}

for every y\in R-\left \{ 1 \right \}\exists x\in R-\left \{ 1 \right \} such that x=\frac{\sqrt[3]{y}+1}{\sqrt[3]{y}-1}

Hence f is onto

since f is both one -one and onto so it is a bijective

8 0

3 years ago