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3 years ago

What is f(5)? –8 –1 1 8

Mathematics

1 answer:

ValentinkaMS [17]3 years ago

7 0

The function of 5 (or y when x is 5) can not be determined with this little information.

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Create an equation using the table of values below.

djyliett [7]

Answer:

1. y = 2x - 1

Step-by-step explanation:

See attached image.

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2 years ago

The length of shoes is 25 centimeters. how long is the shoe meters?

OlgaM077 [116]

Answer:

The shoe is 0.25 or 1/4 meters long.

Step-by-step explanation:

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3 years ago

Expanding brackets (2x + 4)(x - 6)

Natalija [7]

Answer:

=2x²-8x-24 is the answer

Step-by-step explanation:

=(2x+4)(x-6)

opening brackets by multiplying

=2x(x-6)+4(x-6)

=2x²-12x+4x-24

adding like terms

=2x²-8x-24

i hope this will help you :)

5 0

3 years ago

Read 2 more answers

Help me asaaaaappppppppppp please

Alenkinab [10]

Answer:

-29

Step-by-step explanation:

2(b-4)=3(b+7)

2b-8=3b+21

2b-3b=21+8

-b=29

b=-29

5 0

3 years ago

Can you prove it??<br>it's hard,I tried but couldn't solve this.

BARSIC [14]

I'll abbreviate $s=\sin\theta$ and $c=\cos\theta$ , so the identity to prove is

$\dfrac{s+c+1}{s+c-1}-\dfrac{1+s-c}{1-s+c}=2\left(1+\dfrac1s\right)$

On the left side, we can simplify a bit:

$\dfrac{s+c+1}{s+c-1}=\dfrac{s+c-1+2}{s+c-1}=1+\dfrac2{s+c-1}$

$\dfrac{1+s-c}{1-s+c}=-\dfrac{-2+1-s+c}{1-s+c}=-1+\dfrac2{1-s+c}$

Then

$\dfrac{s+c+1}{s+c-1}-\dfrac{1+s-c}{1-s+c}=2\left(1+\dfrac1{s+c-1}-\dfrac1{1-s+c}\right)$

So the establish the original equality, we need to show that

$\dfrac1{s+c-1}-\dfrac1{1-s+c}=\dfrac1s$

Combine the fractions:

$\dfrac{(1-s+c)-(s+c-1)}{(s+c-1)(1-s+c)}=\dfrac{2-2s}{c^2-s^2+2s-1}$

We can rewrite the denominator as

$c^2-s^2+2s-1=c^2+s^2-2s^2+2s-1$

then using the fact that $c^2+s^2=\cos^2\theta+\sin^2\theta=1$ , we get

$1-2s^2+2s-1=2s-2s^2$

so that we have

$\dfrac1{s+c-1}-\dfrac1{1-s+c}=\dfrac{2-2s}{2s-2s^2}=\dfrac1s$

as desired.

6 0

4 years ago