3 years ago

Find the lengths of the missing sides. Write your answers as integers or as decimals rounded to the nearest tenth.

Mathematics

1 answer:

garik1379 [7]3 years ago

4 0

The correct answer would be c

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The table below represents a function. Which of the following equations could be its function rule? PLS HELP!!

Sati [7]

Answer:

y = -3x

Step-by-step explanation:

It is easiest to try the answers for x=1.

a) y = 1/-3 = -1/3 ≠ -3

b) y = 1 ≠-3

c) y = 3(1) = 3 ≠ -3

d) y = -3(1) = -3 . . . . . this rule matches the table

8 0

2 years ago

Please help. No link answers. Will mark brainliest.

wel

The answer is -6/7 or -0.857142

5 0

3 years ago

Read 2 more answers

Which equation would have real zero(s) corresponding to the x-intercept(s) of the graph below?

MA_775_DIABLO [31]

Answer:

Choice A.

$y = - {2}^{x} + 4$

Step-by-step explanation:

Use graphing calculator

7 0

3 years ago

Can someone help?

nalin [4]

No no no no no no no no no jk its c your welcome

7 0

4 years ago

Find the value of x² - x+1 when x=√3 -1, in the form a+b√3 where a and b are integers

lidiya [134]

Answer:

$6-3\sqrt{3}$

If you compare this to $a+b\sqrt{3}$ , then $a=6$ and $b=-3$ .

Step-by-step explanation:

We are given $x=\sqrt{3}-1$ and we are asked to find the value of $x^2-x+1$ .

So we will need to find $x^2$ .

$x=\sqrt{3}-1$

Square both sides:

$x^2=(\sqrt{3}-1)^2$

Expand using identity: $(x-a)^2=x^2-2ax+a^2$ .

$x^2=(\sqrt{3})^2-2(\sqrt{3})(1)+1^2$

$x^2=3-2\sqrt{3}+1$

$x^2=4-2\sqrt{3}$

Let's go back to the full $x^2-x+1$ .

$x^2-x+1$

$(4-2\sqrt{3})-(\sqrt{3}-1)+1$

$4-2\sqrt{3}-\sqrt{3}+1+1$

$6-3\sqrt{3}$

If you compare this to $a+b\sqrt{3}$ , then $a=6$ and $b=-3$ .

5 0

4 years ago