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3 years ago

Which sentence describes why polygon MNOP is congruent to polygon JKLP?

1 answer:

7 0

The segment MN is equal in size to the segment JK, the segment NO is equal in size to the segment KL, the segment OP is equal in size to the segment LP.

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How to find (c,d,e)<br>Please do a step by step working.Thank you.

Natasha2012 [34]

(a)
The inverse is when you swap the variables and solve for y.
g(t) = 2t - 1 (Note: g(t) represents y)
rewrite as: y = 2t - 1
swap the variables: t = 2y - 1
solve for y: t + 1 = 2y
   $\frac{t + 1}{2}$ = y
Answer for (a): $g^{-1}(t)$ = $\frac{t + 1}{2}$

(b)
Same steps as part (a) above:
h(t) = 4t + 3
rewrite as: y = 4t + 3
swap the variables: t = 4y + 3
solve for y: $y =\frac{t - 3}{4}$

Answer for (b): $h^{-1}(t)$ = $\frac{t - 3}{4}$

(c)
$g^{-1} ( h^{-1}(t)) = g^{-1} (\frac{t - 3}{4})$
replace all t's in the $g^{-1}(t)$ equation with $\frac{t - 3}{4}$
$g^{-1} (\frac{t - 3}{4})$ = $\frac{ \frac{t-3}{4} + 1}{2}$
= $\frac{ \frac{t-3}{4} + \frac{4}{4}}{2}$ = $\frac{ \frac{t - 3 + 4}{4}}{2}$ = $\frac{ \frac{t + 1}{4}}{2} = \frac{t + 1}{8}$
Answer for (c): $g^{-1} ( h^{-1}(t))$ = $\frac{t + 1}{8}$

(d)
h(g(t)) = h(2t - 1) = 4(2t - 1) + 3 = 8t - 4 + 3 = 8t - 1
Answer for (d): h(g(t)) = 8t - 1

(e)
h(g(t)) = 8t - 1
   y = 8 t - 1
   t = 8y - 1
t + 1 = 8y
$\frac{t + 1}{8}$ = y
Answer for (e): inverse of h(g(t)) = $\frac{t + 1}{8}$

8 0

3 years ago

Find the x- and y-intercepts of the equation below. Write your answers as ordered pairs.

Airida [17]

Answer:

x intercept: none

y intercept=: (0,-12)

5 0

3 years ago

Help me answer question 1 .

kolezko [41]

C. is the right answer I think so at least

7 0

3 years ago

Please explain with working!!! Find the set of values of x that satisfy the inequality 9x^2-15x<6

Oksi-84 [34.3K]

Answer:

$-1/3$

Step-by-step explanation:

When solving a quadratic inequality, first solve it normally like you would for a normal quadratic equation. We have:

$9x^2-15x$

Ignore the less than sign and replace it with an equal sign and solve the quadratic for its zeros:

$9x^2-15x=6$

Subtract 6 from both sides:

$9x^2-15x-6=0$

Divide everything by 3:

$3x^2-5x-2=0$

Factor. Find two numbers that equal (3)(-2)=-6 that add up to -5.

-6 and 1 works. Thus:

$3x^2-6x+x-2=0\\3x(x-2)+1(x-2)=0\\(3x+1)(x-2)=0$

Find the x using the Zero Product Property:

$3x+1=0 \text{ or }x-2=0\\x=-1/3\text{ or }x=2$

Now, we need to replace the equal signs with symbols again. To do so, we need to test which symbol to place. Let's do the first zero first.

So, the first zero is:

$x=-1/3$

Assume that the correct symbol is >. Thus,

$x>-1/3$

Now, pick any number that is greater than -1/3. I'll pick 0 since it's the easiest. Now, plug 0 back into the very original inequality. If it works, then the sign is correct, if it doesn't, then simply use the opposite one. Therefore:

$9x^2-15x$