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2 years ago

Can someone help me with this math question? First to answer will get Brainliest! :)

Mathematics

2 answers:

antoniya [11.8K]2 years ago

6 0

Answer:

Step-by-step explanation:

svetlana [45]2 years ago

4 0

Answer:

Step-by-step explanation:

yiu have to rotate it in order to make that

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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

2 years ago

Can someone check if this is correct please

xenn [34]

Answer:

it is correct you are right

3 0

2 years ago

Solve for y; 1/y = a/w + w/a

AVprozaik [17]

Do it yourself GET OFF THIS APPPPP just kidding idk the answer

8 0

3 years ago

Read 2 more answers

Triangle XYZ was dilated by a scale factor of 2 to create triangle ACB and cos ∠X = 2.5 over 5 and 59 hundredths. Triangles XYZ

adell [148]

Dilation involves changing the size of a shape

<h3>How to determine the relationship between the trigonometry ratios</h3>

From the question, we understand that:

The dilation of triangle XYZ gives triangle ACB

This means that, the following are corresponding points

Points X and A
Points Y and C
Points Z and B

The above means that:

The relationships between the trigonometry ratios of both triangles are:

cos(X) = cos(A), cos(Y) = cos(C) and cos(Z) = cos(B)

<h3>How to determine the measures of segments AC and AB</h3>

In (a), segments XY and XZ corresponds to the segments AC and AB

So, the product of the lengths of XY and XZ and the scale factor will result in the lengths of AC and AB,

i.e.

AC = 2 * XY and AB = 2 * XZ