Hey can you please help me posted picture of question

Mathematics

2 answers:

77julia77 [94]3 years ago

7 0

The correct answer is option B.

Solution is shown below

$y= 4 x^{3} \\ \\ 
 \frac{y}{4} = x^{3} \\ \\ 
 \sqrt[3]{\frac{y}{4}} =x \\ \\ 
f^{-1}(y)= \sqrt[3]{\frac{y}{4}} \\ \\ 
f^{-1}(x)= \sqrt[3]{\frac{x}{4}} \\ \\ 
y^{-1}= \sqrt[3]{\frac{x}{4}}$

natima [27]3 years ago

3 0

The correct answer is B

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In the diagram below, GH is parallel to DE . If GH is 3 more than EH FH=10, and DE=12 find the length of EH. Figures are not nec

steposvetlana [31]

Answer: 5

Step-by-step explanation:

Since $\overline{GH} \parallel \overline{DE}$ , by the corresponding angles theorem, $\angle FGH \cong \angle FDE, \angle FHG \cong \angle FED$ . This means $\triangle FGH \sim \triangle FDE$ by AA.

As corresponding sides of similar triangles are proportional,

$\frac{10}{x+3}=\frac{10+x}{12}\\(10)(12)=(10+x)(x+3)\\120=x^{2}+13x+30\\x^{2}+13x-90=0\\(x+18)(x-5)=0\\x=-18, 5$
However, as distance must be positive, we consider the positive solution, x=5.