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koban [17]
3 years ago
14

What set of transformations are applied to parallelogram ABCD to create A'B'C'D'?

Mathematics
1 answer:
allochka39001 [22]3 years ago
3 0

Answer:

  Reflection across the x-axis

Step-by-step explanation:

The only apparent transformation is negation of the y-coordinate, corresponding to reflection across the x-axis.

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Evaluate the expression when y=17 and z=7.
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(PLEASE ANSWER!!) a college student completed some courses worth 4 credits and some courses worth 3 credits. the student earned
Ierofanga [76]

The college student took 13 courses of 3 credit hours.

Step-by-step explanation:

Given,

Total credits = 59

Total courses = 18

Let,

Number of 3 credit hour courses = x

Number of 4 credit hour courses = y

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Subtracting Eqn 2 from Eqn 3

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The college student took 13 courses of 3 credit hours.

Keywords: linear equation, subtraction

Learn more about linear equations at:

  • brainly.com/question/10708697
  • brainly.com/question/10710410

#LearnwithBrainly

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3 years ago
Find the value of x and y
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Step-by-step explanation:

\triangle ABC \sim\triangle DEF..(Given)  \\ \therefore \frac{AB}{DE} =\frac{BC}{EF}=\frac{AC}{DF}..(CSST)\\\\\therefore \frac{6}{4} =\frac{7}{x}=\frac{8}{y}\\\\ \therefore \frac{6}{4} =\frac{7}{x} \\ \therefore \: x =  \frac{7 \times 4}{6}  \\  \\\therefore \: x =  \frac{28}{6} =  \frac{14}{3}    \\  \\ \huge \red{ \boxed{ \therefore \: x =  \frac{14}{3} }} \\   \\  \because \: \frac{6}{4} =\frac{8}{y} \\  \\ \therefore \:y =  \frac{4 \times 8}{6}  \\  \\ \therefore \:y =  \frac{32}{6}  \\  \\ \huge \purple{ \boxed{ \therefore \:y =  \frac{16}{3}  }}\\  \\

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Derivative of y=cos²x-sin²x
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One thing to bear in mind with trigonometric expressions is that, the exponent location is next to the function name, instead of the whole expression, though it applies to the whole thing, namely cos²(x), is really [ cos(x) ]², and that matters for using the chain rule.

\bf y=cos^2(x)-sin^2(x)\implies y=[cos(x)]^2-[sin(x)]^2
\\\\\\
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\cfrac{dy}{dx}=-2cos(x)sin(x)-2cos(x)sin(x)
\\\\\\
\cfrac{dy}{dx}=-4cos(x)sin(x)
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