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Andrews
1 month ago
12

On a coordinate plane, 2 rectangles are shown. Rectangle P Q R S has points (1, 1), (1, 5), (3, 5), (1, 5). Rectangle P double-p

rime Q double-prime R double-prime S double-prime has points (negative 1, negative 1), (negative 5, negative 1), (negative 5, negative 3), (negative 1, negative 3). Which rule describes a composition of transformations that maps pre-image PQRS to image P"Q"R"S"? 270 degree rotation about point 0 composition. Translation of negative 2 units x, 0 units y. Translation of negative 2 units x, 0 units y composition 270 degree rotate about point 0. 270 degree rotation about point 0 composition reflected across the y-axis Reflected across the y-axis composition 270 degree rotation about point 0.

Mathematics
2 answers:
Inessa [12.5K]1 month ago
5 0

Answer:

The attached image provides the answer.

The previous response led me to a mistake, so I want to ensure you have the correct answer. ):)

Step-by-step explanation:

AnnZ [12.3K]1 month ago
5 0

Answer:

C.) T0,2 ∘ R0,270°(x, y)

Explanation:

I succeeded with this answer on Edgenuity; if this information was beneficial, please tap the heart button or leave a comment confirming it helped. Have a great day!:)

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Which graph represents the solution set of the compound inequality?<br><br> −5 &lt; a − 4 &lt; 2
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the third graph
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Which function in vertex form is equivalent to f(x) = x2 + 6x + 3?
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<span>f(x)=x^2+6x+3
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In ΔHIJ, the measure of ∠J=90°, JI = 4, HJ = 3, and IH = 5. What ratio represents the tangent of ∠I?
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Answer:

The ratio  \frac{3}{4} corresponds to the tangent of ∠I.

Step-by-step explanation:

Let’s revisit the trigonometric ratios:

  • sin Ф =  \frac{opposite}{hypotenuse}
  • cos Ф =  \frac{adjacent}{hypotenuse}
  • tan Ф =  \frac{opposite}{adjacent}

For triangle HIJ

∵ m∠J = 90°

- The hypotenuse is the side opposite the right angle.

So, HI is the hypotenuse.

∵ HJ = 3 units

∵ IH = 5 units

- We’ll apply the Pythagorean Theorem to solve for HJ.

∵ (HJ)² + (IJ)² = (IH)²

∴ 3² + (IJ)² = 5²

∴ 9 + (IJ)² = 25

- Subtract 9 from both sides.

∴ (IJ)² = 16

- Taking the square root on both sides gives:

∴ IJ = 4 units

To determine the tangent of ∠I, identify the sides that are opposite and adjacent to it.

∵ HJ is opposite to ∠I

∵ IJ is adjacent to ∠I

- Utilizing the rule of tan above:

∴ tan(∠I) = \frac{HJ}{IJ}

∴ tan(∠I) = \frac{3}{4}

The ratio \frac{3}{4} indicates the tangent of ∠I.

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