The distance from the center of dilation, P, to.the image vertice S' is; 6 units.
<h3>What is the distance from the center of dilation, P, to the image S'?</h3>
It follows from the task content that the center of dilation of the triangle QRS is point P and the length of segment PS in the pre-image is; 8 units.
Hence, since the dilation factor as given in the task content is; three-fourths, it therefore follows that the distance of point P to S' in the image is; (3/4) × 8 = 6units.
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Answer:
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Step-by-step explanation:
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If you sketch the man and the building on paper, you'll have a
right triangle. The right angle is the point where the wall of
the building meets the ground. The height of the building
is one leg of the triangle, the line on the ground from the
building to the man's feet is the other leg, and the line
from his feet to the top of the building is the hypotenuse.
We need to find the angle at his feet, between the hypotenuse
and the leg of the triangle.
Well, the side opposite the angle is the height of the building -- 350ft,
and the side adjacent to the angle is the distance from him to the
building -- 1,000 ft.
The tangent of the angle is (opposite) / (adjacent)
= (350 ft) / (1,000 ft) = 0.350 .
To find the angle, use a book, a slide rule, a Curta, or a calculator
to find the angle whose tangent is 0.350 .
tan⁻¹(0.350) = 19.29° . (rounded)
let's first off take a look at the <u>tickmarks</u>, three <u>side tickmarks</u>, so all those 3 sides are equal, all have a length of y - 25, so is an equilateral triangle.
there are two <u>angle tickmarks</u>, meaning those two angles are equal, wait a second! if those two angles are equal, that means is an isosceles triangle.
now, in an equilateral triangle, all sides are equal, but also all angles are equal, since the sum of all interior angles is 180°, then each angle is really 60°.
let's notice that angle on the upper-left-corner, is a right-angle, but 60° are on the equilateral triangle, and so the remaining 30° must be on the isosceles triangle.
the isosceles triangle has then a vertex of 30°, and twin angles, the twin angles let's say are each a° so then
30° + a° + a° = 180°
30 + 2a = 180
2a = 150
a = 75° = y
now, let's recall, the isosceles triangle has twin angles but it also has twin sides, so the side "x" and the side with the tickmark are equal.
well, we know that y = 75, so the sides with the tickmark are then (75) - 25 = 50 = x.
![\sqrt[3]{4}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B4%7D)
