Hey there!
Step By Step:
Well so we know that 12 is about half of an 25. So what we are gonna do is to buy 3 of 12$ and 2 of 25$.
3x12=36 and 2x25=50.
With adding those two last numbers, you’ll get $86.
Graph would be down below.
Hope it helps :D
Two circles<span> of </span>radius<span> 4 are </span>tangent<span> to the </span>graph<span> of y^</span>2<span> = </span>4x<span> at the </span>point<span> (</span>1<span>, </span>2<span>). ... I know how to </span>find<span> the </span>tangent<span> line from a circle and a given </span>point<span>, but ... </span>2a2=42. a2=8. a=±2√2. Then1−xc=±2√2<span> and </span>2−yc=±2√2. ... 4 from (1,2<span>), so you could </span>find these<span> centers, and from there the</span>equations<span> of the circle
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Answer:
The perimeter (to the nearest integer) is 9.
Step-by-step explanation:
The upper half of this figure is a triangle with height 3 and base 6. If we divide this vertically we get two congruent triangles of height 3 and base 3. Using the Pythagorean Theorem we find the length of the diagonal of one of these small triangles: (diagonal)^2 = 3^2 + 3^2, or (diagonal)^2 = 2*3^2.
Therefore the diagonal length is (diagonal) = 3√2, and thus the total length of the uppermost two sides of this figure is 6√2.
The lower half of the figure has the shape of a trapezoid. Its base is 4. Both to the left and to the right of the vertical centerline of this trapezoid is a triangle of base 1 and height 3; we need to find the length of the diagonal of one such triangle. Using the Pythagorean Theorem, we get
(diagonal)^2 = 1^2 + 3^2, or 1 + 9, or 10. Thus, the length of each diagonal is √10, and so two diagonals comes to 2√10.
Then the perimeter consists of the sum 2√10 + 4 + 6√2.
which, when done on a calculator, comes to 9.48. We must round this off to the nearest whole number, obtaining the final result 9.
