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IRISSAK [1]
2 years ago
14

Figure A is a scale image of Figure B

Mathematics
1 answer:
valina [46]2 years ago
5 0

Answer:

x = 30

Step-by-step explanation:

Put in proportions \frac{x}{45} = \frac{18}{27}

Cross multiply and get 27x = 45 x 18

OR...

27x = 810.

THEN...

divide by 27 on both sides to get 30

x = 30

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Helpp please 100 points
omeli [17]

Answer:

z=7 and x=60

Step-by-step explanation:

well from the square that is combining the angles 1 and 2, they tell you that that is a right angle and you know that angle 2 is 38 so 90-38=52 which is the measurement of angle 1 so to find z, you would have to divide 38/-2 which is -19 and then add 12 so 12-19=-7 so z=7. so you know that one side =90 degrees so angle 3 must also =90 degrees because they are on the same line and a line is equal to 180, so all you have to do is reverse the operations in 1 1/2x to find x, so 1 1/2 or 3/2 divided by 90 is 60.

pls this took a lot of work and i kindly ask for brainliest.

8 0
2 years ago
Read 2 more answers
A helicopter makes a round trip flight that lasts 4.5 hrs. if the average rate out was 80 miles per hour and the average rate in
zlopas [31]

Answer:

Let d = the one-way trip

Write a time equation: time = dist/speed

time out + time back = 4.5 hrs

d/80 + d/100 = 4.5

Multiply by 400 to clear the denominators

400*d/80 + 400*d/100 = 400(4.5)

Cancel the denominators

5d + 4d = 1800

d = 1800/9

d = 200 miles max distance

3 0
3 years ago
A circle passes through points A(7,4), B(10,6), C(12,3). Show that AC must be the diameter of the circle.
Artist 52 [7]

so we have three points, A, B and C, if indeed AC is the diameter of the circle, then half the distance of AC is its radius, and the midpoint of AC is the center of the circle, morever, since B is also on the circle, the distance from B to the center must be the same radius distance.

in short, half the distance of AC must be equals to the distance of B to the midpoint of AC, if indeed AC is the diameter.

\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ A(\stackrel{x_1}{7}~,~\stackrel{y_1}{4})\qquad C(\stackrel{x_2}{12}~,~\stackrel{y_2}{3}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{12+7}{2}~~,~~\cfrac{3+4}{2} \right)\implies \left( \cfrac{19}{2}~~,~~\cfrac{7}{2} \right)=M\impliedby \textit{center of the circle}

now, let's check the distance from say A to the center, and check the distance of B to the center, if it's indeed the center, they'll be the same and thus AC its diameter.

\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ A(\stackrel{x_1}{7}~,~\stackrel{y_1}{4})\qquad M(\stackrel{x_2}{\frac{19}{2}}~,~\stackrel{y_2}{\frac{7}{2}})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ AM=\sqrt{\left( \frac{19}{2}-7 \right)^2+\left( \frac{7}{2}-4 \right)^2} \\\\\\ AM=\sqrt{\left( \frac{5}{2}\right)^2+\left( -\frac{1}{2} \right)^2}\implies \boxed{AM\approx 2.549509756796392} \\\\[-0.35em] ~\dotfill

\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ B(\stackrel{x_1}{10}~,~\stackrel{y_1}{6})\qquad M(\stackrel{x_2}{\frac{19}{2}}~,~\stackrel{y_2}{\frac{7}{2}}) \\\\\\ BM=\sqrt{\left( \frac{19}{2}-10 \right)^2+\left( \frac{7}{2}-6 \right)^2} \\\\\\ BM=\sqrt{\left( -\frac{1}{2}\right)^2+\left( -\frac{5}{2} \right)^2}\implies \boxed{BM\approx 2.549509756796392}

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