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WINSTONCH [101]
3 years ago
14

I need help please thank you

Mathematics
1 answer:
Lelu [443]3 years ago
3 0
I know y is a 90 degree angle. but i dont know what x is sorry.
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What is the distance from the school to the theater?
defon
The "horizontal" distance is 8-2 = 6.
The "vertical" distance is 7-3 = 4.
The Pythagorean theorem tells you the straight line distance is then
  √(6² + 4²) = √52 = 2√13 ≈ 7.21
8 0
4 years ago
15.5 decimeters but in meters
Liula [17]

Answer:

1.55 meters

Step-by-step explanation:

Used google conversion. But there are 10 meters in a decimeters, so just divide.

4 0
3 years ago
Read 2 more answers
25, 87, 90, 25, 67
arlik [135]

Answer:

mean number 1 25+87+90+25+67=294 294/5 58.8 question 2 500+100+300+300+500+100=1800 1800/6 =300

8 0
3 years ago
6.7 = 9.7 -0.5x<br><br> How do I solve this?
ra1l [238]

Answer:

Simplifying x + 6.7 = -9.7 Reorder the terms:

6.7 + x = -9.7  Solving 6.7 + x = -9.7 Solving for variable 'x'. Move all terms containing x to the left, all other terms to the right. Add '-6.7' to each side of the equation.

6.7 + -6.7 + x = -9.7 + -6.7 Combine like terms: 6.7 + -6.7 = 0.0

0.0 + x = -9.7 + -6.7

x = -9.7 + -6.7 Combine like terms: -9.7 + -6.7 = -16.4

x = -16.4 Simplifying x = -16.4

Step-by-step explanation:

Hope this helps

6 0
3 years ago
In △ABC, point M is the midpoint of
yulyashka [42]

Answer: The area of triangle BMC is 28 yd² . The area of triangle AMD is 8 yd². The area of CMD is 20 yd².

Explanation:

It is given that the M is the midpoint of the side AB. The line MC is the median of the triangle ABC.

A median divides the area of triangle in two equal parts, therefore the area of triangle BMC is half of the area of triangle ABC and area of triangle BMC and area of triangle AMC is equal.

\text{ Area of }\triangle BMC =\frac{1}{2}\times \text{ Area of }\triangle ABC}

\text{ Area of }\triangle BMC =\frac{1}{2}\times 56}

\text{ Area of }\triangle BMC =28

Therefore the area of triangle BMC and triangle AMC is 28 yd².

Draw a perpendicular on AD from M as shown in the figure.

\frac{\text{ Area of }\triangle AMD}{\text{ Area of }\triangle AMC}= \frac{\frac{1}{2}\times AD\times ME}{\frac{1}{2}\times AC\times ME} =\frac{AD}{AC}= \frac{2}{7}

Therefore the area of AMD is  \frac{2}{7}th  part of the area of AMC.

\text{ Area of }\triangle AMD =\frac{2}{7}\times \text{ Area of }\triangle AMC}

\text{ Area of }\triangle AMD =\frac{2}{7}\times 28

\text{ Area of }\triangle AMD =8

Therefore the area of triangle AMD is 8 yd².

\text{ Area of }\triangle CMD=\text{ Area of }\triangle ABC-\text{ Area of }\triangle AMD-\text{ Area of }\triangle BMC

\text{ Area of }\triangle CMD=56-8-28=20

Therefore the area of triangle CMD is 20 yd².

4 0
4 years ago
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