Given:
square with sides measuring 7 cm.
3 triangles attached to three sides of the square. A line bisecting one triangle is measured at 4 cm.
Area of a square = s² = (7cm)² = 49 cm²
Area of a triangle = hb/2 = (4cm*7cm)/2 = 14 cm²
Area of the 3 triangles = 14 cm² x 3 = 42 cm²
Total area of the logo = 49 cm² + 42 cm² = 91 cm²
Answer: 377
1,400 - 1,023 = 377
1,023 + 377 = 1,400
Step-by-step explanation: easy
Hello!
Trapezoid ABCD is shown. A is at negative 5, 1. B is at negative 4, 3. C is at negative 2, 3. D is at negative 1, 1.
A) x-axis, y=x, y-axis, x-axis
b) x-axis, y-axis, x-axis
c) y=x, x-axis, x-axis
d) y-axis, x-axis, y-axis, x-axis
The best answer is C180 rotation wud take that point to 4th quadrant
reflection in x-axis takes that to 1st quadrant
<span>reflection in y-axis brings it back to 2nd quadrant again. So, the sequence of transformations will bring A back to where it started
</span>
Hope this Helps! :)
Answer:
45
Step-by-step explanation:
To find the number you just add 6 to 39 which gets you 45.
Answer:
Step-by-step explanation:
Given that :
The enormous sinkhole that appeared in the middle of Guatemalan neighborhood in the year 2010 swallowed a three story building above it and the sinkhole has an estimated depth of about 100 feet
How much material is needed to fill the sinkhole?
Then , the material needed to fill the sinkhole will be dependent on the volume of the sinkhole.
Determine what information is needed to answer the question.
The information that is needed to answer this question are:
the base area of the sinkhole will be required; &
the height should also be known
if the base area is determined then we can proceed to determine how much material is needed to be calculated.
Do you think your estimate is more likely to be too high or too low
No, the estimate is not likely to be too high or too low rather the estimate of the material required will be almost equal to the volume of the sinkhole, but that does not implies that it will be exactly the same since the sinkhole is not uniform and regular in shape.
