Help I need answers it's due today please help as fast as possible
1 answer:
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SA= Lateral Area + Area (of base)
Lateral area = perimeter of base x height
An isosceles triangle has two congruent base segments. The hypotenuse of an isosceles right triangle is (length of base segment) x (the square root of 2). I did the math for you (shown above) to get 5 as a base.
So 5 + 5 + square root of 50 is your perimeter
Multiply the perimeter times your height (8) and you get the lateral area.
Add your lateral area to the base area (triangle = 1/2 bh)
LA = 136.568...
Area of Triangle base = 12.5
Surface area = 149.068... square cm
Lateral area = perimeter of base x height
An isosceles triangle has two congruent base segments. The hypotenuse of an isosceles right triangle is (length of base segment) x (the square root of 2). I did the math for you (shown above) to get 5 as a base.
So 5 + 5 + square root of 50 is your perimeter
Multiply the perimeter times your height (8) and you get the lateral area.
Add your lateral area to the base area (triangle = 1/2 bh)
LA = 136.568...
Area of Triangle base = 12.5
Surface area = 149.068... square cm
Answer:
Step-by-step explanation:
Given
The ratio of mixture of sand and cement is 5:3
If the total volume of mixture is
Suppose there is 5x and 3x cubic centimeter of sand and cement respectively i.e.
Sand
So, there is of sand in the mixture
First of all, remember what the equation of a line is:
Where:
m=slope
b=y-intercept.
Given this information, we can start by plugging in what we know, we are given the slope which is 2 so we can fill that in, remember m=slope
Because we were given the point (14,3)
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We need to find the value of b. The 2 is already set and x and y are variables. We want the equation for the line that specifically passes through the given point (14,3).
<span>Plug in for x the number 14 and for y the number 3:</span>(14,3).
y=mx+b or 3=2 × 14+b,
solving for b: b=3-(2)(14). b=-25.<span>Plug that in for b:
</span><span>y=2x-25
<span>Final answer:
</span><span>y=2x-25</span></span><span><span>
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