Answer:22/
9
= 2 4/
9
≅ 2.4444444
Step-by-step explanation:
Add: -19 + 9 = -10
Absolute value: abs(the result of step No. 1) = abs(-10) = 10
Divide: the result of step No. 2 / 3 = 10 / 3 = 3.3333333333333 = 10/
3
Subtract: 14 - 22 = -8
Absolute value: abs(the result of step No. 4) = abs(-8) = 8
Divide: the result of step No. 5 / 9 = 8 / 9 = 0.88888888888889 = 8/
9
Subtract: the result of step No. 3 - the result of step No. 6 = 10/
3
- 8/
9
= 10 · 3/
3 · 3
- 8/
9
= 30/
9
- 8/
9
= 30 - 8/
9
= 22/
9
Answer:
8710 units
Step-by-step explanation:
<em>Step 1: Write all the data</em>
Fixed cost: $9000
Average variable cost: 9.3 per unit
Total cost: 90,000
Total units: x
<em>Step 2: Find the total variable cost</em>
Average variable cost is per unit so it has to be multiplied by the number of units to find the total variable cost.
Total variable cost = average variable cost per unit x number of units
Total variable cost = 9.3x
<em>Step 3: Make the formula for finding x</em>
Total cost = total fixed cost + total variable cost
90,000 = 9000 + 9.3x
81000 = 9.3x
x = 8709.67
Rounded off to 8710 units
!!
Answer:
10-5![\sqrt{2}](https://tex.z-dn.net/?f=%5Csqrt%7B2%7D)
Step-by-step explanation:
As per the attached figure, right angled
has an inscribed circle whose center is
.
We have joined the incenter
to the vertices of the
.
Sides MD and DL are equal because we are given that ![\angle M = \angle L = 45 ^\circ.](https://tex.z-dn.net/?f=%5Cangle%20M%20%3D%20%5Cangle%20L%20%3D%2045%20%5E%5Ccirc.)
Formula for <em>area</em> of a
As per the figure attached, we are given that side <em>a = 10.</em>
Using pythagoras theorem, we can easily calculate that side ML = 10![\sqrt{2}](https://tex.z-dn.net/?f=%5Csqrt%7B2%7D)
Points P,Q and R are at
on the sides ML, MD and DL respectively so IQ, IR and IP are heights of
MIL,
MID and
DIL.
Also,
![\text {Area of } \triangle MDL = \text {Area of } \triangle MIL +\text {Area of } \triangle MID+ \text {Area of } \triangle DIL](https://tex.z-dn.net/?f=%5Ctext%20%7BArea%20of%20%7D%20%5Ctriangle%20MDL%20%3D%20%5Ctext%20%7BArea%20of%20%7D%20%5Ctriangle%20MIL%20%2B%5Ctext%20%7BArea%20of%20%7D%20%5Ctriangle%20MID%2B%20%5Ctext%20%7BArea%20of%20%7D%20%5Ctriangle%20DIL)
![\dfrac{1}{2} \times 10 \times 10 = \dfrac{1}{2} \times r \times 10 + \dfrac{1}{2} \times r \times 10 + \dfrac{1}{2} \times r \times 10\sqrt2\\\Rightarrow r = \dfrac {10}{2+\sqrt2} \\\Rightarrow r = \dfrac{5\sqrt2}{\sqrt2+1}\\\text{Multiplying and divinding by }(\sqrt2 +1)\\\Rightarrow r = 10-5\sqrt2](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B2%7D%20%5Ctimes%2010%20%5Ctimes%2010%20%3D%20%5Cdfrac%7B1%7D%7B2%7D%20%5Ctimes%20r%20%5Ctimes%2010%20%2B%20%5Cdfrac%7B1%7D%7B2%7D%20%5Ctimes%20r%20%5Ctimes%2010%20%2B%20%5Cdfrac%7B1%7D%7B2%7D%20%5Ctimes%20r%20%5Ctimes%2010%5Csqrt2%5C%5C%5CRightarrow%20r%20%3D%20%5Cdfrac%20%7B10%7D%7B2%2B%5Csqrt2%7D%20%5C%5C%5CRightarrow%20r%20%3D%20%5Cdfrac%7B5%5Csqrt2%7D%7B%5Csqrt2%2B1%7D%5C%5C%5Ctext%7BMultiplying%20and%20divinding%20by%20%7D%28%5Csqrt2%20%2B1%29%5C%5C%5CRightarrow%20r%20%3D%2010-5%5Csqrt2)
So, radius of circle = ![10-5\sqrt2](https://tex.z-dn.net/?f=10-5%5Csqrt2)
Answer:
1/3 or 0.33
Step-by-step explanation:
Hope it helps
LMK if it does and weather it is wrong or right
Answer:if you want grams than use metric if pounds use standard
Step-by-step explanation: