Remark
This is quite a nice little problem. It takes a minute or three to figure out the answer, and when you do, you will be certain that you have been tricked. It is a little like the egg of Columbus.
Solution
The Base of Triangle ABN is AB
The Base of Triangle CDM is CD
The height of both given triangles is h. That is the distance between the two parallel lines.
Area ABN = 1/2*AB * h = 23 cm^2
Area CDM = 1/2*CD * h = 18 cm^2
Now the Area of the trapezoid is
Area_Trapezoid = 1/2 * h (AB + CD) Using the distributive property Remove the brackets.
Area_Trapezoid = 1/2*AB*h + 1/2*CD*h Did you notice something? Those terms are just the area of the triangles (written above.)
Area Trapezoid = 23 + 18 = 41 cm^2 <<<< Answer
Answer:
70°
Step-by-step explanation:
The total angle in a triangle is 180°. The total angle in a straight line is 180°. From the image attached, the ∠DCA must be greater than 45° because ∠BCA is greater than 90°. This means that ∠DCA is not 36° or 38° or 40° but it is 70°.
Therefore ∠DCA = 70°.
∠BCA + ∠DCA = 180° (angle on a straight line)
∠BCA = 180 - ∠DCA = 180 - 70 = 110°
∠BCA + ∠BAC + 40 = 180° (angle in a triangle)
∠BAC = 180 - 110 - 40 = 30°
Answer:
D (7=30x)
Step-by-step explanation:
Answer:
8
Step-by-step explanation:
<h2>Answer-Average rate of change(A(x)) of f(x) over a interval [a,b] is given by:</h2><h2 /><h2>A(x) = \frac{f(b)-f(a)}{b-a}A(x)= </h2><h2>b−a</h2><h2>f(b)−f(a)</h2><h2> </h2><h2> </h2><h2 /><h2>Given the function:</h2><h2 /><h2>f(x) = 20 \cdot(\frac{1}{4})^xf(x)=20⋅( </h2><h2>4</h2><h2>1</h2><h2> </h2><h2> ) </h2><h2>x</h2><h2> </h2><h2 /><h2>We have to find the average rate of change from x = 1 to x= 2</h2><h2 /><h2>At x = 1</h2><h2 /><h2>then;</h2><h2 /><h2>f(x) = 20 \cdot(\frac{1}{4})^1 = 5f(x)=20⋅( </h2><h2>4</h2><h2>1</h2><h2> </h2><h2> ) </h2><h2>1</h2><h2> =5</h2><h2 /><h2>At x = 2</h2><h2 /><h2>then;</h2><h2 /><h2>f(x) = 20 \cdot(\frac{1}{4})^2=20 \cdot \frac{1}{16} = 1.25f(x)=20⋅( </h2><h2>4</h2><h2>1</h2><h2> </h2><h2> ) </h2><h2>2</h2><h2> =20⋅ </h2><h2>16</h2><h2>1</h2><h2> </h2><h2> =1.25</h2><h2 /><h2>Substitute these in above formula we have;</h2><h2 /><h2>A(x) = \frac{f(2)-f(1)}{2-1}A(x)= </h2><h2>2−1</h2><h2>f(2)−f(1)</h2><h2> </h2><h2> </h2><h2 /><h2>⇒A(x) = \frac{1.25-5}{1}=-3.75A(x)= </h2><h2>1</h2><h2>1.25−5</h2><h2> </h2><h2> =−3.75</h2><h2 /><h2>therefore, average rate of change of the function f(x) from x = 1 to x = 2 is, -3.75</h2>
<h2>Please Mark me as brainlist. </h2>
