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Mars2501 [29]
3 years ago
15

The area of a trapezoid is calculated using the formula below, where A is the area of the trapezoid, b1 and b2 are the bases of

the trapezoid, and h is the height of the trapezoid. Rewrite the formula to find the base b2.
Mathematics
2 answers:
MA_775_DIABLO [31]3 years ago
8 0

2(\frac{Area}{h} ) - b1 = b2

Step-by-step explanation:

The formula for calculating the area of a trapezoid is the following.

Area = \frac{b1+b2}{2} * h

What we are asked to find is the formula for finding b2. We can do this by rearranging the Area formula and getting b2 by itself on one side.

Area = \frac{b1+b2}{2} * h

\frac{Area}{h} = \frac{b1+b2}{2} .....divide h on both sides

2(\frac{Area}{h}) = b1+b2 .... multiply 2 on both sides

2(\frac{Area}{h} ) - b1 = b2 ....subtract b1 on both sides

Now we have b2 by itself on one side and the formula is rewritten to solve for b2.

andre [41]3 years ago
7 0

Answer:

b2=2(A)/h -b1

Step-by-step explanation:

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Dennis_Churaev [7]
Fruit Punch served = 4/9 x 4 = 1 7/9 liters

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Answer: 2 2/9 liters
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A blank is a fraction with a denominator of 1
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A whole number is a fraction with a denominator of one
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14 sandwiches and 3 drinks cost €33.75. 6 sandwiches and 17 drinks cost €26.25 . How much cost 1 sandwich and 1 drink
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8 0
2 years ago
A. Draw the graph of the function y = |x - 2| - 1. B. Label the vertex and all intercepts. Upload a copy of your handwritten gra
hjlf

Step-by-step explanation:

y = |x - 2| - 1

Given equation is in the form of y=|x-h|+k

Where (h,k) is the vertex

h = 2  and k =-1

So (2, -1) is the vertex for the given function

To find x intercepts we plug in 0 for y

we will get two values for x because of absolute function

0 = |x-2| -1

add 1 on both sides

|x-2| =1

x-2 =1                          and x-2 =-1

x= 3                          and x= 1

So x intercepts are (1,0)  and (3,0)

the graph is attached below


6 0
3 years ago
HELP ASAP I WILL GIVE BRAINLIEST!!
Alexxx [7]

Consider the function F(x)= x³ +½x² −2x +3. Use calc and algebra to find any and all stationary points on the graph of the function.Precalc need help ASAP

Solution:

the above function has more than one root or more than one solution. The derivative of the function represents the slope of the tangent line to the curve

F(x)= x³ +½x² −2x +3, F'(x) = 3(x3-1)+(1/2)(2)(x2-1)-2(x1-1)+0

F'(x)=3x2+x-2 is the slope of the tangent line

setting the slope =0 and solving the quadratic equations, you find all the

critical points of interest

3x2+x-2=0, solve by factoring (3x-2)(x+1)=0, then 3x-2=0, x= 2/3

and x+1=0, x=-1 substituting these values of x in the original equation

you get the corresponding values of F(x) such that;

F(x)= x³ +½x² −2x +3, F(x)=(2/3)3+1/2(2/3)2-2(2/3)+3

= 8/27+4/18-4/3+3 = 8/27+2/9-4/3+3/1 = 59/27 and the first point is

(2/3, 59/27)

F(x)= x³ +½x² −2x +3, F(x)=(-1)3+1/2(-1)2-2(-1)+3

= -1+1/2+2+3 = 9/2 and the 2nd point is ( -1, 9/2)

these two points represents points where the slope of the tangent line to the curve is 0 and could be either maximum or minimum points

You can also find the y intercepts where x = 0 F(x)= x³ +½x² −2x +3, setting x = 0, y= F(x)=3 and the point ( 0,3) is a y intercept setting y = 0, then x intercepts can be found upon solving the equation x³ +½x² −2x +3 =0

3 0
2 years ago
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