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2 years ago

What is the area of the trapezoid shown below?

Mathematics

2 answers:

Arada [10]2 years ago

6 0

Answer: where is the trapezoid

Step-by-step explanation:

Alika [10]2 years ago

3 0

Area of a trapezoid = 1/2(base1 +base2)*h

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Help please now i cant figure it out

12345 [234]

Answer:

a. $\frac{-11}{24}$

b. $\frac{-3}{2} \\$

c. $\frac{31}{7}$

d. $\frac{11}{16}$

11, 3, 31, 11 = code

Step-by-step explanation:

a. $\frac{-1}{6}\\$ * 2 $\frac{3}{4}$ =

2 $\frac{3}{4}$ = $\frac{4*2+3 }{4}$ = $\frac{11}{4}$

$\frac{-1}{6}\\$ * $\frac{11}{4}$ = $\frac{-1*11}{6*4}$ = $\frac{-11}{24}$

b. $\frac{3}{16}$ / $\frac{-1}{8}$ =

$\frac{3}{16}$ * $\frac{8}{-1}$ = $\frac{3*8}{16*-1} \\$ = $\frac{24}{-16}$ = $\frac{-3}{2} \\$

c. $\frac{-31}{56}$ * $\frac{-8}{1}$ = $\frac{31}{7}$ (you can reduce 8 & 56 by dividing by 8: 56/8=7 & -8/8= -1)

d. 1 $\frac{7}{8}$ / 2 $\frac{8}{11}$ =

1 $\frac{7}{8}$ = $\frac{8*1+7}{8}$ = $\frac{15}{8}$

2 $\frac{8}{11}$ = $\frac{11*2+8}{11}$ = $\frac{30}{11}$

$\frac{15}{8}$ * $\frac{11}{30}$ = $\frac{11}{16}$ (you can reduce 15 & 30 by dividing by 15: 15/15=1 & 30/15 = 2)

4 0

2 years ago

The height H of an ball that is thrown straight upward from an initial position 3 feet off the ground with initial velocity of 9

mariarad [96]

Answer:

The ball will be 84 feet above the ground 1.125 seconds and 4.5 seconds after launch.

Step-by-step explanation:

Statement is incorrect. Correct form is presented below:

The height $h(t)$ of an ball that is thrown straight upward from an initial position 3 feet off the ground with initial velocity of 90 feet per second is given by equation $h(t) = 3 +90\cdot t -16\cdot t^{2}$ , where $t$ is time in seconds. After how many seconds will the ball be 84 feet above the ground.

We equalize the kinematic formula to 84 feet and solve the resulting second-order polynomial by Quadratic Formula to determine the instants associated with such height:

$3+90\cdot t -16\cdot t^{2} = 84$