Can someone answer this please

Mathematics

2 answers:

Nana76 [90]3 years ago

6 0

Area is trapezium =
A=1/2h(a+b)

84=1/2(6)(7+b)

84=3(7+b)

84=21+3b

3b=84-21

B=63/3

B= 21 cm

I’m sorry if my answer turned out to be wrong but I tried my best so best of luck !!

LiRa [457]3 years ago

5 0

Answer:

x = 14

Step-by-step explanation:

Please note, the word trapezium is a synonym for the word trapezoid.

This problem gives one the area of the trapezoid, a well as one of the measurements of a base and the height of the figure. One is asked to find the length of the other base. This can be done by using the formula to find the area of a trapezoid. This formula is the following,

$A=(h)(\frac{b_1+b_2}{2})$

Where (A) represents the area of a trapezoid, ( $b_1$ ) and ( $b_2$ ) represents the bases and (h) represents the height. Substitute in the given values and solve for the unknown base.

$b_1=7\\h=6\\A=84$

$A=(h)(\frac{b_1+b_2}{2})\\$

Substitute,

$84=6(\frac{7+b_2}{2})\\$

Inverse operations,

$84=6(\frac{7+b_2}{2})$

$14=\frac{7+b_2}{2}$

$28=7+b_2$

$14=b_2$

Read more about end behaviour at

brainly.com/question/1365136

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7 0

1 year ago

I need help please!<br> Ap calculus

Vika [28.1K]

9514 1404 393

Answer:

Step-by-step explanation:

For a lot of these summation problems it is worthwhile to learn to use a calculator or spreadsheet to do the arithmetic. Here, the ends of the intervals are 1 unit apart, so we only need to evaluate the function for integer values of x.

Almost any of these numerical integration methods involve some sort of weighted sum. For <em>trapezoidal</em> integration, the weights of all of the middle function values are 1. The weights of the first and last function values are 1/2. The weighted sum is multiplied by the interval width, which is 1 for this problem.

The area by trapezoidal integration is about 1.649 square units.

In the attached, we have shown the calculation both by computing the area of each trapezoid (f1 does that), and by creating the weighted sum of function values.