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4 years ago

What is the area of the trapezoid and how do you find it?

Mathematics

2 answers:

seraphim [82]4 years ago

8 0

You can use the formula 2(a+b $2(\frac{a+b}{2} )$
This will get you 161.5 ft

:)

Furkat [3]4 years ago

6 0

A= a+b/2 H Find it by take the sum of its bases, multiply the sum by the height of the trapezoid, and then divide the result by 2.

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2(b+3c) equivalent expressions Options: 3(b+2c) (b+3c) +(b+3c) None of the above

Lana71 [14]

2(b + 3c) → We first need to simplify this.

Simplify.

2b + 6c

1st Option :

3(b + 2c)

Simplify.

3b + 6c

This is INCORRECT, as it is not equal to 2b + 6c.

2nd Option :

(b + 3c) + (b + 3c)

Simplify.

2b + 6c

This is CORRECT because 2b+6c = 2b+6c

(b + 3c) + (b + 3c) → Answer

~Hope I helped!~

5 0

3 years ago

Write -12 7/8 as a decimal. (NEED ASAP WITH WORK)

seropon [69]

7/8 can be converted to .875 because 1/8 is .125 and .125*7=.875. Add -12 to it so the answer is -12.875

5 0

3 years ago

7-10/2 2 -13 12 -1.5

Luden [163]

Answer:

=14

Step-by-step explanation:

3 0

4 years ago

Write an equation in point-slope form of the line having the given slope that contains the given point. then graph the line

Natali5045456 [20]

Answer:

$y=-\frac{5}{3}x-3$

Step-by-step explanation:

Equation of straight line in point slope form is given a s

$(y-y_1)=m(x=x_1)$

Here

$m=-\frac{5}{2} \ and\ (x_1, y_1)=(0, -3)$

$(y+3)=-\frac{5}{3}(x-0)$

$y=-\frac{5}{3}x-3$

6 0

3 years ago

HELP ASAP Find the 15th term or the geometric sequence 7, -21, 63,

mihalych1998 [28]

<h3>Answer:</h3>

33480783

<h2>Step-by-step explanation:</h2>

Finding the Common Ratio:

We know that the formula for the common ratio of a GP is:

r = aₙ / (aₙ₋₁) (Where n is any integer)

replacing the values, taking n=2

r = -21 / 7

r = -3

Solving for the 15th Term:

We know that the Formula for the nth term of a GP:

G(n) = a * r ⁿ⁻¹ (where a is the first term and r is the common ratio)

replacing the values (taking n = 15 since we need the 15th term)

G(15) = 7 * (-3)¹⁴

G(15) = 33480783

Hence, the 15th term of the given GP is 33480783

5 0

3 years ago