The elevations of five points below sea level are given here. Which elevation is closest to sea level?

What is the area of this trapezoid? 175 in² 140 in² 129 in² 85 in²

aksik [14]

Area of trapezoid = (B1 + B2)h/2

B1 = 15 in. + 4 in. + 6 in. = 25 in.
B2 = 15 in.
h = 7 in.

Area = (25 in. + 15 in.)(7 in.)/2 = 140 in.^2

3 0

2 years ago

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Please help asap!! Graph (x, y): x + 2y 6

sergiy2304 [10]

The answer is (, )=(6−2, )

5 0

3 years ago

Expand the given power by using Pascal’s triangle. (9a - 10b)^6

xxMikexx [17]

Answer:

$531441a^6-3542940a^5b+9841500a^4b^2-14580000a^3b^3+12150000a^2b^4-5400000ab^5+1000000b^6$

Step-by-step explanation:

1 n=0

1 1 n=1

1 2 1 n=2

1 3 3 1 n=3

1 4 6 4 1 n=4

1 5 10 10 5 1 n=5

1 6 15 20 15 6 1 n=6

This is where n is the exponent in

$(x+y)^n$ .

$(x+y)^6=1x^6+6x^5y+15x^4y^2+20x^3y^3+15x^2y^4+6xy^5+1y^6$

Now we want to expand:

$(9a-10b)^6$ or we we can rewrite as $(9a+(-10b))^6$ .

Let's replace $x$ with $(9a)$ and $y$ with $(-10b)$ in the expansion:

$(x+y)^6=1x^6+6x^5y+15x^4y^2+20x^3y^3+15x^2y^4+6xy^5+1y^6$

$((9a)+(-10b))^6$

$=1(9a)^6+6(9a)^5(-10b)+15(9a)^4(-10b)^2+20(9a)^3(-10b)^3+15(9a)^2(-10b)^4+6(9a)(-10b)^5+1(-10b)^6$

Let's simplify a bit:

$=9^6a^6-60(9)^5a^5b+15(-10)^2(9)^4a^4b^2+20(9)^3(-10)^3a^3b^3+15(9)^2(-10)^4a^2b^4+6(9)(-10)^5ab^5+(-10)^6b^6$

$=531441a^6-3542940a^5b+9841500a^4b^2-14580000a^3b^3+12150000a^2b^4-5400000ab^5+1000000b^6$

8 0

2 years ago

How to find the median

KengaRu [80]

We can find the median by using the formula:

(N+1/2) th item

For example:

Given data: 4,6,3,5,7

We have to arrange the data into ascending order:

=3,4,5,6,7

N(total no of items)=5

Now,

Median=(N+1/2)th item

 =(5+1/2) th item

 =(6/2) th item

 = 3rd item

Median=5

Hope it helps

Good luck on your assignment

3 0

2 years ago

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Point b is between points a and

Kisachek [45]

2x-3+7x-10=32
9x-13=32
9x=32+13
9x=45
x=5

7 0

2 years ago