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Arisa [49]
2 years ago
7

The hole for a support needs to be 6 feet deep. It is currently 2 feet 5 inches deep. How much deeper must the hole​ be? Use pen

cil and paper. Show two ways to solve this problem. Use the conversion factor
12 inches
________
1 foot.
The hole needs to be nothing inches deeper.
Mathematics
2 answers:
BigorU [14]2 years ago
6 0

Answer:

43 inches deeper and it will be 6 feet deep

weeeeeb [17]2 years ago
6 0

Answer:

3feet 7inches

Step-by-step explanation:

2feet + 3feet=5feet  5inches+7inches =1 feet

5feet + 1 feet=6feet

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nekit [7.7K]

Answer:

10ft

Step-by-step explanation:

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2 years ago
777 watermelons cost \$8.96$8.96dollar sign, 8, point, 96.
julia-pushkina [17]

Answer:

a)

Step-by-step explanation:

x/5 = $8.96/7 a)

5 0
3 years ago
Read 2 more answers
Please answer number 29 ASAP
Effectus [21]

h = 5b - 2

\frac 1 2 b h= 36

b(5b - 2) = 72

5b^2 - 2b - 72 = 0

(b - 4)(5b + 18) = 0

b = 4 \textrm{ or }b = -18/5

We can rule out the negative length.

h = 72/b = 72/4 = 18

Answer: b=4, h=18


3 0
3 years ago
Read 2 more answers
What is 8 2/5 as a decimal number
Firdavs [7]

Answer:

8.4

Step-by-step explanation:

8 is a whole njmber, so it stays the same.

2/5 is equivalent to .4, so you put that at the end.

Your answer is 8.4 Hope this helps!

7 0
3 years ago
Unsure how to do this calculus, the book isn't explaining it well. Thanks
krok68 [10]

One way to capture the domain of integration is with the set

D = \left\{(x,y) \mid 0 \le x \le 1 \text{ and } -x \le y \le 0\right\}

Then we can write the double integral as the iterated integral

\displaystyle \iint_D \cos(y+x) \, dA = \int_0^1 \int_{-x}^0 \cos(y+x) \, dy \, dx

Compute the integral with respect to y.

\displaystyle \int_{-x}^0 \cos(y+x) \, dy = \sin(y+x)\bigg|_{y=-x}^{y=0} = \sin(0+x) - \sin(-x+x) = \sin(x)

Compute the remaining integral.

\displaystyle \int_0^1 \sin(x) \, dx = -\cos(x) \bigg|_{x=0}^{x=1} = -\cos(1) + \cos(0) = \boxed{1 - \cos(1)}

We could also swap the order of integration variables by writing

D = \left\{(x,y) \mid -1 \le y \le 0 \text{ and } -y \le x \le 1\right\}

and

\displaystyle \iint_D \cos(y+x) \, dA = \int_{-1}^0 \int_{-y}^1 \cos(y+x) \, dx\, dy

and this would have led to the same result.

\displaystyle \int_{-y}^1 \cos(y+x) \, dx = \sin(y+x)\bigg|_{x=-y}^{x=1} = \sin(y+1) - \sin(y-y) = \sin(y+1)

\displaystyle \int_{-1}^0 \sin(y+1) \, dy = -\cos(y+1)\bigg|_{y=-1}^{y=0} = -\cos(0+1) + \cos(-1+1) = 1 - \cos(1)

7 0
1 year ago
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