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Eddi Din [679]
2 years ago
9

(9t + 1) + 7(1 + 9t) Complete the missing expressions!!

Mathematics
1 answer:
gavmur [86]2 years ago
8 0

Answer:

Step-by-step explanation:

Distribute:

=9t+1+(7)(1)+(7)(9t)

=9t+1+7+63t

Combine Like Terms:

=9t+1+7+63t

=(9t+63t)+(1+7)

=72t+8

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4=2(c-6)-4 solve for c
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4=2c-12-4
4=2c-16
20=2c
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3 years ago
The volume of a rectangular box can be found using the formula lwh where l represents the length, w represents the width, and h
Dafna1 [17]

Answer:

120 cm^3

Step-by-step explanation:

volume = lwh

I = 4 centimeters  

w = 5 centimeters  

h = 6 centimeters

volume = (4 cm)(5 cm)(6 cm)

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6 0
2 years ago
The submarine is underwater. It’s position is 645 feet below sea level. It rises in the water 100 feet and then descends 100 fee
kolbaska11 [484]

Hello!

This submarine rose 100 feet (+100) and then descended (-100). Usually in an equation this would cancel out and make 0. Therefore, our answer is 0.

I hope this helps!

5 0
2 years ago
Read 2 more answers
the sum of the interior angle in a polygon is 40,500° what is the number of sides in the regular polygon
larisa86 [58]

Answer:

227 sides

Step-by-step explanation:

The sum of the interior angles of a convex polygon is S=180(n-2) where n is the number of sides. If S=40500, then we have our equation 40500=180(n-2) and we can solve for n:

40500 = 180(n-2)

225 = n-2

227 = n

Therefore, the number of sides in the regular polygon is 227.

4 0
2 years ago
The line integral of (2x+9z) ds where the curve is given by the parametric equations x=t, y=t^2, z=t^3 for t between 0 and 1. Pl
Naya [18.7K]
Let r = (t,t^2,t^3)

Then r' = (1, 2t, 3t^2)

General Line integral is:
\int_a^b f(r) |r'| dt

The limits are 0 to 1
f(r) = 2x + 9z = 2t +9t^3
|r'| is magnitude of derivative vector \sqrt{(x')^2 + (y')^2 + (z')^2}

\int_0^1 (2t+9t^3) \sqrt{1+4t^2 +9t^4} dt

Fortunately, this simplifies nicely with a 'u' substitution.

Let u = 1+4t^2 +9t^4

du = 8t + 36t^3  dt

\int_0^1 \frac{2t+9t^3}{8t+36t^3} \sqrt{u}  du \\  \\ \int_0^1 \frac{2t+9t^3}{4(2t+9t^3)} \sqrt{u}  du \\  \\  \frac{1}{4} \int_0^1 \sqrt{u}  du

After integrating using power rule, replace 'u' with function for 't' and evaluate limits:
=\frac{1}{4} |_0^1 (\frac{2}{3}) (1+4t^2 +9t^4)^{3/2} \\  \\ =\frac{1}{6} (14^{3/2} - 1)
7 0
3 years ago
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