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laila [671]
3 years ago
6

20p + 9t=44.4 solve for p

Mathematics
2 answers:
lana66690 [7]3 years ago
7 0
20p + 9t = 44.4

Subtract 9t from both sides:

20p = 44.4 - 9t

Divide both sides by 20:

p = 2.22 - 0.45t
horsena [70]3 years ago
5 0
Divide both sides by 20
p = 2.22 - 0.45t
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There are three consecutive even integers. If twice the first integer added to the second is 268, 244 find all three integers.
Rudik [331]

Answer:

The numbers would be 89,414 , 89,416, and 89,418

Step-by-step explanation:

In order to find this, start by setting the lowest number as x. Then we can determine that the next two even integers are x + 2 and x + 4. From this we can write the following equation.

2(x) + x + 2 = 268,244

3x + 2 = 268,244

3x = 268,242

x = 89,414

Since the first number is 89,414, we know the next two are 89,416 and 89,418.

6 0
3 years ago
find the first fourth and tenth terms of the arithmetic sequence described by the given rule a (n)=5+(n-1)(1/2)
gulaghasi [49]
An=a1+d(n-1)
a1=first term
d=common differnce

easy

inptu 4 for n and 10

a4=5+(4-1)1/2
a4=5+3(1/2)
a4=5+3/2
a4=6 and 1/2
4th term is 6 and 1/2


10th term
a10=5+(10-1)1/2
a10=5+(9)1/2
a10=5+9/2
a10=9 and 1/2




4th term is 6 and 1/2
10th term is 9 and 1/2

5 0
4 years ago
NEED HELP ASAP!!
pantera1 [17]

Answer:

t=1/30 h = 2 min

Step-by-step explanation:

d = vt

system of equations:

d = 13t

d - 2/5 = t       ->     d = t+2/5

Solve

13t = t +2/5

t = 1/30 h = 60 min / 30 = 2 min

8 0
3 years ago
(sinx-1)(sinx+cos^2x) multiply and simplify
sveta [45]
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sin^2(x)+sin(x)cos^2(x)-sin(x)-cos^2(x)
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[sin^2(x)-cos^2(x)]+[sin(x)cos^2(x)-sin(x)]
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-[cos^2(x)-sin^2(x)]-[sin(x)-sin(x)cos^2(x)]
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-[\stackrel{cos(2x)}{cos^2(x)-sin^2(x)}]-sin(x)\stackrel{sin^2(x)}{[1-cos^2(x)]}
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-cos(2x)-sin(x)sin^2(x)\implies -cos(2x)-sin^3(x)
7 0
3 years ago
Kindly solve this question 6 too !!
Shkiper50 [21]

Answer:

x=1

Step-by-step explanation:

~~~~~x = \dfrac{1}{2 - \dfrac 1{2-\dfrac{1}{2-x}}}\\\\\\\implies x = \dfrac{1}{2 - \dfrac 1{\dfrac{4-2x-1}{2-x}}}\\\\\\\implies x= \dfrac{1}{2 - \dfrac{2-x}{3-2x}}\\\\\\\implies x = \dfrac{1}{\dfrac{6-4x-2+x}{3-2x}}\\\\\\\implies x = \dfrac{3-2x}{4-3x}\\\\\\\implies 4x -3x^2 = 3-2x\\\\\\\implies 3x^2 -4x +3-2x = 0\\\\\\\implies 3x^2 -6x +3 = 0\\\\\\\implies 3(x^2 -2x +1) =0\\\\\\\implies x^2 -2x +1 = 0\\\\\\\implies (x-1)^2 = 0\\\\\\\implies x -1 = 0\\\\\\\implies x = 1

7 0
3 years ago
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