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Eddi Din [679]
3 years ago
11

OeowowowkwiHELP ASaPieisowowiiwiwiwiiwiwisnsnsnxxj

Mathematics
1 answer:
puteri [66]3 years ago
6 0

Answer:

5

Step-by-step explanation:

40÷[20-4*(7-4)]

Start with the inner most parentheses

40÷[20-4*(3)]

Then the brackets, multiply first

40÷[20-12]

Then subtract

40÷[8]

We are now left with the division

5

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Can someone help me?
Kitty [74]

Answer:

see explanation

Step-by-step explanation:

Given the width is \frac{1}{3} of the length, then

y - 4 = \frac{1}{3}(2y + 6)

Multiply through by 3 to clear the fraction

3y - 12 = 2y + 6 ( subtract 2y from both sides )

y - 12 = 6 ( add 12 to both sides )

y = 18

Thus

length = 2y + 6 = 2(18) + 6 = 36 + 6 = 42 in

width = y - 4 = 18 - 4 = 14 in

6 0
2 years ago
In a circle with a radius of 5cm and is centered at point O, angle AOB intercepts arc AB. Arc AB has a length of 10cm. What is t
Elenna [48]
<span>circumference of circle = 2*pi*r given r = 5 so circumference = 31.42 now we know 360 degree is uniformly distributed on this circumference so x/360=10/31.42 so x= 114.65 so option C.) 114.65 is the answer</span>
4 0
2 years ago
(1,-7) with a slope of -5 as an equation
Liono4ka [1.6K]

Answer:

y=-5x-2

Step-by-step explanation:

y=mx+b

where m is the slope

y=-5x+b

to solve for b, which is the y intercept, plug in (1,-7)

-7=-5(1)+b

-7=-5+b

-2=b

y=-5x-2

7 0
2 years ago
A pile of cards contains 3 diamonds,
Shtirlitz [24]

Answer:G

Step-by-step explanation:

4/9 times 2/8= 1/9

6 0
1 year ago
Read 2 more answers
Solve the following differential equations or initial value problems. In part (a), leave your answer in implicit form. For parts
shepuryov [24]

Answer:

(a) (y^5)/5 + y^4 = (t^3)/3 + 7t + C

(b) y = arctan(t(lnt - 1) + C)

(c) y = -1/ln|0.09(t + 1)²/t|

Step-by-step explanation:

(a) dy/dt = (t^2 + 7)/(y^4 - 4y^3)

Separate the variables

(y^4 - 4y^3)dy = (t^2 + 7)dt

Integrate both sides

(y^5)/5 + y^4 = (t^3)/3 + 7t + C

(b) dy/dt = (cos²y)lnt

Separate the variables

dy/cos²y = lnt dt

Integrate both sides

tany = t(lnt - 1) + C

y = arctan(t(lnt - 1) + C)

(c) (t² + t) dy/dt + y² = ty², y(1) = -1

(t² + t) dy/dt = ty² - y²

(t² + t) dy/dt = y²(t - 1)

(t² + t)/(t - 1)dy/dt = y²

Separating the variables

(t - 1)dt/(t² + t) = dy/y²

tdt/(t² + t) - dt/(t² + t) = dy/y²

dt/(t + 1) - dt/(t(t + 1)) = dy/y²

dt/(t + 1) - dt/t + dt/(t + 1) = dy/y²

Integrate both sides

ln(t + 1) - lnt + ln(t + 1) + lnC = -1/y

2ln(t + 1) - lnt + lnC = -1/y

ln|C(t + 1)²/t| = -1/y

y = -1/ln|C(t + 1)²/t|

Apply y(1) = -1

-1 = ln|C(1 + 1)²/1|

-1 = ln(4C)

4C = e^(-1)

C = (1/4)e^(-1) ≈ 0.09

y = -1/ln|0.09(t + 1)²/t|

8 0
3 years ago
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