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3 years ago

One number is eight less than a second number. Twice the first is 12 more than 4 times the second. Find the numbers.

Mathematics

1 answer:

tia_tia [17]3 years ago

6 0

Answer:

Step-by-step explanation:

"One number is eight less than a second number."

x = y-8

"Twice the first is 12 more than 4 times the second."

2x = 4y+12

substitution

2(y-8) = 4y + 12

2y - 16 = 4y + 12

2y = -28

y = -14

x = y-8 = -22

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A collection of 77 quarters and dimes is worth $12.50. How many quarters are in the collection? PLS HELP ILL GIVE YOU BRAINLEST

mrs_skeptik [129]

Answer:

Step-by-step explanation:

77 quarters and dimes

q + d = 77

subtract q from both sides

d = 77 - q

---------------------------

A collection of quarters and dimes is worth $12.50 (1250 cents)

25q + 10d = 1250

substitute for d

25q + 10(77 - q) = 1250

distribute

25q + 770 - 10q = 1250

Combine like terms

15q + 770 = 1250

subtract 770 from both sides

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divide both sides by 15

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3 years ago

The area (A) of a circle with a radius of r is given by the formula and its diameter (d) is given by . Arrange the equations in

ella [17]

Answer:

see explanation

Step-by-step explanation:

A = πr² ← r is the radius and r = $\frac{1}{2}$ d ← d is the diameter

Hence

π ( $\frac{1}{2}$ d)² = A, that is

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7 0

3 years ago

5 cakes cost £4 how much do 7 cakes cost

Nostrana [21]

You can solve this using proportion.

5 = 4

7 = ?

7(4)/5= 5.6

7 cakes cost £5.6

4 0

3 years ago

Read 2 more answers

A factory packed 9600kg of flour in to small bags and large bags. Each small bag contained 3kg of flour and each large bag conta

yawa3891 [41]

Answer:

696 large bags of flour were packed.

Step-by-step explanation:

Let large bags of flour be 'y'

Let small bags of flour be 'x'. That is x=2040 (according to the question)

$(3 \times x) + (5 \times y) = 9600$

$(3 \times 2040) + ( 5 \times y) = 9600$

$(6120) + (5 \times x) = 9600$

$(5 \times x) = 9600 - 6120$

$(5 \times x) = 3480$

$x = 3480 \div 5$

$x = 696$

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#PEACE

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3 years ago

Am i correct? calculus

Advocard [28]

Check the picture below.

now, the "x" is a constant, the rocket is going up, so "y" is changing and so is the angle, but "x" is always just 15 feet from the observer. That matters because the derivative of a constant is zero.

now, those are the values when the rocket is 30 feet up above.

$\bf tan(\theta )=\cfrac{y}{x}\implies tan(\theta )=\cfrac{y}{15}\implies tan(\theta )=\cfrac{1}{15}\cdot y
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\stackrel{chain~rule}{sec^2(\theta )\cfrac{d\theta }{dt}}=\cfrac{1}{15}\cdot \cfrac{dy}{dt}\implies \cfrac{1}{cos^2(\theta )}\cdot\cfrac{d\theta }{dt}=\cfrac{1}{15}\cdot \cfrac{dy}{dt}
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\boxed{\cfrac{d\theta }{dt}=\cfrac{cos^2(\theta )\frac{dy}{dt}}{15}}\\\\
-------------------------------\\\\
$

$\bf cos(\theta )=\cfrac{adjacent}{hypotenuse}\implies cos(\theta )=\cfrac{15}{15\sqrt{5}}\implies cos(\theta )=\cfrac{1}{\sqrt{5}}\\\\
-------------------------------\\\\
\cfrac{d\theta }{dt}=\cfrac{\left( \frac{1}{\sqrt{5}} \right)^2\cdot 11}{15}\implies \cfrac{d\theta }{dt}=\cfrac{\frac{1}{5}\cdot 11}{15}\implies \cfrac{d\theta }{dt}=\cfrac{\frac{11}{5}}{15}\implies \cfrac{d\theta }{dt}=\cfrac{11}{5}\cdot \cfrac{1}{15}
\\\\\\
\cfrac{d\theta }{dt}=\cfrac{11}{75}$

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3 years ago