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Dafna1 [17]
3 years ago
5

Suppose that the weight (in kilograms) of an airplane is a linear function of the amount of fuel (in liters) in its tank. When c

arrying 20 liters of fuel, the airplane weighs 2016 kilograms. When carrying 35 liters of fuel, it weighs 2028 kilograms. How much does the airplane weigh if it is carrying 45 liters of fuel?
Mathematics
1 answer:
Aleks [24]3 years ago
6 0
Bsjkanvshksndvkka vrkke henneknsvysm send
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HELP ME OUT PLEASE!!!
Natali [406]

Answer:

I think it's D) 3

Step-by-step explanation:

hope I helped

8 0
3 years ago
At the store, you buy a backpack for $30.42, a binder for $9.89 and a combination lock for $5.75. If you give the cashier $50.00
Shkiper50 [21]

Answer:

$3.94

Step-by-step explanation:

$30.42 + $9.89 + $5.75 = £46.06

$50.00 - $46.06 = $3.94

7 0
3 years ago
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tiny-mole [99]

Answer:

B

Step-by-step explanation:

this is the answer to this question

8 0
3 years ago
Find the first term and the common ratio <br>third term =6 and seventh term= 96​
mamaluj [8]

Answer:

First term a₁ = 3/2 and common ratio r = 2

Step-by-step explanation:

We need to find the first term and common ratio while we are given third term = 6 and seventh term = 96

Since common ratio is required so, the sequence is geometric sequence

The formula used is: a_n= a_1r^{n-1}

We are given: third term = 6 i,e

a_3=a_1r(3-1)\\6=a_1r^2----eq(1)

Seventh term = 96

a_7=a_1r(7-1)\\96=a_1r^6----eq(2)

Dividing eq(2) and eq(3)

\frac{96}{6}=\frac{a_1r^6}{a1r2}\\16=\frac{r^6}{r^2}\\16=r^{6-2}\\=> r^4=16\\Taking \ fourth \ root\\ r=2,-2,2i,-2i\\ \ We \ will \ consider \ only \ positive \ value \ of \ r \ i.e \ \textbf{ r= 2}

So, Common Ratio r = 2

Finding First term using eq(1)

a_3=a_1r^2\\6=a_1(2)^2\\6=a_1(4)\\a_1= \frac{6}{4}\\a_1= \frac{3}{2}

So, First term a₁ = 3/2 and common ratio r = 2

6 0
3 years ago
X = 10<br><br> x = 2<br><br> x = two divided by five.<br><br> x = 11
KengaRu [80]

Answer:

x = 10

Step-by-step explanation:

You can try the answers to see which works. (The first one does.)

Or, you can solve for the variable:

Divide by 75

... (1/5)^(x/5) = 3/75 = 1/25

Recognize that 25 = 5^2, so ...

... (1/5)^(x/5) = (1/5)^2

Equating exponents, you have

... x/5 = 2

... x = 10 . . . . . multiply by 5

_____

You can also start by taking logarithms:

... log(75) +(x/5)log(1/5) = log(3)

... (x/5)log(1/5) = log(3) -log(75) = log(3/75) = log(1/25) . . . . simplify the log

... x/5 = log(1/25)/log(1/5) = 2 . . . . . simplify (or evaluate) the log expression

... x = 10 . . . . . multiply by 5

_____

"Equating exponents" is essentially the same as taking logarithms.

5 0
3 years ago
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