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Mazyrski [523]
3 years ago
8

Suppose that the height (in centimeters) of a candle is a linear function of the amount of time (in hours) it has been burning.

After 11 hours of burning, a candle has a height of 27.8 centimeters. After 26 hours of burning, its height is 24.8 centimeters. What is the height of the candle after 23 hours?
Mathematics
1 answer:
andrey2020 [161]3 years ago
8 0

Answer:

Hey! The answer is 25,4 centimeters.

Step-by-step explanation:

In a linear function you can use y=mx+n

So if x = hours and y= height.

You know that

\left \{ {{27,8=11m +n} \atop {24,8=26m +n}} \right.

So:

If you multiply (24,8=26m +n) x-1

It is: -24,8 = -26m -n

Adding up the other equation

-24,8 = -26m -n

<u>27,8 = 11m + n</u>

<u>3 = -15m</u>

<u>m= -0,2</u>

So

27,8 = 11 x 0,2 + n

27,8 = -2,2 + n

30 = n

This results that te linear function for this equation is:

y = -0,2x + 30

The answer is

y=23(-0,2) + 30

y=-4,6+30

y=25,4

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Anton [14]
5 quarters, 15 dimes

q + 10 = d
0.25q + 0.1d = 2.75

0.25q + 0.1(q + 10) = 2.75
0.25q + 0.1q + 1 = 2.75
0.35q = 1.75
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Check:

0.25(5)+ 0.1(15)= 2.75
1.25 + 1.5 = 2.75
2.75 = 2.75 :)
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2 years ago
Which expression is equivalent to the one in the picture?
Ne4ueva [31]

Answer:

d.

Step-by-step explanation:

To convert a root to a fraction in the exponent, remember this rule:

\sqrt[n]{a^{m}}=a^{\frac{m}{n}}

The index becomes the denominator in the fraction. (The index is the little number in front of the root, "n".) The original exponent remains in the numerator.

In this question, the index is 4.

The index is applied to every base in the equation under the root. The bases are 16, 'x' and 'y'.

\sqrt[4]{16x^{15}y^{17}} = (\sqrt[4]{16})(\sqrt[4]{x^{15}})(\sqrt[4]{y^{17}}) = (2)(x^{\frac{15}{4}}})(y^{\frac{17}{4}}) = 2x^{\frac{15}{4}}}y^{\frac{17}{4}}

To find the quad root of 16, input this into your calculator. Since 2⁴ = 16, \sqrt[4]{16} = 2.

For the "x" and "y" bases, use the rule for converting roots to exponent fractions. The index, 4, becomes the denominator in each fraction.

2x^{\frac{15}{4}}y^{\frac{17}{4}}

5 0
3 years ago
Determine whether the improper integral converges or diverges, and find the value of each that converges.
Brrunno [24]

Answer:

Converges at -1

Step-by-step explanation:

The integral converges if the limit exists, if the limit does not exist or if the limit is infinity it diverges.

We will make use of integral by parts to determine:

let:

u=x             dv=e^(2x)\cdot{dx}

du=dx         v=2\cdot{e^(2x)}

\int\limits^a_b {u} \, dv = uv -\int\limits^a_b {v} \, du

\int\limits^a_b {x\cdot{e^2^x} \, dx =2xe^2^x- \int\limits^a_b {2e^2^x} \, dx

\int\limits^a_b {xe^2^x} \, dx = 2xe^2^x-2e^2^x-C

We can therefore determine that if x tends to 0 the limit is -1

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5 0
3 years ago
5x-4y+-13 and 3x-4y+-11
drek231 [11]
5x + -4y = 13

Solving
-5x + -4y = 13

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '4y' to each side of the equation.
-5x + -4y + 4y = 13 + 4y

Combine like terms: -4y + 4y = 0
-5x + 0 = 13 + 4y
-5x = 13 + 4y

Divide each side by '-5'.
x = -2.6 + -0.8y

Simplifying
x = -2.6 + -0.8y
Simplifying
3x + -4y + -11 = 0

Reorder the terms:
-11 + 3x + -4y = 0

Solving
-11 + 3x + -4y = 0

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '11' to each side of the equation.
-11 + 3x + 11 + -4y = 0 + 11

Reorder the terms:
-11 + 11 + 3x + -4y = 0 + 11

Combine like terms: -11 + 11 = 0
0 + 3x + -4y = 0 + 11
3x + -4y = 0 + 11Combine like terms: 0 + 11 = 11
3x + -4y = 11

Add '4y' to each side of the equation.
3x + -4y + 4y = 11 + 4y

Combine like terms: -4y + 4y = 0
3x + 0 = 11 + 4y
3x = 11 + 4y

Divide each side by '3'.
x = 3.666666667 + 1.333333333y

Simplifying
x = 3.666666667 + 1.333333333y
3 0
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Varvara68 [4.7K]
Its 2.456*10=11

hope this helps and please rate below and also i am great with helping people who are going through a lot like family or school or friendship problems so if u want help or tips about those things or even depression feel free to ask me through message and i will see what i can do to help you. :)
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