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Tju [1.3M]
3 years ago
7

the sum of three numbers is 144. the third number is 4 times the first. the second number is 6 more than the first. what are the

numbers?
Mathematics
1 answer:
Law Incorporation [45]3 years ago
7 0
First I would change the descriptions of the numbers into expressions.

first number is n
second number is n + 6
third number is 4n (4 x n)

Then you would insert these expressions into an equation and isolate n.

n + n + 6 + 4n = 144
n + n + 4n = 144 - 6
6n = 138
n = 138/6
n = 23

Lastly, you would plug in this value into all of the expressions.

first number is 23
second number is 23 + 6 = 29
third number is 4(23) = 92

Therefore, the numbers are 23, 29, and 92.
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Read 2 more answers
Find the derivative.
Aleksandr [31]

Answer:

Using either method, we obtain:  t^\frac{3}{8}

Step-by-step explanation:

a) By evaluating the integral:

 \frac{d}{dt} \int\limits^t_0 {\sqrt[8]{u^3} } \, du

The integral itself can be evaluated by writing the root and exponent of the variable u as:   \sqrt[8]{u^3} =u^{\frac{3}{8}

Then, an antiderivative of this is: \frac{8}{11} u^\frac{3+8}{8} =\frac{8}{11} u^\frac{11}{8}

which evaluated between the limits of integration gives:

\frac{8}{11} t^\frac{11}{8}-\frac{8}{11} 0^\frac{11}{8}=\frac{8}{11} t^\frac{11}{8}

and now the derivative of this expression with respect to "t" is:

\frac{d}{dt} (\frac{8}{11} t^\frac{11}{8})=\frac{8}{11}\,*\,\frac{11}{8}\,t^\frac{3}{8}=t^\frac{3}{8}

b) by differentiating the integral directly: We use Part 1 of the Fundamental Theorem of Calculus which states:

"If f is continuous on [a,b] then

g(x)=\int\limits^x_a {f(t)} \, dt

is continuous on [a,b], differentiable on (a,b) and  g'(x)=f(x)

Since this this function u^{\frac{3}{8} is continuous starting at zero, and differentiable on values larger than zero, then we can apply the theorem. That means:

\frac{d}{dt} \int\limits^t_0 {u^\frac{3}{8} } } \, du=t^\frac{3}{8}

5 0
3 years ago
A rectangular garden has an area of 24 square meters. The width is 8 meters less than twice the length. Find the dimensions of t
OLEGan [10]

Answer:

Length: 6 m

Width: 4 m

Step-by-step explanation:

Area: 24 m^{2}

Width: 2L - 8 m

Length: ?

Formula: A = W * L

Replace the variables:

24 = (2L - 8) * L

24 = 2L^{2} - 8L

0 = 2L^{2} - 8L - 24

or

2L^{2} - 8L - 24 = 0

Solve; if you don't know how to solve it using this method, let me know. It's the expression Ax^{2} +Bx+C

1. 2 (\frac{(2L^2 - 8L - 24)}{2})

2. \frac{(2L)^2 - 8(2L) - 48}{2}

3. \frac{(2L - 12)(2L + 4)}{2}

4. 2(\frac{(L - 6)(2L + 4)}{2})

5. (L-6) (2L+4)

6. L-6 = 0 AND 2L+4 = 0

L - 6 =0

L = 6

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

2L = -4

L = \frac{-4}{2}

L = -2

Since the length cannot be negative, the only real value is 6.

Area: 24 m^{2}

Width: 2L - 8 m

Length: 6 m

Calculate width:

W = 2L - 8

W = 2(6) - 8

W = 12 - 8

W = 4 m

3 0
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