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Aleksandr-060686 [28]
3 years ago
10

What is the what is the lowest common multiple of 7 and 4

Mathematics
2 answers:
GalinKa [24]3 years ago
8 0
2x2x7=28
I hope this is correct
:)
Lyrx [107]3 years ago
5 0

28,

7*1=7/4=nope:(

7*2=14/4=nope

7*3=21/4=nope

7*4=28/4=YES

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Please help this is due today at 2pm Thanks!
Alja [10]

The formula for simple interest is:

A = P(1+rt)

Where P = amount borrowed

r = interest rate

t = time of loan

A = 5000(1 + 0.125*2)

A =6,250

The total amount she paid back was $6,250

For her monthly payment divide total paid back by 24 months ( 2 years = 24 months)

6250 / 24 = 260.42

Her monthly payment was $260.42

7 0
3 years ago
How can I write 2,351 and standard form
Ludmilka [50]
Isn't 2,351 the standardized form already?
there is word form: two thousand, three hundred fifty one
there is expanded form: (2* 1000)+ (3*100)+ (5* 10)+ (1* 1)= 2,351
3 0
3 years ago
Read 2 more answers
Consider the function below. f(x) = ln(x4 + 27) (a) Find the interval of increase. (Enter your answer using interval notation.)
andrezito [222]

Answer:

a) The function is constantly increasing and is never decreasing

b) There is no local maximum or local minimum.

Step-by-step explanation:

To find the intervals of increasing and decreasing, we can start by finding the answers to part b, which is to find the local maximums and minimums. We do this by taking the derivatives of the equation.

f(x) = ln(x^4 + 27)

f'(x) = 1/(x^2 + 27)

Now we take the derivative and solve for zero to find the local max and mins.

f'(x) = 1/(x^2 + 27)

0 = 1/(x^2 + 27)

Since this function can never be equal to one, we know that there are no local maximums or minimums. This also lets us know that this function will constantly be increasing.

6 0
3 years ago
Taylor tried to solve an equation step by step.
polet [3.4K]
B: step 2
You need to multiply the 7 with the 4.5
3 0
3 years ago
2. Consider the circle x^2−4x+y^2+10y+13=0. There are two lines tangent to this circle having a slope of 2/3.
likoan [24]

Answer:

a) The coordinates are (0.431, -2.646) and (3.568,-7.354)

b) The tangent lines are

l₁ = (x-0.431)*2/3 - 2.646

l₂ = (x-3.568)2/3 - 7.354

Step-by-step explanation:

First, lets complete squares, by taking for each cordinate the square of a linear expression

0= x²−4x+y²+10y+13 = (x-2)²+ 4  + (y+5)²-25+13 = (x-2)²+(y+5)² - 8

Hence (x-2)² + (y+5)² = 8

Lets put y in function of x. We should obtain 2 functions f and g that represent the circle.

(y+5)² = 8 - (x-2)² = -x² + 4x + 4

y = ^+_- \sqrt{-x^2+4x+4} -5

Thus

f(x) = \sqrt{-x^2+4x+4} - 5

g(x) = -\sqrt{-x^2+4x+4} - 5

Lets find the derivate of each function and the points in which they reach the value 2/3. In those points the tangent line will have a slope of 2/3.

We may just find the values of x which the derivate of f is either 2/3 or -2/3.

\frac{-x+2}{\sqrt{-x^2+4x+4}} = ^+_-\frac{2}{3} \Leftrightarrow \frac{x^2-4x+4}{-x^2+4x+4} = \frac{4}{9} \Leftrightarrow 9(x^2-4x+4) = 4(-x^2+4x+4) \\\Leftrightarrow 13x^2-52x+20 = 0

The quadratic has roots

\frac{52 ^+_-\sqrt{1664}}{26}

one root is 3.568, which corresponds with g, and the other root is 0.431, corresponding to f.

Also g(3.568) = - 7.354 and f(0.431) = -2.646

This means that the coordinates are (0.431 , -2.646), (3.568 , -7.354) and the tangent lines are

l₁ = (x-0.431)*2/3 - 2.646

l₂ = (x-3.568)2/3 - 7.354

5 0
3 years ago
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