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3 years ago

What is the sum of numbers as a product of their GCF 45+60

1 answer:

5 0

Find the prime factorization

45=3*3*5
60=2*2*3*5
GCF=3*5=15

45=3*15
60=4*15

remember
ab+ac=a(b+c) so
45+60=15(3)+15(4)=15(3+4)=15(7)=105

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I need to solve the whole thing

kipiarov [429]

1.) 40, 2.) 8, 3.) 18, 4.) 15, 8.) 12, 9.) 30, 10.) 72, 11.) 36

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3 years ago

A square has an area of 144 sq. units. how long is its perimeter?

Arada [10]

Formula for Area of a square:

$\boxed {\boxed {\text {Area = Length x Length}}}$

Given that the area is 144 unit²:

$\text {Length}^2 = 144$

$\text {Length} = \sqrt{144} = 12 \text { units}$

Formula for Perimeter of a square:

$\boxed {\boxed {\text {Perimeter = 4 x Length }}}$

$\text {Perimeter = } 4 \times 12 = 48 \text { units}$

$\boxed { \boxed { \text {Answer: The perimeter is 48 units.}}}$

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3 years ago

Write a definite integral that represents the area of the region. (Do not evaluate the integral.) y1 = x2 + 2x + 3 y2 = 2x + 12F

Svet_ta [14]

Answer:

$A = \int\limits^3__-3}{9}-{x^{2}} \, dx = 36$

Step-by-step explanation:

The equations are:

$y = x^{2} + 2x + 3$

$y = 2x + 12$

The two graphs intersect when:

$x^{2} + 2x + 3 = 2x + 12$

$x^{2} = 0$

$x_{1} = 3\\x_{2} = -3$

To find the area under the curve for the first equation:

$A_{1} = \int\limits^3__-3}{x^{2} + 2x + 3} \, dx$

To find the area under the curve for the second equation:

$A_{2} = \int\limits^3__-3}{2x + 12} \, dx$

To find the total area:

$A = A_{2} -A_{1} = \int\limits^3__-3}{2x + 12} \, dx -\int\limits^3__-3}{x^{2} + 2x + 3} \, dx$

Simplifying the equation:

$A = \int\limits^3__-3}{2x + 12}-({x^{2} + 2x + 3}) \, dx = \int\limits^3__-3}{9}-{x^{2}} \, dx$

Note: The reason the area is equal to the area two minus area one is that the line, area 2, is above the region of interest (see image).

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3 years ago

What is x+2y=6 in point slop form

olasank [31]

Y=x+3 (I did this in my head so I'm not sure)

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4 years ago

Which function grows the fastest for large values of x? f(x)=8x f(x)=3x f(x)=4x2+3 f(x)=1.5x 20 points

Aleonysh [2.5K]

The 4 functions are:
$f_1 (x) = 8x$
$f_2(x)=3x$
$f_3(x)=4x^2+3$
$f_4(x)=1.5 x$

Let's keep in mind that for large values of x, a quadratic function grows faster than a linear function:
$ax^2 \ \textgreater \ kx$ for large values of x

In this problem, we can see that the only quadratic function is $f_3(x)$ , while all the others are linear functions, so the function that grows faster for large values of x is
$f_3(x) = 4x^2 +3$

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3 years ago