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cluponka [151]
3 years ago
9

What is 10√273s · 3√14st³u³

Mathematics
1 answer:
tensa zangetsu [6.8K]3 years ago
6 0

Answer:

its 210stu\sqrt{78tu}

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Factor completely and then place the factors in the proper location on the grid. x 2 - 10x + 24
dem82 [27]

Answer:

<h2>(x - 6)(x - 4)</h2><h2 />

Step-by-step explanation:

x² - 10x + 24 =  (x² - 6x) + (24 - 4x) = x(x - 6) + 4(6 - x) = (x - 6)(x - 4)

7 0
3 years ago
Select two ordered pairs that satisfy the function y = –4x – 1.
Nimfa-mama [501]
X = -0.25 and x = -2^-2
I’m not sure if this helps
5 0
3 years ago
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What is this polynomial
GenaCL600 [577]
What question is it?

4 0
3 years ago
A Baker Had 1/3 Cups Of Suger . he Added 1/9 Cups To It Find The Total Number Of Sugar . ( Can Someone Help Me Out)
ella [17]

The total number of sugar will be 9/4.

<h3>What is the unitary method?</h3>

The unitary method is a method for solving a problem by the first value of a single unit and then finding the value by multiplying the single value.

A Baker Had 1/3 Cups Of Sugar.

he Added 1/9 Cups To It

Let the number of sugar be x

x (1/3)  + 1/9x = 1

3x + 1x = 9

4x = 9

x = 9/4

Hence, the total number of sugar will be 9/4.

Learn more about the unitary method;

brainly.com/question/23423168

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3 0
2 years ago
The perimeter of a rectangle is 30.8 km and it’s diagonal length is 11 km. Find it’s length and width
blsea [12.9K]

Answer:

Length of the rectangle is 15.0325 km and width is 0.3765 km.

Explanation:

Given:

Perimeter of a rectangle = 30.8 km

Length of diagonal of rectangle = 11 km

To find:

The length and width of rectangle=?

Solution:

Lets assume length of the rectangle = x km

And assume width of the rectangle = y km

Lets first create equation using given  perimeter

perimeter of rectangle = 2 ( length +  width )

=> 30.8 km = 2 ( x + y )  

=>x + y = \frac{30.8}{2}

=> y = 15.4 – x             ------(1)

As diagonal and two sides of rectangle forms right angle triangle whose hypoteneus is diagonal ,  

=> length^2 + width^2 = diagonal^2

=> x^2 + y^2 = 11^2

=> x^2 + y^2 = 121

On substituting value of y from (1) in above equation we get

=> x^2 + (15.4-x)^2 = 121

=>x^2 + (15.4)^2 + x^2 – 2 x 15.4 \times x   = 121

=> 2x^2-30.8x + 237.16 -121  = 0

=> 2x^2-30.8x + 116.16 = 0

Solving above quadratic equation using quadratic formula

General form of quadratic equation is  

ax^2 +bx +c = 0

And quadratic formula for getting roots of quadratic equation is  

x= \frac{ -b\pm\sqrt{(b^2-4ac)}}{2a}

As equation is 2x^2-30.8x + 116.16 = 0, in our case

a = 2 ,  b = -30.8 and c = 116.16

Calculating roots of the equation we get

x=\frac{ -(-30.8)\pm\sqrt{(-30.8)^2-4(2)( 11)} } {(2\times2)}

x=\frac{30.8\pm\sqrt{(948.64-88)}}{4}

x=\frac{30.8\pm\sqrt{860.64}}{4}

x=\frac{30.8\pm\sqrt{860.64}}{4}

x=\frac{(30.8\pm29.33)}4

x=\frac{(30.8+29.33)}{4}

x=\frac{(30.8-29.33)}{4}

=> x = 15.0325 or x = 0.3675

As generally length is longer one ,  

So x = 1.0325

From equation (1) y = 15.4 – x = 0.3765

Hence length of the rectangle is 15.0325 km and width is 0.3765 km.

6 0
3 years ago
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