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statuscvo [17]
3 years ago
9

Someone help quick!!!

Mathematics
2 answers:
n200080 [17]3 years ago
4 0

Answer:

B, E

Step-by-step explanation:

J, K, Q are all on the same line: collinear

You can draw only one line through L and P

Sav [38]3 years ago
4 0

Answer:

b,e

Step-by-step explanation:

e2020

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What is the solution for 3x-4=-1/2x+3
prohojiy [21]

Answer:

2

Step-by-step explanation:

3x-4 =-1/2x+3

3x+1/2x=3+4

6x+x/2=7

6x+x=14

7x=14

x=14/2

x= 2

4 0
3 years ago
Read 2 more answers
For each order. Determine weather is a solution to x = -5
Yakvenalex [24]
First number is x and second number is y
(x,y)

So first and last are correct
4 0
3 years ago
y 5 1.055x, where x is the sales price in dollars and y is the price including sales tax, in dollars.
lakkis [162]

Answer:

I am sorry I am not able to help you but I have no idea what that problem is you might have to just ask someone else or look it up on another website.

8 0
3 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
Plz help ASAP
DENIUS [597]
A+c = 150
10.25a+7.75c = 1470
4 0
4 years ago
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