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igor_vitrenko [27]
3 years ago
10

10) 17x-2=7x+8

Mathematics
1 answer:
STatiana [176]3 years ago
6 0
10.x=1
11.3/4y+4 , x=1/5y+1/5
12.x=2

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How does the area of the triangle relate to the area of the square? Please Explain!!! I really don't get it!
kolbaska11 [484]
A square has 4 right angles and 4 equal sides.
If you draw a diagonal through the square then you get two right triangles. So the area relation between a square and a right triangle is going to be half. The area of the triangle is half the area of the square. 

Also the formula for area of a square is length times width.
And the formula for the area of a right triangle is 0.5 times length times width.

So if you set up the equation (let's use * as multiplication symbol ): 

Area of a Square : Area of a Right Triangle
length * width : 0.5 * length * width
(Divide by length * width on both sides)
1:0.5 or 2:1
4 0
3 years ago
Which symbol makes the statement true? 125.32 ____ 125.093<br><br> &lt;<br> &gt;<br> =<br> ≥
Alexandra [31]
Hello : 
125.32  > 125.093 .. because : 3 <span>> 0</span>
4 0
4 years ago
Read 2 more answers
cate and elena were playing a card game the stack of cards in the middle had 24 cards in it to begin wiht. Cate added 8 cards to
Kaylis [27]
24+8 = 30

30 - 12 = 18

18-9 = 8

There are 8 cards left in the stack
7 0
3 years ago
Read 2 more answers
What is the answer?<br> Please and thanks
Paha777 [63]

59 is prime number so there's only 1 factor pair  of 59

1 and 59

Answer

1 and 59

7 0
3 years ago
Read 2 more answers
Please help me on this problem
Anna71 [15]

Answer:

The pairs of integer having two real solution forax^{2} -6x+c = 0 are

  1. a = -4, c = 5
  2. a = 1, c = 6
  3. a = 2, c = 3
  4. a = 3, c = 3

Step-by-step explanation:

Given

ax^{2} -6x+c = 0

Now we will solve the equation by putting all the 6 pairs so we get the  following

-3x^{2} -6x-5 = 0 for a = -3 , c=-5

-4x^{2} -6x+5 = 0 for a = -4 , c=5

1x^{2} -6x+6 = 0 for a = 1 , c=6

2x^{2} -6x+3 = 0 for a = 2 , c=3

3x^{2} -6x+3 = 0 for a = 3 , c=3

5x^{2} -6x+4 = 0 for a = 5 , c=4

The above  all are Quadratic equations inn general form ax^{2} +bx+c=0

where we have a,b and c constant values

So for a real Solution we must have

Disciminant , b^{2} -4\timesa\timesc \geq 0

for a = -3 , c=-5 we have

Discriminant =-24 which is less than 0 ∴ not a real solution.

for a = -4 , c=5 we have

Discriminant = 116 which is greater than 0 ∴ a real solution.

for a = 1 , c=6 we have

Discriminant =12 which is greater than 0 ∴ a real solution.

for a = 2 , c=3 we have

Discriminant =12 which is greater than 0 ∴ a real solution.

for a = 3 , c=3 we have

Discriminant =0 which is equal to 0 ∴ a real solution.

for a = 5 , c=4 we have

Discriminant =-44 which is less than 0 ∴ not a real solution.

7 0
3 years ago
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