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Fudgin [204]
3 years ago
8

What are the real numbers that are greater than -6 but less than 6

Mathematics
2 answers:
Lisa [10]3 years ago
6 0
-5,-4,-3,-3,-2,-1,1,2,3,4,5
hopefully that helps
maksim [4K]3 years ago
6 0
The numbers that are greater than -6 but less than 6 are:

-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5
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Solve this question​
Maru [420]

Answer:

∠ RQU = 25°

Step-by-step explanation:

The opposite angles of an inscribed circle are supplementary, thus

∠ URS + ∠ UTS = 180°, that is

∠ URS + 124° = 180° ( subtract 124° from both sides )

∠ URS = 56°

The external angle of a triangle is equal to the sum of the 2 opposite interior angles.

∠ URS is an external angle to Δ QUR, that is

∠ RQU + 31° = 56° ( subtract 31° from both sides )

∠ RQU = 25°

5 0
3 years ago
The diameter of a circular garden is 18 feet. What is the approximate area of the garden?
nlexa [21]
Area of a circle = πr²

Radius = 18/2 = 9

A = (3.14)(9)²
A = 254.34

Hope this helps :)
4 0
3 years ago
Read 2 more answers
Solve the system of equations by graphing: 2x + 3y = 2 6x + 18y = 9
cluponka [151]
The answer is  A: x=1/2 , y= 1/3

Solve the following system:
{2 x + 3 y = 2 | (equation 1)
{6 x + 18 y = 9 | (equation 2)

Swap equation 1 with equation 2:
{6 x + 18 y = 9 | (equation 1)
{2 x + 3 y = 2 | (equation 2)

Subtract 1/3 × (equation 1) from equation 2:
{6 x + 18 y = 9 | (equation 1)
{0 x - 3 y = -1 | (equation 2)

Divide equation 1 by 3:
{2 x + 6 y = 3 | (equation 1)
{0 x - 3 y = -1 | (equation 2)

Multiply equation 2 by -1:
{2 x + 6 y = 3 | (equation 1)
{0 x+3 y = 1 | (equation 2)

Divide equation 2 by 3:
{2 x + 6 y = 3 | (equation 1)
{0 x+y = 1/3 | (equation 2)

Subtract 6 × (equation 2) from equation 1:
{2 x+0 y = 1 | (equation 1)
v0 x+y = 1/3 | (equation 2)

Divide equation 1 by 2:
{x+0 y = 1/2 | (equation 1)
v0 x+y = 1/3 | (equation 2)

Collect results:
Answer:  {x = 1/2, y = 1/3

7 0
2 years ago
V is greater than or equal to 5
Citrus2011 [14]

Answer:

greater

Step-by-step explanation:

3 0
3 years ago
Hi everyone I was wondering if someone could please help me out with this problem and explain it to me
PSYCHO15rus [73]
Join the centre O to the chord (let it be MN) & let OH be the perpendicular to the chord

OH bisects MN into 2 equal parts (each one is x/2)
OMH is a right triangle with one side =8, the second leg =x/2 & the hypotenuse = 12 (Radius)
Apply Pythagoras:

12² = 8² +(x/2)² ==>144=64 + x²/4 ==> x²=4(144-64) =320

x²=320==> x=√320 =17.88 ≈17.9



3 0
3 years ago
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