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rewona [7]
3 years ago
6

Which of the following graphs shows all the possible values for a number that is at least 6? Number line with closed circle on 6

and shading to the left. Number line with closed circle on 6 and shading to the right. Number line with open circle on 6 and shading to the left. Number line with open circle on 6 and shading to the right.

Mathematics
1 answer:
Ahat [919]3 years ago
4 0
The answer is

B.

100% Verified

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Find y, at and tu I need help please help
lesantik [10]

Answer:

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8 0
2 years ago
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8_murik_8 [283]

$138.00

Step-by-step explanation:

Steven earns in 4 hours -$23.00

he earns in 6 hours-$34.50

there fore in 10 hours -23.00+34.50=$57.50

in 20 hours -57.50+57.50=$115

and in 4 hours-$23.00+115.00=138.00$

then he earn $138.00 one day

8 0
2 years ago
Are the graphs of the lines in the pair parallel? Explain.
jonny [76]
You need to convert the second equation to slope/intercept form. The first equation is in that form already. Then you can compare  slopes.

-6x + 8y = 14
8y = 6x + 14
y = (3/4)x + 14/8

SO THE SLOPE IS 3/4 which is the same as slope of equation 1

Therfore they are parallel.
7 0
3 years ago
if 1 gallon is equivelent to 16 cups and addibg 1 cup of vinegar makes solution. what is the ratio fraction
dolphi86 [110]

Answer:1 and 1/16

Step-by-step explanation:

7 0
3 years ago
Solve the system of eqautions <br> y=x2-5x+7 , y=2x+1
IgorC [24]

Answer:

Step-by-step explanation:

The system of equations is given as

y=x^2-5x+7 - - - - - - - - - - 1

y=2x+1 - - - - - - - - - - - - - - 2

We would equate equation 1 to equation 2, it becomes

x^2-5x+7 = 2x+1

x^2-5x+7- 2x - 1 = 0

x^2-5x+7- 2x - 1 = 0

x^2 - 5x - 2x + 7 - 1 = 0

x^2 - 7x + 6 = 0

We would use the factorization method of solving quadratic equations.

x^2 - 6x - x + 6 = 0

x(x - 6) - 1(x - 6)

(x - 6)(x - 1) = 0

x - 6 = 0 or x - 1 = 0

x = 6 or x = 1

Substituting both values of x into equation 2, it becomes

For x = 6,

y=2×6 + 1 = 12 + 1 = 13

y = 13

For x = 1,

y=2 × 1 +1 = 2 + 1 = 3

y = 3

3 0
2 years ago
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