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kakasveta [241]
3 years ago
5

Find the values of x when y =1

Mathematics
1 answer:
Ivahew [28]3 years ago
8 0
Graph the rationale functions f(x)x2+3X-1/x+1 by identifying its key features.
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Suppose a + b = 0.
Digiron [165]
I'm pretty sure that it's b. That's assuming that you find the negative of b^2 and not the square of negative b.

5 0
3 years ago
30t - 20=70 two step equations
Alenkasestr [34]

Answer:

Hi!

To solve this we can begin by adding 20 to both sides

30t = 90

Next we divide both sides by 30


t = 3


~CoCo

8 0
3 years ago
Read 2 more answers
6+b/b is an integer number. find all possible integer values of b​
Andrei [34K]

Step-by-step explanation:

(6+b)/b = x (= an integer number)

6 + b = bx (= still an integer number)

6 = bx - b = b(x - 1) (= still an integer number)

b = 6/(x - 1) (= still an integer number)

now for b to be integer, (x-1) must be a factor of 6 (6 must be divisible by x-1 with 0 remainder).

and integer includes negative numbers too.

the factors of 6 are therefore

1, 2, 3, 6, -1, -2, -3, -6

so, b can be 6/±1 = ±6

b can be 6/±2 = ±3

b can be 6/±3 = ±2

b can be 6/±6 = ±1

so, all possible values for b are also the listed factors of 6 :

-6, -3, -2, -1, 1, 2, 3, 6

5 0
2 years ago
I will mark you brainiest if you can answer this
saul85 [17]

Answer:

B - the graph of the function does not form a straight line

Step-by-step explanation:

linear functions have to form a straight line, and that function in the question is a quadratic because x is to the power of 2.

3 0
2 years ago
If x = t^3 and y = t^2 - 2 , what is y in terms of x?
alekssr [168]

Answer:

E

Step-by-step explanation:

You have to get t^2 in the first equation.

x = t^3                         Take the cube root of both sides

\sqrt[3]{x} = \sqrt[3]{t^3}

\sqrt[3]{x}=t                         Now square both sides

(\sqrt[3]{x})^2 = t^2

x^\frac{2}{3} = t^2

Now you can substitute into the second equation

y = t^2 - 2

y = x^\frac{2}{3} - 2

E

7 0
2 years ago
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