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2 years ago

Find the x-intercept of the line whose equation is 8x + 2y = 4.

Mathematics

1 answer:

horrorfan [7]2 years ago

8 0

Answer: Hope this helps you

1/2

Step-by-step explanation:

To find the x-int of any equation, you will need to replace the y with a 0 since when the x-int is when y is 0.

8x+2y = 4 ➨ 8x+2*0 = 4

Now I will do 2*0 which anything times 0 is equal to 0. And we are left with the equation of:

8x = 4

Now I will divide both sides by 8, which isolates our x.

x = 1/2

The x-int is 1/2.

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What happens to the value of the expression 2t/t as t decreases

anyanavicka [17]

Answer:

it stays the same until t=0, where it is undefined

Step-by-step explanation:

2t/t = 2 . . . . when t ≠ 0

2t/t = undefined when t = 0.

The value of t can increase or decrease and the value of the ratio remains the same. The value has a "hole" at t=0, where it is undefined. Otherwise, its value is always 2.

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3 years ago

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kozerog [31]

Answer:

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Step-by-step explanation:

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3 years ago

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andreev551 [17]

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3 years ago

Assume that you plan to use a significance level of α = 0.05 to test the claim that p1 = p2, use the given sample sizes and numb

11Alexandr11 [23.1K]

Since the data is insufficient. Let us denote that n1 = 100, n2 = 100X1 = 38. X2 = 40
So looking for the p value:
p = (38+ 40) / (100 + 100) = 0.39
z = (38/100 - 40/100)/√ (0.39*(1-0.39)*(1/100 + 1/100))= -0.2899
P-value = P (|z| > 0.2899) = 0.7718

7 0

3 years ago

Could the inverse of a non-function be a function? Explain or give an example.

Kitty [74]

Answer:

The inverse of a non-function mapping is not necessarily a function.

For example, the inverse of the non-function mapping $\lbrace (0,\, 0),\, (0,\, 1),\, (1,\, 0),\, (1,\, 1) \rbrace\!$ is the same as itself (and thus isn't a function, either.)

Step-by-step explanation:

A mapping is a set of pairs of the form $(a,\, b)$ . The first entry of each pair is the value of the input. The second entry of the pair would be the value of the output.

A mapping is a function if and only if for each possible input value $x$ , at most one of the distinct pairs includes $x\!$ as the value of first entry.

For example, the mapping $\lbrace (0,\, 0),\, (1,\, 0) \rbrace$ is a function. However, the mapping $\lbrace (0,\, 0),\, (1,\, 0),\, (1,\, 1) \rbrace$ isn't a function since more than one of the distinct pairs in this mapping include $1$ as the value of the first entry.

The inverse of a mapping is obtained by interchanging the two entries of each of the pairs. For example, the inverse of the mapping $\lbrace (a_{1},\, b_{1}),\, (a_{2},\, b_{2})\rbrace$ is the mapping $\lbrace (b_{1},\, a_{1}),\, (b_{2},\, a_{2})\rbrace$ .

Consider mapping $\lbrace (0,\, 0),\, (0,\, 1),\, (1,\, 0),\, (1,\, 1) \rbrace\!$ . This mapping isn't a function since the input value $0$ is the first entry of more than one of the pairs.

Invert $\lbrace (0,\, 0),\, (0,\, 1),\, (1,\, 0),\, (1,\, 1) \rbrace\!$ as follows:

$(0,\, 0)$ becomes $(0,\, 0)$ .
$(0,\, 1)$ becomes $(1,\, 0)$ .
$(1,\, 0)$ becomes $(0,\, 1)$ .
$(1,\, 1)$ becomes $(1,\, 1)$ .

In other words, the inverse of the mapping $\lbrace (0,\, 0),\, (0,\, 1),\, (1,\, 0),\, (1,\, 1) \rbrace\!$ would be $\lbrace (0,\, 0),\, (1,\, 0),\, (0,\, 1),\, (1,\, 1) \rbrace\!$ , which is the same as the original mapping. (Mappings are sets. There is no order between entries within a mapping.)

Thus, $\lbrace (0,\, 0),\, (0,\, 1),\, (1,\, 0),\, (1,\, 1) \rbrace\!$ is an example of a non-function mapping that is still not a function.

More generally, the inverse of non-trivial ellipses (a class of continuous non-function $\mathbb{R} \to \mathbb{R}$ mappings, including circles) are also non-function mappings.

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2 years ago