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3 years ago

What are the intercepts for the equation? Select all that apply. x - 3y = 6 *

Mathematics

1 answer:

EleoNora [17]3 years ago

8 0

Answer: get a tutor or search it up

Step-by-step explanation: or something

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Solve the compound inequality 3x + 4 ≤ 7

Ierofanga [76]

Answer:

x ≤ 1

Step-by-step explanation:

3x+4≤ 7

3x + 4 - 4 ≤ 7 - 4

3x≤3

x ≤ 1

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3 years ago

What the expanded form of 8.65?

Stells [14]

Answer:

8 .60 .05

Hope it helps :D

btw pls mark brainlest

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3 years ago

Help me with math please

Firdavs [7]

Answer is A..... hope it helps

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3 years ago

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Find the distance between the two points. Round to the nearest tenth if necessary.

prohojiy [21]

Answer:

the answer to the nearest tenths is 7.8

8 0

4 years ago

Find the range of p in equation p(x+1) (x-3)=x-4p-2 has no real roots.

Tanya [424]

Answer:

$\displaystyle p > \frac{1}{4}$ .

Step-by-step explanation:

Expand the left-hand side of this equation:

$p\, (x + 1)\, (x - 3) = p\, \left(x^2 - 2\, x - 3\right) = p\, x^2 - 2\, p\, x - 3\, p$ .

Collect the terms, so that this quadratic equation is in the form $a\, x^2 + b\, x+ c = 0$ :

$p\, x^2 - 2\, p\, x - 3\, p = x - 4\, p - 2$ .

$p\, x^2 - (2\, p + 1)\, x + (p + 2) = 0$ .

In this equation:

$a = p$ .
$b = -(2\, p + 1)$ .
$c = p + 2$ .

Calculate the quadratic discriminant of this quadratic equation:

$\begin{aligned}b^2 - 4\, a\, c &= (-(2\, p + 1))^2 - 4\, p\, (p + 2) \\ &= 4\, p^2 + 4\, p + 1 - 4\, p^2- 8\, p = -4\, p + 1\end{aligned}$ .

A quadratic equation has no real root if its quadratic discriminant is less then zero. As a result, this quadratic equation will have no real root when $-4\, p + 1 < 0$ . Solve for the range of $p$ :

$\displaystyle p > \frac{1}{4}$ .

4 0

4 years ago