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3 years ago

Whats the x and y intercept for x + 4y = 36

Mathematics

1 answer:

sergij07 [2.7K]3 years ago

3 0

Answer:

x intercept (36, 0)

y intercept (0, 9)

Step-by-step explanation:

x intercept when y = 0 so x = 36

y intercept when x = 0 so 4y = 36; y = 9

Answer

x intercept (36, 0)

y intercept (0, 9)

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Which of the following creates 90° angles at the points of intersection and also cuts a segment into two congruent pieces?

MA_775_DIABLO [31]

Perpendicular bisected .-.

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4 years ago

Please help I need to finish my winter packet and no one is answering my questions

Damm [24]

Answer:

$\sqrt[7]{x^{4}}$

$(x^{\frac{1}{7}})^{4}$

$(\sqrt[7]{x})^{4}$

Step-by-step explanation:

we have

$x^{\frac{4}{7}}$

Remember the properties

$\sqrt[n]{a^{m}}=a^{\frac{m}{n}}$

$(a^m)^{n}=a^{m*n}$

Verify each case

Part 1) we have

$\sqrt[4]{x^{7}}$

we know that

$\sqrt[4]{x^{7}}=x^{\frac{7}{4}}$

Compare with the given expression

$x^{\frac{7}{4}} \neq x^{\frac{4}{7}}$

Part 2) we have

$\sqrt[7]{x^{4}}$

we know that

$\sqrt[7]{x^{4}}=x^{\frac{4}{7}}$

Compare with the given expression

$x^{\frac{4}{7}} = x^{\frac{4}{7}}$

therefore

Is equivalent to the given expression

Part 3) we have

$(x^{\frac{1}{7}})^{4}$

we know that

$(x^{\frac{1}{7}})^{4}=x^{\frac{4}{7}}$

Compare with the given expression

$x^{\frac{4}{7}} = x^{\frac{4}{7}}$

therefore

Is equivalent to the given expression

Part 4) we have

$(x^{\frac{1}{4}})^{7}$

we know that

$(x^{\frac{1}{4}})^{7}=x^{\frac{7}{4}}$

Compare with the given expression

$x^{\frac{7}{4}} \neq x^{\frac{4}{7}}$

Part 5) we have

$(\sqrt[4]{x})^{7}$

we know that

$(\sqrt[4]{x})^{7}=(x^{\frac{1}{4}})^{7}=x^{\frac{7}{4}}$

Compare with the given expression

$x^{\frac{7}{4}} \neq x^{\frac{4}{7}}$

Part 6) we have

$(\sqrt[7]{x})^{4}$

we know that

$(\sqrt[7]{x})^{4}=(x^{\frac{1}{7}})^{4}=x^{\frac{4}{7}}$

Compare with the given expression

$x^{\frac{4}{7}} = x^{\frac{4}{7}}$

therefore

Is equivalent to the given expression

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3 years ago

What is the simplified form of x+2/x^2-3x-10 * x-3/x^2+x-12

Vinvika [58]

I mean based on your possible answers, the third one is correct,but you can go further that that, in the steps i will note when you've came to the "answer"

$\frac{x + 2}{ {x}^{2} - 3x - 10 } \times \frac{x - 3}{ {x}^{2} + x - 12 }$
$\frac{x + 2}{ {x}^{2} + 2x - 5x - 10 } \times \frac{x - 3}{ {x}^{2} + 4x - 3x - 12 }$
$\frac{x + 2}{x(x + 2) - 5(x + 2)} \times \frac{x - 3}{x(x + 4) - 3(x + 4)}$
$\frac{x + 2}{(x + 2) \times (x - 5)} \times \frac{x - 3}{(x + 4) \times (x - 3)}$
$\frac{1}{(x + 4) \times (x + 5)}$
and basically that's the "answer", but you can simplify more.

$\frac{1}{ {x}^{2} + 4x - 5x - 20}$
$\frac{1}{ {x}^{2} - x - 20 }$
Hope that's helpful

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3 years ago

Pls answer quickly

Katarina [22]

Answer: B

Step-by-step explanation: This answer can be found by just doing process of elimination. A has the numbers switched around, C and D has 2 adding by ones and 7 multipying. Mathmaticlly finding the answer for B, 2 is multiplying my 2, 3 and 4. First, you have 4, then you have 6, and last you have 8. 7 is doing the same thing. 7 multiplied by 2 gives you 14, by 3 gives you 21 and by 4 gives you 28.

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3 years ago

Substitution method 4a=4-b 4a=5-b How to do this

Rufina [12.5K]

A "solution" to this pair of equations is a pair of numbers, for 'a' and 'b',
that make the equations true statements.

I can tell you right now that this system of equations has no solution.
Here, let me show you why:

 4a = 4 - b
 4a = 5 - b

Subtract the bottom equation from the top one:

 0 = -1

Can you think of any numbers for 'a' and 'b' that could
make 0 equal to -1 ?

I can't either.

There are no numbers for 'a' and 'b' that can make both equations true
statements. That's just another way to say that the system of equations
has no solution.

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3 years ago