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3 years ago

Find the equation of the line parallel to y=-11-5, that passes through the point

Mathematics

2 answers:

Elanso [62]3 years ago

7 0

Answer:

y=41−11x

Step-by-step explanation:

The equation of the line in the slope-intercept form is y=5−11x.

The slope of the parallel line is the same: m=−11.

So, the equation of the parallel line is y=−11x+a.

To find a, we use the fact that the line should pass through the given point: −3=(−11)⋅(4)+a.

Thus, a=41.

Therefore, the equation of the line is y=41−11x.

RSB [31]3 years ago

3 0

Answer:

y=41−11x

Step-by-step explanation:

The equation of the line in the slope-intercept form is y=5−11x.

The slope of the parallel line is the same: m=−11.

So, the equation of the parallel line is y=−11x+a.

To find a, we use the fact that the line should pass through the given point: −3=(−11)⋅(4)+a.

Thus, a=41.

Therefore, the equation of the line is y=41−11x.

Answer: y=41−11x.

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The graphs of f(x) and g(x) are shown below: graph of function f of x open upward and has its vertex at negative 7, 0. Graph of

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<h3>What is a Function ?</h3>

A function is a mathematical statement that defines relation between two variables.

The f(x) = (x+7)²

The equation for g(x) is ?

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Graph of function g of x opens upward and has its vertex at (-7,2).

g(x) is formed by the translation of the function f(x) such that it's vertex is (-7,-2)

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Option D is the correct answer.

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2 years ago

Match the numerical expressions to their simplest forms.

Aloiza [94]

Answer:

$(a^6b^1^2)^\frac{1}{3}$ = $a^2b^4$

$\frac{(a^5b^3)^\frac{1}{2}}{(ab)^-^\frac{1}{2}}$ = $a^3b^2$

$(\frac{a^5}{a^-^3b^-^4})^\frac{1}{4}$ = $a^2b$

$(\frac{a^3}{ab^-^6})^\frac{1}{2}$ = $ab^3$

Step-by-step explanation:

Simplify each of the expressions:

$(a^6b^1^2)^\frac{1}{3}$

Distribute the exponent. Multiply the exponent of the term outside of the parenthesis by the exponents of the variable.

$(a^6b^1^2)^\frac{1}{3}$

$a^6^*^\frac{1}{3}b^1^2^*^\frac{1}{3}$

Simplify,

$a^2b^4$

Use a similar technique to solve this problem. Remember, a fractional exponent is the same as a radical, if the denominator is (2), then the operation is taking the square root of the number.

$\frac{(a^5b^3)^\frac{1}{2}}{(ab)^-^\frac{1}{2}}$

Rewrite as square roots:

$\frac{\sqrt{a^5b^3}}{\sqrt{(ab)}^-^1}$

A negative exponent indicates one needs to take the reciprocal of the number. Apply this here:

$\frac{\sqrt{a^5b^3}}{\frac{1}{\sqrt{ab}}}$

Simplify,

$\sqrt{a^5b^3}*\sqrt{ab}$

Since both numbers are under a radical, one can rewrite them such that they are under the same radical,

$\sqrt{a^5b^3*ab}$

Simplify,

$\sqrt{a^6b^4}$

Since this operation is taking the square root, divide the exponents in half to do this operation:

$a^3b^2$

$(\frac{a^5}{a^-^3b^-^4})^\frac{1}{4}$

Simplify, to simplify the expression in the numerator and the denominator, the base must be the same. Remember, the base is the number that is being raised to the exponent. One subtracts the exponent of the number in the denominator from the exponent of the like base in the numerator. This only works if all terms in both the numerator and the denominator have the operation of multiplication between them:

$(\frac{a^8}{b^-^4})^\frac{1}{4}$

Bring the negative exponent to the numerator. Change the sign of the exponent and rewrite it in the numerator,

$(a^8b^4)^\frac{1}{4}$

This expression to the power of the one forth. This is the same as taking the quartic root of the expression. Rewrite the expression with such,

$\sqrt[4]{a^8b^4}$

SImplify, divide the exponents by (4) to simulate taking the quartic root,

$a^2b$

$(\frac{a^3}{ab^-^6})^\frac{1}{2}$

Using all of the rules mentioned above, simplify the fraction. The only operation happening between the numbers in both the numerator and the denominator is multiplication. Therefore, one can subtract the exponents of the terms with the like base. The term in the denomaintor can be rewritten in the numerator with its exponent times negative (1).

$(a^3^-^1b^(^-^6^*^(^-^1^)^))^\frac{1}{2}$