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3 years ago

which of the following is the equation of a line parallel to the line y = -x + 1, passing through the point (4,1)

Mathematics

1 answer:

fomenos3 years ago

4 0

Answer:

A. x + y = 5

Step-by-step explanation:

The slope of the equation is equal for parallel lines,

; Slope = (-1)

;Use the formula

(y - y(coordinate of the given point)) = slope × [(x - x(coordinate of the given point)]

; y - 1 = (-1)(x - 4)

; y - 1 = -x + 4...then move the variable (x) to the left hand side and then move (-1) to the right hand side

; x + y = 4 + 1

; Hence,

; x + y = 5

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Which ordered pairs are solutions to the inequality 3x-4y>5

Sloan [31]

Note: you did not provide the answer options, so I am, in general, solving this query to solve your concept, which anyways would clear your concept.

Answer:

Please check the explanation.

Step-by-step explanation:

Given the inequality

$3x-4y>5$

All we need is to find any random value of 'x' and then solve the inequality.

For example, putting x=3

$3\left(3\right)-4y>5$

$9-4y>5$

$-4y>-4$

$\mathrm{Multiply\:both\:sides\:by\:-1\:\left(reverse\:the\:inequality\right)}$

$\left(-4y\right)\left(-1\right)$

$4y$

$\mathrm{Divide\:both\:sides\:by\:}4$

$\frac{4y}{4}$

$y$

So, at x = 3, the calculation shows that the value of y must be less

than 1 i.e. y<1 in order to be the solution.

Let us take the random y value that is less than 1.

As y=0.9 < 1

so putting y=0.9 in the inequality

$3\left(3\right)-4\left(0.9\right)$

$=9-3.6$

$=5.4$

As 5.4 > 5

Means at x=3, and y=0.9, the inequality is satisfied.

Thus, (3, 0.9) is one of the many ordered pairs solutions to the inequality 3x-4y>5.

6 0

3 years ago

Hi! May I please have some help?

max2010maxim [7]

Steps:
18 x 2
What is that?
You can also think of it as 9 x 4 if that is easier
Then whatever you got for that, add 21 then subtract 12 and add 1

6 0

3 years ago

Read 2 more answers

The expression 3 x 7 - 4x8+2 is equivalent to which of the following?

ELEN [110]

Answer: b

explanation: only choice B equals -9 like the given equation.

8 0

4 years ago

A homogeneous rectangular lamina has constant area density ρ. Find the moment of inertia of the lamina about one corner

frozen [14]

Answer:

$I_{corner} =\frac{\rho _{ab}}{3}(a^2+b^2)$

Step-by-step explanation:

By applying the concept of calculus;

the moment of inertia of the lamina about one corner $I_{corner}$ is:

$I_{corner} = \int\limits \int\limits_R (x^2+y^2) \rho d A \\ \\ I_{corner} = \int\limits^a_0\int\limits^b_0 \rho(x^2+y^2) dy dx$

where :

(a and b are the length and the breath of the rectangle respectively )

$I_{corner} = \rho \int\limits^a_0 {x^2y}+ \frac{y^3}{3} |^ {^ b}_{_0} \, dx$

$I_{corner} = \rho \int\limits^a_0 (bx^2 + \frac{b^3}{3})dx$

$I_{corner} = \rho [\frac{bx^3}{3}+ \frac{b^3x}{3}]^ {^ a} _{_0}$

$I_{corner} = \rho [\frac{a^3b}{3}+ \frac{ab^3}{3}]$

$I_{corner} =\frac{\rho _{ab}}{3}(a^2+b^2)$

Thus; the moment of inertia of the lamina about one corner is $I_{corner} =\frac{\rho _{ab}}{3}(a^2+b^2)$

7 0

3 years ago

Nigel tried to solve an equation step by step

kaheart [24]

Answer:

Step-by-step explanation:

8 0

3 years ago