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3 years ago

Solve the system of equations 2r - 6y= 28 -x-y = 14 PLEASE HELP

Mathematics

1 answer:

o-na [289]3 years ago

3 0

Answer:

$\displaystyle [-7, -7]$

Step-by-step explanation:

{2x - 6y = 28

{−x - y = 14

−⅙[2x - 6y = 28]

{−⅓x + y = −4⅔

{−x - y = 14

____________

$\displaystyle \frac{-1\frac{1}{3}x}{-1\frac{1}{3}} = \frac{9\frac{1}{3}}{-1\frac{1}{3}} \\ \\$

$x = -7$ [Plug this back into both equations above to get the y-coordinate of −7]; $\displaystyle -7 = y$

I am joyous to assist you anytime.

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Please help ASAP thanks

hjlf

Answer:

x = 2.5 inches. Or

x = 5/2

Step-by-step explanation:

Length: 7 + 2x

Width: 5 + 2x

Note that you have to add x at each end of the picture and onto each part of the width.

The entire area is 120 square inches.

Formula

L*W = Area

(2x + 7)(2x + 5) = 120

Solution

4x^2 + 10x + 14x + 35 = 120 Subtract 120 from both sides. Add xs

4x^2 + 24x + 35 - 120 = 0

4x^2 + 24x - 85 = 0 Factor

(2x + 17)(2x - 5)

2x + 17 has no meaning. It will give - 8.5 which is not possible.

2x - 5 =0

2x = 5 Divide by 2

x = 5/2

x = 2.5

Note: I don't know which one the computer is looking for.

8 0

2 years ago

Read 2 more answers

Consider the line y= -7/4x+8

VARVARA [1.3K]

Perpendicular line has slope that multiplies to -1
paralel line has same slope

y=mx+b
slope=-7/4
perpendicular
4/7
(-7,-2)
-2=4/7(-7)+b
-2=-4+b
add 4 both sides
2=b
y=4/7x+2

paralel
y=-7/4x+b
-2=-7/4(-7)+b
-2=49/4+b
minus 49/4 from both sides (-2=-8/4)
-57/4=b
y=-7/4x-14.25

perpendicular
y=4/7x+2
paralel
y=-7/4x-14.25

8 0

2 years ago

Convert the complex number z = -12 + 5i from rectangular form to polar form.

Paha777 [63]

Answer:

$z=13(\cos 157\degree +i\sin157\degree)$

Step-by-step explanation:

The given complex number is

$z=-12+5i$

The polar form of this complex number is;

$z=r(\cos \theta +i\sin \theta)$ , where

$r=\sqrt{(-12)^2+5^2}$

$r=\sqrt{144+25}=\sqrt{169}=13$

and

$\theta =\tan^{-1}(\frac{5}{-12})$

This implies that;

$\theta=157\degree$ to the nearest degree.

Hence the polar form is

$z=13(\cos 157\degree +i\sin157\degree)$

7 0

3 years ago

Find the least squares regression line for the data points. (Let x be the independent variable and y be the dependent variable.)

solniwko [45]

Answer:

y = - 0.9167 - 0.75x

Step-by-step explanation:

Given the data:

(−1, 0), (1, −2), (3, −3)

X : - 1, 1, 3

Y : 0, - 2, - 3

The regression model is written in the form:

y = C + mx

Where ;

C = intercept ; m = slope / gradient

Using the regression model calculator on the data values given above :

Slope (m) = - 0.75

Intercept (c) = - 0.9167

Hence,

y = - 0.9167 - 0.75x

6 0

2 years ago

Evaluate the triple integral ∭ExydV where E is the solid tetrahedon with vertices (0,0,0),(5,0,0),(0,9,0),(0,0,4).

Elan Coil [88]

Answer: $\int\limits^a_E {\int\limits^a_E {\int\limits^a_E {xy} } \, dV$ = 1087.5

Step-by-step explanation: To evaluate the triple integral, first an equation of a plane is needed, since the tetrahedon is a geometric form that occupies a 3 dimensional plane. The region of the integral is in the attachment.

An equation of a plane is found with a point and a normal vector. Normal vector is a perpendicular vector on the plane.

Given the points, determine the vectors:

P = (5,0,0); Q = (0,9,0); R = (0,0,4)

vector PQ = (5,0,0) - (0,9,0) = (5,-9,0)

vector QR = (0,9,0) - (0,0,4) = (0,9,-4)

Knowing that cross product of two vectors will be perpendicular to these vectors, you can use the cross product as normal vector:

n = PQ × QR = $\left[\begin{array}{ccc}i&j&k\\5&-9&0\\0&9&-4\end{array}\right]$ $\left[\begin{array}{ccc}i&j\\5&-9\\0&9\end{array}\right]$