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4 years ago

10

6x5+2÷9-4help me please

1 answer:

igomit [66]4 years ago

6 0

6x5+2÷9-4
30x+2/9-4
2/9-4
-34/9
30x-34/9

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3<br> 5<br> y = mx + c<br> Find the value of y when m= -2, r = -7 and c= -3.

mario62 [17]

Answer:

<u>This assume the equation is 35y = mx + c, and that r is meant to be x.</u>

Step-by-step explanation:

35y = mx + c

35y = -2(-7) - 3

35y = 14 - 3

35y = 11

y = (11/35)

Please check the formatting of your question.

6 0

2 years ago

A quadrilateral has vertices at $(0,1)$, $(3,4)$, $(4,3)$ and $(3,0)$. Its perimeter can be expressed in the form $a\sqrt2+b\sqr

seraphim [82]

Answer:

a + b = 12

Step-by-step explanation:

Given

Quadrilateral;

Vertices of (0,1), (3,4) (4,3) and (3,0)

$Perimeter = a\sqrt{2} + b\sqrt{10}$

Required

$a + b$

Let the vertices be represented with A,B,C,D such as

A = (0,1); B = (3,4); C = (4,3) and D = (3,0)

To calculate the actual perimeter, we need to first calculate the distance between the points;

Such that:

AB represents distance between point A and B

BC represents distance between point B and C

CD represents distance between point C and D

DA represents distance between point D and A

Calculating AB

Here, we consider A = (0,1); B = (3,4);

Distance is calculated as;

$Distance = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$

$(x_1,y_1) = A(0,1)$

$(x_2,y_2) = B(3,4)$

Substitute these values in the formula above

$Distance = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$

$AB = \sqrt{(0 - 3)^2 + (1 - 4)^2}$

$AB = \sqrt{( - 3)^2 + (-3)^2}$

$AB = \sqrt{9+ 9}$

$AB = \sqrt{18}$

$AB = \sqrt{9*2}$

$AB = \sqrt{9}*\sqrt{2}$

$AB = 3\sqrt{2}$

Calculating BC

Here, we consider B = (3,4); C = (4,3)

Here,

$(x_1,y_1) = B (3,4)$

$(x_2,y_2) = C(4,3)$

Substitute these values in the formula above

$Distance = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$

$BC = \sqrt{(3 - 4)^2 + (4 - 3)^2}$

$BC = \sqrt{(-1)^2 + (1)^2}$

$BC = \sqrt{1 + 1}$

$BC = \sqrt{2}$

Calculating CD

Here, we consider C = (4,3); D = (3,0)

Here,

$(x_1,y_1) = C(4,3)$

$(x_2,y_2) = D (3,0)$

Substitute these values in the formula above

$Distance = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$

$CD = \sqrt{(4 - 3)^2 + (3 - 0)^2}$

$CD = \sqrt{(1)^2 + (3)^2}$

$CD = \sqrt{1 + 9}$

$CD = \sqrt{10}$

Lastly;

Calculating DA

Here, we consider C = (4,3); D = (3,0)

Here,

$(x_1,y_1) = D (3,0)$

$(x_2,y_2) = A (0,1)$

Substitute these values in the formula above

$Distance = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$

$DA = \sqrt{(3 - 0)^2 + (0 - 1)^2}$

$DA = \sqrt{(3)^2 + (- 1)^2}$

$DA = \sqrt{9 + 1}$

$DA = \sqrt{10}$

The addition of the values of distances AB, BC, CD and DA gives the perimeter of the quadrilateral

$Perimeter = 3\sqrt{2} + \sqrt{2} + \sqrt{10} + \sqrt{10}$

$Perimeter = 4\sqrt{2} + 2\sqrt{10}$

Recall that

$Perimeter = a\sqrt{2} + b\sqrt{10}$

This implies that

$a\sqrt{2} + b\sqrt{10} = 4\sqrt{2} + 2\sqrt{10}$

By comparison

$a\sqrt{2} = 4\sqrt{2}$

Divide both sides by $\sqrt{2}$

$a = 4$

By comparison

$b\sqrt{10} = 2\sqrt{10}$

Divide both sides by $\sqrt{10}$

$b = 2$

Hence,

a + b = 2 + 10

a + b = 12

3 0

3 years ago

A garden is 53 feet long and 24 feet wide. how many square feet are in the garden?(set up an equation for this Problem but do no

9966 [12]

I think your talking about area so just 53 x 24= your answer

7 0

3 years ago

[Functions] I just need yes's and no's.......a lot of them, functions just really don't click with me and I don't understand why

Semenov [28]

Answer:

uh-

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

4. Which of the following lines is perpendicular to y = -2X + 8?

Alenkinab [10]

For this case we have to by definition, if two lines are perpendicular then the product of its slopes is -1.

That is to say:

$m_ {1} * m_ {2} = - 1$

We have the following equation:

$y = -2x + 8$

So:

$m_ {1} = - 2$

Thus:

$m_ {2} = \frac {-1} {m_ {1}}\\m_ {2} = \frac {-1} {- 2}\\m = \frac {1}{2}$

Thus, a line perpendicular to the given line must have slope $m = \frac {1} {2}.$

Option A:

$x + 2y = 8\\2y = -x + 8\\y = - \frac {1} {2} x + 4$

It is not perpendicular!

Option B:

$x-2y = 6\\2y = x-6\\y = \frac {1} {2} x-3$

If it is perpendicular!

Option C:

$2x + y = 4\\y = -2x + 4$

It is not perpendicular!

Option D:

$2x-y = 1\\y = 2x-1$

It is not perpendicular!

The correct option is option B

ANswer:

Option B

5 0

3 years ago