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3 years ago

Pls help!! 5+3x=8 Must show work

Mathematics

2 answers:

Andrej [43]3 years ago

7 0

Answer:

x = 1

Step-by-step explanation:

5 - 5 + 3x = 8 - 5

3x = 3

3x/3 = 3/3

x = 1

Annette [7]3 years ago

5 0

Here’s your answer please tel me if right

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Each side of a square is lengthened by 5 inches. the area of this new, larger square is 36 square inches. find the length of a

butalik [34]

The formula of the area is:

$A = s^{2}$

Where s stands for side.

So if the area is 36, the lenght of the side must be:

$s = \sqrt{36}$

$\bf~s = 6$

So if 6 is the side of the original square plus 5, than the original lenght of the side is:

6 - 5 = 1 inch

Hope it helped,

BioTeacher101

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4 years ago

Find an equation of the plane through the point(1, 5,-1) and perpendicular to the vector (1, 5, 1). Do this problem in the stand

Vladimir79 [104]

Answer:

Step-by-step explanation:

Given a plane passing through the point(1, 5,-1) and perpendicular to the vector (1, 5, 1), the equation of the plane can be expressed generally as;

a(x-x₀)+b(y-y₀)+c(z-z₀) = 0 where (x₀, y₀, z₀) is the point on the plane and (a, b,c) is the normal vector perpendicular to the plane i.e (1,5,1)

Given the point P (1, 5, -1) and the normal vector n(1, 5, 1)

x₀ = 1, y₀ =5, z₀ = -1, a = 1, b = 5 and c = 1

Substituting this point in the formula we will have;

a(x-x₀)+b(y-y₀)+c(z-z₀) = 0

1(x-1)+5(y-5)+1(z-(-1)) = 0

(x-1)+5(y-5)+(z+1) = 0

x-1+5y-25+z+1 = 0

x+5y+z-1-25+1 = 0

x+5y+z-25 = 0

x+5y+z = 25

The final expression gives the equation of the plane.

8 0

3 years ago

What is the factored form of 2/3x + 4

I am Lyosha [343]

Answer: <em>(x+1)(x−4). </em>

Explanation: <em>x2−3x−4 x2−4x+x−4 x(x−4)+1(x−4) (x+1)(x−4). </em>

4 0

3 years ago

please solve this question

____ [38]

A, c, and e are the correct answers

3 0

3 years ago

Read 2 more answers

Write the general equation for the circle that passes through the points (1, 1), (1, 3), and (9, 2).

Burka [1]

Step-by-step explanation:

As we know that

The circle center is equidistant from all three points, the distance being the circle radius.

Any point equidistant from two points must lie on the perpendicular bisector of the line segment which join those two points.

Which is, on the line through the midpoint of the line segment, perpendicular to the line segment.

The perpendicular bisector of the line segment joining the points (1, 1) and (1, 3) will be:

$\:y=\:\frac{1+3}{2}=\frac{4}{2}=2$

The perpendicular bisector of the line segment joining the points (1, 3), and (9, 2) will be:

$x=\:\frac{1+9}{2}=\frac{10}{2}=5$

These intersect at the center of the circle (5, 2).

The distance between (1, 1) and (5, 2) will be:

$\mathrm{Compute\:the\:distance\:between\:}\left(x_1,\:y_1\right),\:\left(x_2,\:y_2\right):\quad \sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$

$\mathrm{The\:distance\:between\:}\left(1,\:1\right)\mathrm{\:and\:}\left(5,\:2\right)\mathrm{\:is\:}$

$=\sqrt{\left(5-1\right)^2+\left(2-1\right)^2}$

$=\sqrt{4^2+1}$

$=\sqrt{16+1}$

$=\sqrt{17}$

So the equation of the circle can be written as:

$\left(x-5\right)^2+\left(y-2\right)^2=17$

$x^2-10x+y^2+29-4y=17$

$x^2-10x+y^2-4y+12\:=0$

$x^2+y^2-10x-4y+12=0$

8 0

3 years ago