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4 years ago

Solving 12-(x+3)=10

1 answer:

4 0

12-(x+3)=10

Simplify the equation.

12-x-3=10

Combine like terms.

(12-3)-x=10

Simplify.

9-x=10

Subtract 9 from both sides to get x by itself.

-x=1

Divide both sides by -1 to get x by itself.

x=-1

~Hope I helped!~

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The area of a triangle with a base length of 6 units in the height of 7 units

Karolina [17]

<h3>- - - - - - - - - - - - - - - ~Hello There!~ - - - - - - - - - - - - - - - </h3>

➷ base x height / 2 = area

6 x 7 / 2 = 21

answer is 21 units

➶ Hope This Helps You!

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➶ Have A Great Day ^-^

↬ ʜᴀɴɴᴀʜ ♡

6 0

4 years ago

Read 2 more answers

What is the slope of a line that is perpendicular to the line y = 2x – 6?

Zolol [24]

Answer:

The slope of a line that is perpendicular to the line y = 2x – 6 is -1/2.

Step-by-step explanation:

The slope-intercept form of the line equation

$y = mx+b$

where

$m$ is the slope

$b$ is the y-intercept

Step 1:

Finding the slope of the given equation

The given equation is

$y = 2x - 6$

comparing with the slope-intercept form of the line equation

The slope of the equation is: m = 2

Step 2:

Determining the slope of the perpendicular line

In Mathematics, a line perpendicular to another line has a slope that is the negative reciprocal of the slope of the other line.

As the slope of the equation is: m = 2

Therefore, the slope of the new perpendicular line:

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

Hence, the slope of a line that is perpendicular to the line y = 2x – 6 is -1/2.

Important Tip:

The product of the slopes of perpendicular lines is -1.

Verification:

The slope of the given line:

$m_1=2$

The slope of the perpendicular line:

$m_2=-\frac{1}{2}\:\:\:\:$

The product of the slopes is:

$\:m_1\times \:m_2\:=2\times -\frac{1}{2}\:\:=-1$

As the product of the slopes of perpendicular lines is -1, therefore, the lines are perpendicular.

Thus, the slope of a line that is perpendicular to the line y = 2x – 6 is -1/2.

7 0

3 years ago

Solve f - 6 = 3f + 1

LUCKY_DIMON [66]

Simplifying f + -6 = 3f + 1 Reorder the terms: -6 + f = 3f + 1 Reorder the terms: -6 + f = 1 + 3f Solving -6 + f = 1 + 3f Solving for variable 'f'. Move all terms containing f to the left, all other terms to the right. Add '-3f' to each side of the equation. -6 + f + -3f = 1 + 3f + -3f Combine like terms: f + -3f = -2f -6 + -2f = 1 + 3f + -3f Combine like terms: 3f + -3f = 0 -6 + -2f = 1 + 0 -6 + -2f = 1 Add '6' to each side of the equation. -6 + 6 + -2f = 1 + 6 Combine like terms: -6 + 6 = 0 0 + -2f = 1 + 6 -2f = 1 + 6 Combine like terms: 1 + 6 = 7 -2f = 7 Divide each side by '-2'. f = -3.5 Simplifying f = -3.5

7 0

4 years ago

Read 2 more answers

Part 1.] Does the graph have even symmetry, odd symmetry, or neither?

QveST [7]

Part 1
The graph has even symmetry. You can see that because it is symmetric with respect to the y-axis.
Functions that have even symmetry have the following property:
$f(x)=f(-x)$
Part 2
To answer this we can simply check if the property we mentioned earlier holds for this function.
$sin(\frac{\pi}{2})\ne sin(-\frac{\pi}{2})$
We can see that sine does not have even symmetry.
In fact, sine function has the following property:
$sin(x)=-sin(x)$
This is called odd symetry.
Part 3
Take a look at the function that you attached in the picture. We know that function has even symmetry.
Reflection over x-axis and 180° rotation around the origin would give us -f(x). We would not end up with the same function, so these two are out.
90° rotation around the origin would mean we swapped x and y so that one is out too. Reflection over the line y=x is a property of functions that have an odd symmetry.
We are left with reflection around y-axis and, as mentioned before, this is the property of evenly symmetric functions.

4 0

4 years ago

Let's see how good you are at math answer this question (if you think your the best)

stich3 [128]

Answer:

Step-by-step explanation:

for the perimeter we just need the distances from point to point

P1(1,1) (x1,y1)

P2(6,2) (x2,y2)

P3(5,4) (x3,y3)

P4(2,5) (x4,y4)

from P1 to P2 is

dist = sqrt [ (x2-x1)^2 +(y2-y1)^2 ]

dist = $\sqrt{26}$

from P2 to P3 is

dist = sqrt [ (x3-x2)^2 +(y3-y2)^2 ]

dist = $\sqrt{5}$

from P3 to P4

dist = sqrt [ (x4-x3)^2 +(y4-y3)^2 ]

dist = $\sqrt{10}$

dist = sqrt [ (x1-x4)^2 +(y1-y4)^2 ]

dist = $\sqrt{17}$

Perimeter is exactly = $\sqrt{26}$ + $\sqrt{5}$ + $\sqrt{10}$ + $\sqrt{17}$

To find the area we have to use the non-right triangle formula of

1/2*a*b*sin(x)

if we knew the angle at P1 then we could find the area of the triangle that is made up by P1, P2 and P4

P1(1,1) (x1,y1)

P2(6,2) (x2,y2)

P3(5,4) (x3,y3)

P4(2,5) (x4,y4)

let's use that fancy slope formula to find the angle

tan(Ф) = abs [ (m1-m2)/(1+m1.m2) ]

m1 = P1P2

m1 = y2-y1 / x2-x1

m1 = 1/5 or .2

m2 = P1P4

m2 = y4-y1 / x4-x1

m2 = 4

now plug into the formula for the angle of P1

tan(Ф) = abs [(0.2 - 5) / (1+0.2*5)]

Ф = arcTan(2.4)

Ф=-67.380135°

the negative signs just means going from P1P4 to P1P2 is -67.380135° we can take it to be positive as well

now plug that angle into our area formula for a non-right triangle

1/2*a*b*sin(x)

area of triangle P1,P2,P4 = 1/2* $\sqrt{26}$ * $\sqrt{17}$ *sin(67.380135°)

area P1,P2,P4 = 9.703290477 $units^{2}$

now lets find that other triangle up at the top formed by P2, P3, P4

let's find the slope of the two legs we know the lengths of, of the part that looks a little like a "hat" on top of our 1st triangle

P1(1,1) (x1,y1)

P2(6,2) (x2,y2)

P3(5,4) (x3,y3)

P4(2,5) (x4,y4)

m3= P4P3

m3=y3-y4 / x3-x4

m3 = -1/3

m4 = P3P2

m4 = y2-y3 / x2-x3

m4 = -2/1

m4 = -2

now find the angle between them

tan(Ф) = abs [ (m3-m4)/(1+m3.m4) ]

Ф = arcTan(1)

Ф = 45°

this is the other angle on the 180 line so we have 180-45 =135 as the angle inside our triangle at P3, although, sin 135 and 45 are going to be the same.

now plug that info into our special area of a non-right triangle

area of P2P3P4 = 1/2* $\sqrt{5}$ * $\sqrt{10}$ *sin(135)

area of P2P3P4 = 2.5

2.5 + 9.703290477 = 12.20329048 $units^{2}$

area =12.20329048 $units^{2}$

6 0

3 years ago