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3 years ago

Evaluate 12x−3y12x−3y when x=−14x=−14 and y=3 Help.

Mathematics

2 answers:

Citrus2011 [14]3 years ago

5 0

12x-3y12x-3y (x=-14x=-14 or x=1) and( y=3)
12(1)-3(3)12(1)-3(3)
12-9*12-9
12-108-9
96-9
87

Nonamiya [84]3 years ago

4 0

12*-14-3*3*12*-14- 3*3
168+1512-9
1671

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What fraction is £1 of 13p?

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I believe it would be 13p im not for sure though

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4 years ago

Find m∠STU and m∠TUA

PIT_PIT [208]

Answers:

angle STU = 65 degrees
angle TUA = 123 degrees

=========================================================

Explanation:

Let's find the expression for angle TUS in terms of x

Angle TUA has the expression 11x+2. It will add to angle TUS to get 180 degrees

(angle TUS) + (angle TUA) = 180

angle TUS = 180 - (angle TUA)

angle TUS = 180 - (11x + 2)

angle TUS = 180 - 11x - 2

angle TUS = -11x + 178

Now focus on the interior angles of triangle TUS. We have these three angles

T = 5x+10
U = -11x+178
S = 58

For any triangle, the interior angles always add to 180, so,

T+U+S = 180

(5x+10) + (-11x+178) + (58) = 180

(5x-11x) + (10+178+58) = 180

-6x + 246 = 180

-6x = 180-246

-6x = -66

x = -66/(-6)

x = 11

Use this x value to find the angle measures we're after:

angle STU = 5x+10 = 5*11+10 = 55+10 = 65 degrees
angle TUA = 11x+2 = 11*11+2 = 121+2 = 123 degrees

--------------------------

As an alternative route, you can apply the remote interior angle theorem. This says that adding two interior angles leads to the exterior angle that isn't adjacent to either one.

In this case, we would say

(angle STU) + (angle TSU) = angle TUA

that leads to the equation

(5x+10) + (58) = 11x+2

Solving this should lead to x = 11 to generate the angles mentioned earlier.

6 0

3 years ago

An equilateral triangle has a side length of 6. What is the height of the triangle?

vlabodo [156]

Since the height of an equilateral triangle in terms of its side s is s√3/2, the height of the triangle is 6√3/2 = 3√3 and so the area is (1/2)(6)(3√3) = 9√3.

If we draw a horizontal line a height of h from the base of the triangle, the region is split into two regions: the lower region consisting of a trapezoid of height h and the upper region consisting of a triangle of height 3√3 - h. 

Since the upper triangle and the triangle itself are similar triangles, the base and height are proportional. If we let x denote the base of the length of the upper triangle, we have: 

(S. of small triangle)/(S. of big triangle) = (Ht. of small triangle)/(Ht. of big triangle) 
==> x/6 = (3√3 - h)/(3√3) 
==> x = (6√3 - 2h)/√3 

Thus, the area of the upper triangle is: 

A = (1/2)[(6√3 - 2h)/√3](3√3 - h) = [(6√3 - 2h)(3√3 - h)]/(2√3). 
(Made a dumb mistake about the height here for some reason) 

Since we require that the area of this triangle is to be half of the total area (9√3/2), we need to solve: 

[(6√3 - 2h)(3√3 - h)]/(2√3) = 9√3/2 
==> (6√3 - 2h)(3√3 - h) = 27 
==> 54 - 6h√3 - 6h√3 + 2h^2 = 27 
==> 2h^2 - 12h√3 + 27 = 0. 

Solving with the Quadratic Formula gives: 

h = (6√3 + 3√6)/2 ≈ 8.87 units and h = (6√3 - 3√6)/2 ≈ 1.52 units. 

Since h = (6√3 + 3√6)/2 would place the line outside of the triangle, we pick h = (6√3 - 3√6)/2. 

Therefore, the line should be ==> (6√3 - 3√6)/2 units from the base. 

I hope this helps! ^^ Brainliest Please?

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3 years ago

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These are all of the OC's i have made so fare what do you guys think

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nice i like them

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3 years ago

What is the resulting ordered pair if the value of the independent variable is 5?

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