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babunello [35]
3 years ago
6

Using the image below to answer the question. Using the segment addition postulate find the value of X

Mathematics
2 answers:
Marysya12 [62]3 years ago
4 0
X=13 I just answered that question
RSB [31]3 years ago
3 0
Its clear that EF+FG=EG
(7x+9)+(3x+4)=143
10x+13=143
10x=130
X=13
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How do u calculate Pythagorean theorem and what is the formula
PIT_PIT [208]

Answer:

The formula is

Step-by-step explanation:

a^2 + b^2 = c^2

hope this helps

6 0
3 years ago
What is the area?<br> 2 yd<br> 8 yd<br> 5 yd
MrRissso [65]

Answer:

28

Step-by-step explanation:

I assume the polygon is a trapezoid with parallel bases of 2 yd and 5 yd, and an altitude of 8 yd.

area of trapezoid = (base1 + base2)h/2

area = (2 yd + 5 yd)(8 yd)/2

area = 28 yd²

5 0
2 years ago
Triangle ABC has been rotated 90° to create triangle DEF. Write the equation, in slope-intercept form, of the side of triangle A
Ket [755]

the equation in the slope-intercept of the side of triangle ABC that is perpendicular to segment EF is y = x + 1

<h3>How to determine the equation</h3>

From the figure given, we can deduce the coordinates of the sides

For A

A ( 4,2)

For B

B ( 4, 5)

C ( 1, 2)

D ( 2, -4 )

E  ( 5, -4)

F ( 2, -1)

The slope for BC

Slope = \frac{y2 - y1}{x2 - x1}

Substitute the values for both B and C coordinates, we have

Slope = \frac{2- 5}{1 - 4}

Find the difference for both the numerator and denominator

Slope = \frac{-3}{-3}

Slope = 1

We have the rotation for both point ( 0, 1)

y - y1 = m ( x - x1)

The values for y1 and x1 are 1 and 0 respectively and the slope m is 1

Substitute the values

y - 1 = 1 ( x - 0)

y - 1 = x

Make 'y' the subject of formula

y = x + 1

Thus, the equation in the slope-intercept of the side of triangle ABC that is perpendicular to segment EF is y = x + 1

Learn more about linear graphs here:

brainly.com/question/4074386

#SPJ1

4 0
2 years ago
Consider the function f(x)= (x+4)(x-2) what is the equation of the axis of symmetry
almond37 [142]

if we zero out f(x), namely make y = 0, we can get the roots or x-intercepts for this quadratic equation

\bf 0 = (x+4)(x-2)\implies \begin{cases} 0=x+4\implies &-4=x\\ 0=x-2\implies &2=x \end{cases}

now, the equation is in x-terms, meaning is a vertically opening parabola, so the axis of symmetry will be x = something, a vertical line.

well, we have two x-intercepts, one at -4 and another at 2, and the vertex is right half-way between those guys

-4------------(-1)------------2

so the vertex is at x=-1, namely the axis of symmetry is x = -1.

8 0
3 years ago
Find an equation for the line tangent to the curve at the point defined by the given value of d²y/dx².​
mars1129 [50]

Answer:

Step-by-step explanation:

Given:

x = 2cost,

t = (1/2)arccosx

y = 2sint

dy/dx = dy/dt . dt/dx

dy/dt = 2cost

dt/dx = -1/√(1 - x²)

dy/dx = -2cost/√(1 - x²)

Differentiate again to obtain d²y/dx²

d²y/dx² = 2sint/√(1 - x²) - 2xcost/(1 - x²)^(-3/2)

At t = π/4, we have

(√2)/√(1 - x²) - (√2)x(1 - x²)^(3/2)

4 0
3 years ago
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