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otez555 [7]
2 years ago
10

If geometry symbol represented as small triangle with three sides. abc geometry congruent symbol represented as two small horizo

ntal parallel lines and a horizontal symbol s. geometry symbol represented as small triangle with three sides. dec, what is the value of x?
Mathematics
1 answer:
n200080 [17]2 years ago
3 0

The value of x in the congruent triangles abc and dec is 1

<h3>How to determine the value x?</h3>

The question implies that the triangles  abc and dec are congruent triangles.

The congruent sides are:

ab = de

bc = ce = 4

ac = cd = 5

The congruent side ab = de implies that:

4x - 1 = x + 2

Collect like terms

4x - x = 2 + 1

Evaluate the like terms

3x = 3

Divide through by 3

x = 1

Hence, the value of x is 1

Read more about congruent triangles at:

brainly.com/question/12413243

#SPJ1

<u>Complete question</u>

Two triangles, abc and cde, share a common vertex c on a grid. in triangle abc, side ab is 4x - 1, side bc is 4, side ac is 5. in triangle cde, side cd is 5, side de is x + 2, side ce is 4. If Δabc ≅ Δdec, what is the value of x? a. x = 8 b. x = 5 c. x = 4 d. x = 1 e. x = 2

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Please help! .........
stira [4]

Answer:

1) When x is 1, y is 4, and vice versa

2) When x is -6, y is 42, and when x is 7, y is 114

3) When x is -2, y is 2, and when x is 8, y is 42

Step-by-step explanation:

1)

y=x^2-6x+9

y+x=5 which can be rearranged as y=5-x.

Substituting this lone y into the first equation, you get:

5-x=x^2-6x+9

Move everything to one side:

x^2-5x+4=0

Factor:

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y=4, 1

2)

y-30=12x which can be rearranged as y=12x+30.

y=x^2+11x-12

Let's use elimination this time and subtract the first equation from the second:

0=x^2-x-42

Factor:

(x-7)(x+6)=0

x=-6, 7

y= -42, 114

3)

y=x^2-2x+6

y=4x+10

Let's set the two equations equal to each other through substitution:

x^2-2x-6=4x+10

Move everything to one side:

x^2-6x-16=0

Factor:

(x-8)(x+2)=0

x=-2, 8

y= 2, 42

Hope this helps!

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3 years ago
Vertex of f(x) = 1/2(x-2)^2-3
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A teacher summarizes grades on an exam by,Min = 26, Q1 = 67, Q2 = 80, Q3 = 87, Max = 100, Mean = 76, Mode = 100, Standard deviat
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we have

Min=26

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Q2=80

Q3=87

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Mean=76

Mode=100

standard deviation=76

IQR=20

so

In this problem

the mean is equal to the standard deviation

therefore

<h2>The standard deviation is incorrectly</h2>
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