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kati45 [8]
3 years ago
13

Select the correct answer.

Mathematics
1 answer:
likoan [24]3 years ago
6 0

Answer:

x = 20

Step-by-step explanation:

Traingle ABC = Traingle DEC

4x-1×4×5 = x+2×4×5

80x-1 = 2×20x

80x-20x = 2+1

60x = 3

60÷3 = x

x = 20

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if you repeat the perpendicular line segment construction twice using paper folding, you can construct:
ehidna [41]

Answer:

The correct answer is  option C.

The mid point of the line segment.

Step-by-step explanation:

the perpendicular line segment construction twice using paper folding

we have to find the mid point of the given line segment.

We get the midpoint easily when fold the paper correctly

Therefore the correct answer is  option C.

The mid point of the line segment.

6 0
3 years ago
Read 2 more answers
What's the solution to the equation x/3 + x/6 = 7/2
marta [7]

Step-by-step explanation:

x/3+x/6=7/2

We simplify the equation to the form, which is simple to understand

x/3+x/6=7/2

Simplifying:

+ 0.333333333333x+x/6=7/2

Simplifying:

+ 0.333333333333x + 0.166666666667x=7/2

Simplifying:

+ 0.333333333333x + 0.166666666667x=+3.5

We move all terms containing x to the left and all other terms to the right.

+ 0.333333333333x + 0.166666666667x=+3.5

We simplify left and right side of the equation.

+ 0.5x=+3.5

We divide both sides of the equation by 0.5 to get x.

x=7

4 0
2 years ago
Read 2 more answers
Could someone help me with number 19
azamat

Solving #19

<u>Take y-values from the graph</u>

  • a) (g·f)(-1) = g(f(-1)) = g(1) = 4
  • b) (g·f)(6) = g(f(6)) = g(2) = 2
  • c) (f·g)(6) = f(g(6)) = f(5) = 1
  • d) (f·g)(4) = f(g(4)) = f(2) = -2
8 0
3 years ago
A rectangle is twice as long as it is wide. If its length and width are both
puteri [66]

Answer:

a rectangle is twice as long as it is wide . if both its dimensions are increased 4 m , its area is increaed by 88 m squared make a sketch and find its original dimensions of the original rectangle

Step-by-step explanation:

Let l = the original length of the original rectangle

Let w = the original width of the original rectangle

From the description of the problem, we can construct the following two equations

l=2*w (Equation #1)

(l+4)*(w+4)=l*w+88 (Equation #2)

Substitute equation #1 into equation #2

(2w+4)*(w+4)=(2w*w)+88

2w^2+4w+8w+16=2w^2+88

collect like terms on the same side of the equation

2w^2+2w^2 +12w+16-88=0

4w^2+12w-72=0

Since 4 is afactor of each term, divide both sides of the equation by 4

w^2+3w-18=0

The quadratic equation can be factored into (w+6)*(w-3)=0

Therefore w=-6 or w=3

w=-6 can be rejected because the length of a rectangle can't be negative so

w=3 and from equation #1 l=2*w=2*3=6

I hope that this helps. The difficult part of the problem probably was to construct equation #1 and to factor the equation after performing all of the arithmetic operations.

5 0
3 years ago
Translate the sentence into an inequality.<br> The sum of 4 and y is less than 15.
bulgar [2K]
I believe the answer would be: 4+y<15
5 0
3 years ago
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