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VikaD [51]
3 years ago
10

Solve the equation: 4(x - 2) = 12

Mathematics
1 answer:
Natalka [10]3 years ago
5 0

Answer:x= 14

Step-by-step explanation:

84

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The function f(x) and g(x) are graphed of the set of aces below. for which value of x is f(x)≠g(x)
aleksklad [387]

Answer:

f(x)\neq g(x)

when ~x=3

So, the ~third~ option~is ~your~ answer

<h3>-----------------------</h3><h3>hope it helps...</h3><h3>have a great day!!</h3>
6 0
3 years ago
I am a 3-digit number. If rounded off to the nearest hundreds. I become 500. If rounded off to the nearest tens, I become 540. I
dimulka [17.4K]

Answer:

535 is the cirrext answer

6 0
3 years ago
60=F(15) <br><br> Please solve this and show work, please
Lubov Fominskaja [6]

Answer:

f = 4

Step-by-step explanation:

60= 15f

divide both side by 15

f = 4

3 0
3 years ago
PLEASE HELP!! URGENT!! i will mark brainliest if its right!! In the figure below, ∠DEC ≅ ∠DCE, ∠B ≅ ∠F, and DF ≅ BD. Point C is
zvonat [6]

Answer:

See below.

Step-by-step explanation:

This is how you prove it.

<B and <F are given as congruent.

This is 1 pair of congruent angles for triangles ABC and GFE.

<DEC and <DCE are given as congruent.

Using vertical angles and substitution of transitivity of congruence of angles, show that angles ACB and GEF are congruent.

This is 1 pair of congruent angles for triangles ABC and GFE.

Now you need another side to do either AAS or ASA.

Look at triangle DCE. Using the fact that angles DEC and DCE are congruent, opposite sides are congruent, so segments DC and DE are congruent. You are told segments DF and BD are congruent. Using segment addition postulate and substitution, show that segments CB and EF are congruent.

Now you have 1 pair of included sides congruent ABC and GFE.

Now using ASA, you prove triangles ABC and GFE congruent.

7 0
3 years ago
Which of the following is a solution of x2 + 2x + 4?
RideAnS [48]

Answer:

option-D

Step-by-step explanation:

we are given

x^2+2x+4

Let's assume it is equal to 0

x^2+2x+4=0

We can use quadratic formula

Suppose, we are given quadratic equations as

ax^2+bx+c=0

x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

we can compare and find a,b and c

a=1 , b=2 , c=4

now, we can plug values

and we get

x=\frac{-2\pm \sqrt{2^2-4\cdot \:1\cdot \:4}}{2\cdot \:1}

now, we can simplify it

and we get

x=-1+\sqrt{3}i,\:x=-1-\sqrt{3}i


6 0
3 years ago
Read 2 more answers
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