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2 years ago

Solve: |4x + 2| + 6 =0 a.all real number b.1 or -2 c.-1 or 2 d. no solution

1 answer:

6 0

No solution since when you subtract six from both sides you cannot get -6 on the left side since you have use the absolute value

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5/6 N=10 solve this problem. I could not put the six under the five where it belongs. please show the work

Andreyy89

5/6N = 10

Mutiply by 6/5

So N=12

3 0

3 years ago

Unit 4 Test, Part 2: Congruence and Constructions<br> Help with my math please

lakkis [162]

Answer/Step-by-sep explanation:

1. Given:

∆NMK ≅ ∆TRP

$\overline{NM} = 20$

$\overline{MK} = 15$

$\overline{KN} = 25$

$\overline{TR} = 3x - 1$

a. To complete the congruent statement, thus: ∆MNK ≅ ∆RTP

b. The side that is congruent to $\overline{TR}$ is $\overline{NM}$ . Thus:

$\overline{TR}$ ≅ $\overline{NM}$

c. Since $\overline{TR}$ ≅ $\overline{NM}$ , therefore:

$\overline{TR} = \overline{NM}$

$3x - 1 = 20$ (substitution)

Add 1 to both sides

$3x = 20 + 1$

$3x = 21$

Divide both sides by 3

$x = \frac{21}{3}$

$x = 7$

2. a. Slope of LK = $\frac{rise}{run} = \frac{4}{3}$

Slope of LM = $\frac{rise}{run} = -\frac{3}{5}$

b. ✍️Length of LK is the distance between L(-7, 4) and (-4, 8):

$Lk = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

$Lk = \sqrt{(-4 -(-7))^2 + (8 - 4)^2}$

$Lk = \sqrt{(3)^2 + (4)^2}$

$Lk = \sqrt{9 + 16}$

$Lk = \sqrt{25}$

$Lk = 5$

✍️Length of LM is the distance between L(-7, 4) and (-2, 1):

$LM = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

$LM = \sqrt{(-2 -(-7))^2 + (1 - 4)^2}$

$LM = \sqrt{(5)^2 + (-3)^2}$

$LM = \sqrt{25 + 9}$

$LM = \sqrt{34}$

$LM = 5.8$ (nearest tenth)

∆KLM is not an isosceles ∆ because it does not has two equal side lengths. This we can see because LK and LM are not equal.

Therefore, Anthony is incorrect. Am isosceles ∆ has two equal sides.

8 0

3 years ago

I NEED HELP I WILL GIVE BRAINLIEST!!!

sergiy2304 [10]

Answer:

The probability is 2/5.

Step-by-step explanation:

4/5 is the probability of not getting a 5

1/2 is the probability of flipping heads

4/5x1/2 = 4/10 = 2/5

3 0

2 years ago

Plz solve will give brainliest

Levart [38]

Answer:

thank you sa point

Step-by-step explanation:

yun lang sowwey

6 0

2 years ago

1.What are the zeros of the polynomial function?

Lorico [155]

Let's to the first example:

f(x) = x^2 + 9x + 20

Ussing the formula of basckara

a = 1
b = 9
c = 20

Delta = b^2 - 4ac

Delta = 9^2 - 4.(1).(20)

Delta = 81 - 80

Delta = 1

x = [ -b +/- √(Delta) ]/2a

Replacing the data:

x = [ -9 +/- √1 ]/2

x' = (-9 -1)/2 <=> - 5

Or

x" = (-9+1)/2 <=> - 4
_______________

Already the second example:

f(x) = x^2 -4x -60

Ussing the formula of basckara again

a = 1
b = -4
c = -60

Delta = b^2 -4ac

Delta = (-4)^2 -4.(1).(-60)

Delta = 16 + 240

Delta = 256

Then, following:

x = [ -b +/- √(Delta)]/2a

Replacing the information

x = [ -(-4) +/- √256 ]/2

x = [ 4 +/- 16]/2

x' = (4-16)/2 <=> -6

Or

x" = (4+16)/2 <=> 10
______________

Now we are going to the 3 example

x^2 + 24 = 14x

Isolating 14x , but changing the sinal positive to negative

x^2 - 14x + 24 = 0

Now we can to apply the formula of basckara

a = 1
b = -14
c = 24

Delta = b^2 -4ac

Delta = (-14)^2 -4.(1).(24)

Delta = 196 - 96

Delta = 100

Then we stayed with:

x = [ -b +/- √Delta ]/2a

x = [ -(-14) +/- √100 ]/2

We wiil have two possibilities

x' = ( 14 -10)/2 <=> 2

Or

x" = (14 +10)/2 <=> 12
________________

To the last example will be the same thing.

f(x) = x^2 - x -72

a = 1
b = -1
c = -72

Delta = b^2 -4ac

Delta = (-1)^2 -4(1).(-72)

Delta = 1 + 288

Delta = 289

Then we are going to stay:

x = [ -b +/- √Delta]/2a

x = [ -(-1) +/- √289]/2

x = ( 1 +/- 17)/2

We will have two roots

That's :

x = (1 - 17)/2 <=> -8

Or

x = (1+17)/2 <=> 9

Well, this would be your answers.

7 0

3 years ago