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nadya68 [22]
3 years ago
12

How do I do it? I didn't understand the math.

Mathematics
1 answer:
Setler [38]3 years ago
8 0

Answer:

please try do the rest by yourself..Of you didn't get the answer you can ask me through comment..

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Find the domain and range of f(x)= 2|x+2|-8​
nordsb [41]

Answer:

Domain (-∞,∞)

Range: [-8,∞)

3 0
3 years ago
Help please I will give you the brainliest
luda_lava [24]

Answer:

The area of a triangle is defined as the total region that is enclosed by the three sides of any particular triangle. Basically, it is equal to half of the base times height, i.e. A = 1/2 × b × h.

Step-by-step explanation:

6 0
3 years ago
The sum of a number (n) and 14 is 72. Which equation shows this relationship
NemiM [27]
I believe the answer is N+14=72
6 0
3 years ago
Read 2 more answers
Rex, Paulo, and Ben are standing on shore watching for dolphins. Paulo sees one surface directly in front of him about a hundred
Ksivusya [100]

1. m\angle BAC=m\angle CAD,\ m\angle ACB=m\angle ADC=90^{\circ}, then m\angle ABC=m\angle ACD and triangles ADC and ACB are similar by AAA theorem.


2. The ratio of the corresponding sides of similar triangles is constant, so


\dfrac{AC}{AB}= \dfrac{AD}{AC}.


3. Knowing lengths you could state that \dfrac{b}{c}= \dfrac{e}{b}.


4. This ratio is equivalent to b^2=ce.


5. m\angle ABC=m\angle CBD,\ m\angle ACB=m\angle CDB=90^{\circ}, then m\angle BAC=m\angle BCD and triangles BDC and BCA are similar by AAA theorem.


6. The ratio of the corresponding sides of similar triangles is constant, so


\dfrac{BC}{BD}= \dfrac{AB}{BC}.


7. Knowing lengths you could state that \dfrac{a}{d}= \dfrac{c}{a}.


8. This ratio is equivalent to a^2=cd.


9. Now add results of parts 4 and 8:


b^2+a^2=ce+cd.


10. c is common factor, then:


b^2+a^2=c(e+d).


11. Since e+d=c you have a^2+b^2=c\cdot c=c^2.



7 0
3 years ago
Read 2 more answers
Given the function h(x) =1/3 |x-6| +4, evaluate the function when x = - 3, - 2, and 0
Masja [62]

|x| = x for x ≥ 0

examples:

|3| = 3; |0.56| = 0.56; |102| = 102

|x| = -x for x < 0

examples:

|-3| = -(-3) = 3; |-0.56| = -(-0.56) = 0.56; |-102| = 102

--------------------------------------------------------------------------------

Use PEMDAS:

P Parentheses first

E Exponents (ie Powers and Square Roots, etc.)

MD Multiplication and Division (left-to-right)

AS Addition and Subtraction (left-to-right)

--------------------------------------------------------------------------------

h(x)=\dfrac{1}{3}|x-6|+4

Put the values of x to the equation of the function h(x):

x=-3\to h(-3)=\dfrac{1}{3}|-3-6|+4=\dfrac{1}{3}|-9|+4=\dfrac{1}{3}(9)+4=3+4=7\\\\x=-2\to h(-2)=\dfrac{1}{3}|-2-6|+4=\dfrac{1}{3}|-8|+4=\dfrac{1}{3}(8)+4=\dfrac{8}{3}+\dfrac{12}{3}=\dfrac{20}{3}\\\\x=0\to h(0)=\dfrac{1}{3}|0-6|+4=\dfrac{1}{3}|-6|+4=\dfrac{1}{3}(6)+4=2+4=6


6 0
3 years ago
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