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Vesnalui [34]
3 years ago
13

Please help me with this math problem

Mathematics
2 answers:
gogolik [260]3 years ago
4 0

Answer: (z+2)(z-9)

Step-by-step explanation:

z^{2} -7z-18  to solve for this you need to find two numbers that if multiplied together they give you negative 18 and at the same time adds up to -7.

Those two numbers could be  -9 and 2 .

now rewrite the expression like this.

z^{2} +2z - 9z -18   Group them  

((  z^{2} +2z) ( -9z- 18)    Factor it out z in the first grouping

z(z +2)  -9(z + 2)          factor out  x+2

(z +2 ) (z-9)

erik [133]3 years ago
3 0

Answer:

(z-9)(z+2)

Step-by-step explanation:

U have to find numbers which when u multiply u get -18 and subtract u get -7

-18= -9×+2,2-9=-7

Z²+2z-9z-18

Z(z+2)-9(z+2)

(Z-9)(z+2)

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Zielflug [23.3K]

(-3)^2+5(-3)-4

9-15-4

=-10

3 0
3 years ago
2x(8+7)-6/2x(9-3) evaluate
Katarina [22]
Ok so assuming that the 'x' means multiply and not 'x' as in placeolder we do

PEMDAS
parenthaseees
exponents
multipilcation or division
additon or subtraction

parethasees simplify first
(8+7)=15


(9-3)=6

now we have
2x(15)-6/2x(6)
multiply
2x15=30

6/2x6=36/2=18

now we have
30-18
12

answer is 12
6 0
3 years ago
Read 2 more answers
Please help me with this!!
Troyanec [42]

Answer:

As the height increases, the temperature decreases.

It is decreasing 5°C per kilometer.

At the height 8 km the temperature is 0°.

5 0
3 years ago
I need help with this pleaseee!
ololo11 [35]
Probably B

STEP BY STEP
7 0
3 years ago
HELP ASAP,, WILL GIVE BRAINLIEST​
madreJ [45]

Answer:

a = 50°, b = 130°, c = 20°

Step-by-step explanation:

The sum of the 3 angles in a triangle = 180°

Sum the 3 angles in the right triangle

a + 40° + 90° = 180°

a + 130° = 180° ( subtract 130° from both sides )

a = 50°

a and b are adjacent angles and sum to 180° , so

a + b = 180°

50° + b = 180° ( subtract 50° from both sides )

b = 130°

Then sum the angles in the top triangle, that is

30° + 130° + c = 180°

160° + c = 180° ( subtract 160° from both sides )

c = 20°

8 0
3 years ago
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