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Kisachek [45]
2 years ago
13

Can someone please tell me if this is right!

Mathematics
1 answer:
Andrei [34K]2 years ago
6 0

Answer:

the leading coefficient of a polynomial is the coefficient of the term with the highest degree of the polynomial

so the answer should be the second one -7+6x+5x^2

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Through: (6, 2), perpendicular to<br> y = -x-5 <br> in slope intercept form
True [87]

Answer:

y = x-4

Step-by-step explanation:

y = -x-5

Slope of line is -1.

Slope of perpendicular to line is 1.

Point-slope equation for line of slope 1 that passes through (6,2):

y-2 = 1(x-6)

Rearrange equation to slope-intercept form:

y = 1(x-6) + 2

y = x-4

6 0
2 years ago
I have a question for 7th grade math that is do today.The math question is what is 56+(-56)=? I would guess it equals 0 but am I
qaws [65]

Answer:

56 + -56 = 0

Step-by-step explanation:

because when there is a negative and a positive you add them together getting 0

Sorry its hard to explain but yes you are correct

I hope this helps

Have a great day!

3 0
3 years ago
Read 2 more answers
I am a factor of 120, and a common multiple of 3, 4, and 10. The sum of my digits is 6.
Galina-37 [17]

Answer:

At first find the lowest common multiple of 3, 4, 10

so it is 60

and 60is a factor of 120 and also the sum of the digits of 60 = 6+0 = 6

.°. 60 is the answer

7 0
3 years ago
Find the range of the function for the given domain, f(x)=3x+3; (-2,-1,0,1,2)
stich3 [128]

Answer:

Range: (-3,0,3,6,9)

Step-by-step explanation:

We need to evaluate each value in the domain of the function, so

f(-2)=3(-2)+3=-6+3=-3

f(-1)=3(-1)+3=-3+3=0

f(0)=3(0)+3=0+3=3

f(1)=3(1)+3=3+3=6

f(2)=3(2)+3=6+3=9

8 0
3 years ago
Which equation demonstrates the distributive property? A. 36+ 9 = 45 B. 36 +9 =3 (12+3) C. 36 x9 =9 x 36 D. (12 +3) 3 = 3 (12+3
Mazyrski [523]

Answer:

option b

Because 3 can be taken out as a common number which is three vice versa off three the distributive property.

hope it helps ..if not you can reach out to me ....for better defination.

3 0
2 years ago
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