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scoray [572]
3 years ago
6

Define the word factor mathmatically

Mathematics
2 answers:
lukranit [14]3 years ago
5 0
Factor is a integer / number being multiplied by another number
Nataliya [291]3 years ago
4 0
The word factor in mathematics means any two numbers that can be multiplied to get a product.

3 X 4 = 12

3 and 4 are the factors.
12 is the product.
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There are three squares inside the triangle . Find the area of the third triangle with steps , please
pogonyaev

Answer: 49 cm^2

Step-by-step explanation:

The length of the small pink square box will be:

L^2 = 9

L = sqrt( 9 )

L = 3

The length of the big pink square box will be:

L^2 = 16

L = sqrt( 16 )

L = 4

The length of the yellow square box will be:

4 + 3 = 7

The area of the third square box will be

A = 7^2

A = 49 cm^2

6 0
2 years ago
Solve for the compound inequality 6b < 42 or 4b + 12 > 8.
Anuta_ua [19.1K]

Answer:

-1<b<7, or i suppose b.

Step-by-step explanation:

6b<42

b<7

4b+12>8

4b>-4

b>-1

6 0
3 years ago
-3p + 8 = -13 <br><br> HELLLLPP
ella [17]

Answer:

p=-7     :)

Step-by-step explanation:

-13-8=3p

-13+(-8)=3p

-21=3p

  ÷3

-7=p

5 0
2 years ago
(x2−1)(x+1)≥0. pls help me​
docker41 [41]

Step-by-step explanation:

(x²-1)(x+1)≥0

As it has two multiples,

Therefore,

Either,

x²-1≥0

x²≥0+1

x²≥1

x≤√1

x≤1

Or,

x+1≥0

x≥-1

7 0
2 years ago
Solve the following differential equations or initial value problems. In part (a), leave your answer in implicit form. For parts
shepuryov [24]

Answer:

(a) (y^5)/5 + y^4 = (t^3)/3 + 7t + C

(b) y = arctan(t(lnt - 1) + C)

(c) y = -1/ln|0.09(t + 1)²/t|

Step-by-step explanation:

(a) dy/dt = (t^2 + 7)/(y^4 - 4y^3)

Separate the variables

(y^4 - 4y^3)dy = (t^2 + 7)dt

Integrate both sides

(y^5)/5 + y^4 = (t^3)/3 + 7t + C

(b) dy/dt = (cos²y)lnt

Separate the variables

dy/cos²y = lnt dt

Integrate both sides

tany = t(lnt - 1) + C

y = arctan(t(lnt - 1) + C)

(c) (t² + t) dy/dt + y² = ty², y(1) = -1

(t² + t) dy/dt = ty² - y²

(t² + t) dy/dt = y²(t - 1)

(t² + t)/(t - 1)dy/dt = y²

Separating the variables

(t - 1)dt/(t² + t) = dy/y²

tdt/(t² + t) - dt/(t² + t) = dy/y²

dt/(t + 1) - dt/(t(t + 1)) = dy/y²

dt/(t + 1) - dt/t + dt/(t + 1) = dy/y²

Integrate both sides

ln(t + 1) - lnt + ln(t + 1) + lnC = -1/y

2ln(t + 1) - lnt + lnC = -1/y

ln|C(t + 1)²/t| = -1/y

y = -1/ln|C(t + 1)²/t|

Apply y(1) = -1

-1 = ln|C(1 + 1)²/1|

-1 = ln(4C)

4C = e^(-1)

C = (1/4)e^(-1) ≈ 0.09

y = -1/ln|0.09(t + 1)²/t|

8 0
3 years ago
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