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laila [671]
3 years ago
9

What is the length of segment AB?

Mathematics
1 answer:
8_murik_8 [283]3 years ago
8 0

Answer:

-1

Step-by-step explanation:

then the sum of the line come home x axis

-4+3=-1

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What is the slope of the graph of y = -3x?<br><br> -1/3<br> 1/3<br> -3<br> 3
NISA [10]

Answer:

Slope is -3

General Formulas and Concepts:

<u>Algebra I</u>

Slope-Intercept Form: y = mx + b

  • m - slope
  • b - y-intercept<u> </u>

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

y = -3x

↓ Compare to y = mx + b

Slope <em>m</em> = -3

y-intercept <em>b</em> = 0

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3 years ago
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Ratio of 200m and 1km.​
kkurt [141]

Answer:

1 : 5

Step-by-step explanation:

→ Convert 1km into metres

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3 years ago
A ball is dropped from the top of a 83-m-high building. What speed does the ball have in falling 3.6 s?
GalinKa [24]

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23.05

Step-by-step explanation:

divide

7 0
3 years ago
Consider the curve defined by the equation y=6x2+14x. Set up an integral that represents the length of curve from the point (−2,
torisob [31]

Answer:

32.66 units

Step-by-step explanation:

We are given that

y=6x^2+14x

Point A=(-2,-4) and point B=(1,20)

Differentiate w.r. t x

\frac{dy}{dx}=12x+14

We know that length of curve

s=\int_{a}^{b}\sqrt{1+(\frac{dy}{dx})^2}dx

We have a=-2 and b=1

Using the formula

Length of curve=s=\int_{-2}^{1}\sqrt{1+(12x+14)^2}dx

Using substitution method

Substitute t=12x+14

Differentiate w.r t. x

dt=12dx

dx=\frac{1}{12}dt

Length of curve=s=\frac{1}{12}\int_{-2}^{1}\sqrt{1+t^2}dt

We know that

\sqrt{x^2+a^2}dx=\frac{x\sqrt {x^2+a^2}}{2}+\frac{1}{2}\ln(x+\sqrt {x^2+a^2})+C

By using the formula

Length of curve=s=\frac{1}{12}[\frac{t}{2}\sqrt{1+t^2}+\frac{1}{2}ln(t+\sqrt{1+t^2})]^{1}_{-2}

Length of curve=s=\frac{1}{12}[\frac{12x+14}{2}\sqrt{1+(12x+14)^2}+\frac{1}{2}ln(12x+14+\sqrt{1+(12x+14)^2})]^{1}_{-2}

Length of curve=s=\frac{1}{12}(\frac{(12+14)\sqrt{1+(26)^2}}{2}+\frac{1}{2}ln(26+\sqrt{1+(26)^2})-\frac{12(-2)+14}{2}\sqrt{1+(-10)^2}-\frac{1}{2}ln(-10+\sqrt{1+(-10)^2})

Length of curve=s=\frac{1}{12}(13\sqrt{677}+\frac{1}{2}ln(26+\sqrt{677})+5\sqrt{101}-\frac{1}{2}ln(-10+\sqrt{101})

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5 0
3 years ago
Write an indirect proof..if two angles are supplementary, than they both cannot be obtuse angles.
Aleksandr-060686 [28]
Let's assume that both angles are, in fact, obtuse supplements. We know that supplementary angles must add up to 180°. We also know by definition that obtuse angles are greater than 90°. If we were to take the two supplementary, obtuse angles, ∡A=90+x and ∡B=90+y, with x and y equaling positive real numbers, then we should be able to say that 180=m∡A+mB, or 180=90+x+90+y. By simplifying we get that 180=180+x+y. Simplify further and you get that 0=x+y. If we define x in terms of y, then x= -y. If we define y in terms of x, then y= -x. Because either x or y must be negative to make this statement true, one of the angle measures must be less than 90. If one of the angles must be less than 90 while the other is greater than 90, then one angle MUST be acute if the other is obtuse in order for them to be supplements of each other. 
4 0
3 years ago
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