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3 years ago

Help me out with this question (geometry)

Mathematics

1 answer:

nadezda [96]3 years ago

6 0

Answer:

116

Step-by-step explanation:

13x-1 = 2(6x+4)

13x-1 = 12x+8

x=9

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Is there a proportional relationship between time and the cost of the cell phone plan?

belka [17]

Answer:

Step-by-step explanation:

3 0

3 years ago

Math help 30 points

mash [69]

Again here we're using like terms

a) -2x² + 12x² = 10x² (they have the same term that's why I grouped them together)

-4x + 2x = -2x

13 - 25 = -12

So the final answer is 10x² - 2x -12

b) 7x² - (-7x²) = 7x² + 7x² or 14x²

4x - (-3x) = 4x + 3x or 7x

-26 - 15 = -41

The final answer to this one is 14x² + 7x - 41

7 0

3 years ago

Whats the answer to this???

garri49 [273]

Answer:

A: 2/5

B: -0.2

C: -4/5

4 0

3 years ago

a book is 6 inches wide and 9 inches tall A publisher want to scale factor of 1.5 to enlarge a book. what will the area of the f

VladimirAG [237]

Answer:

Step-by-step explanation:

6 inches enlarged by 1.5 = 9 inches

9 inches enlarged by 1.5 = 13.5 inches

Area would be 9x13.5=

121.5 inches for the area

Hope this helps you out.

8 0

3 years ago

The acceleration of an object (in m/s^2) is given by the function a(t) = 9 sin(t). The initial velocity of the object is v(0) =

pentagon [3]

a) Acceleration is the derivative of velocity. By the fundamental theorem of calculus,

$v(t)=v(0)+\displaystyle\int_0^ta(u)\,\mathrm du$

so that

$v(t)=\left(-11\frac{\rm m}{\rm s}\right)+\int_0^t9\sin u\,\mathrm du$

$\boxed{v(t)=-\left(2+9\cos t)\right)\frac{\rm m}{\rm s}}$

b) We get the displacement by integrating the velocity function like above. Assume the object starts at the origin, so that its initial position is $s(0)=0\,\mathrm m$ . Then its displacement over the time interval [0, 3] is

$s(0)+\displaystyle\int_0^3v(t)\,\mathrm dt=-\int_0^3(2+9\cos t)\,\mathrm dt=\boxed{-6-9\sin3}$

c) The total distance traveled is the integral of the absolute value of the velocity function:

$s(0)+\displaystyle\int_0^3|v(t)|\,\mathrm dt$

$v(t)$ for $0\le t$ and $v(t)\ge0$ for $\cos^{-1}\left(-\frac29\right)\le t\le3$ , so we split the integral into two as

$\displaystyle\int_0^{\cos^{-1}\left(-\frac29\right)}-v(t)\,\mathrm dt+\int_{\cos^{-1}\left(-\frac29\right)}^3v(t)\,\mathrm dt$

$=\displaystyle\int_0^{\cos^{-1}\left(-\frac29\right)}(2+9\cos t)\,\mathrm dt-\int_{\cos^{-1}\left(-\frac29\right)}^3(2+9\cos t)\,\mathrm dt$

$\displaystyle=\boxed{2\sqrt{77}-6+4\cos^{-1}\left(-\frac29\right)-9\sin3}$

4 0

3 years ago