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2 years ago

PLEASE HURRY!!! what is the slope of the graph?

Mathematics

2 answers:

hjlf2 years ago

8 0

The slope is 2 i’m pretty sure

Fofino [41]2 years ago

3 0

Step-by-step explanation:

2..................pls follow me

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Find the value of x in the triangle shown below.

DIA [1.3K]

Answer:

x=8

Step-by-step explanation:

We can use the Pythagorean theorem to solve

a^2+b^2 = c^2 where a and b are the legs and c is the hypotenuse

6^2 + x^2 = 10^2

36 + x^2 = 100

Subtract 36 from each side

36-36 +x^2 = 100-36

x^2 = 64

Take the square root of each side

sqrt(x^2) = sqrt(64)

x = 8

7 0

2 years ago

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Which word indicates that the integer is "negative"?

sweet-ann [11.9K]

Loss
gain means you’re getting more, which is positive

7 0

2 years ago

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Is the answer 40? I am kind of confused

klemol [59]

Answer:

Yes

Step-by-step explanation:

29-18=11

c=11

18+2(11)=40

4 0

2 years ago

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The weights of certain machine components are normally distributed with a mean of 5.19 ounces and a standard deviation of 0.05 o

Brums [2.3K]

Answer:

The weight that separate the top 8% by 5.2605 and the weight that separate bottom 8% by 5.1195.

Step-by-step explanation:

We are given that

Mean, $\mu=5.19$

Standard deviation, $\sigma=0.05$

We have to find the two weights that separate the top 8% and the bottom 8%.

Let x1 and x2 the two weights that separate the top 8% and the bottom 8%.

Z-value for p-value =0.08 = $-1.41$

For 8% bottom

$Z=\frac{x_1-\mu}{\sigma}=-1.41$

$\frac{x_1-5.19}{0.05}=-1.41$

$x_1-5.19=-1.41\times 0.05$

$x_1=-1.41\times 0.05+5.19$

$x_1=5.1195$

For 8% top

p-Value=1-0.08=0.92

Z- value=1.41

Now,

$\frac{x_2-5.19}{0.05}=1.41$

$x_2-5.19=1.41\times 0.05$

$x_2=1.41\times 0.05+5.19$

$x_2=5.2605$

(x1,x2)=(5.1195,5.2605)

4 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