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2 years ago

The angle whose sine is 0.39581 is

1 answer:

5 0

The angle whose sine is 0.39581 is 23.32 degrees. This is calculated by obtaining the arcsine of the value given above. The sine function is one of the basic functions in trigonometry together with cosine, tangent, cosecant,secant and cotangent.

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A recent poll in your state shows that 25% of the vehicles are SUV’s and 10% are trucks. The poll counted 5 million SUV’s. How m

rjkz [21]

Answer:

20 million

Step-by-step explanation:

5 million SUVs are 25% of the total number of vehicles.

0.25x = 5,000,000

x = 20,000,000

Answer: 20 million

4 0

1 year ago

On her first day in a hospital, Kiri receives u1 milligrams (mg) of a therapeutic drug. The amount of the drug Kiri receives inc

Valentin [98]

Answer:

a. 21 = u1 + 6·d

b. 29 = u1 + 10·d

c. d = 2, u1 = 9

Step-by-step explanation:

a. The given parameters are;

The amount of therapeutic drug Kiri receives on her first at the hospital = u1 milligrams

The amount of drug increase received by Kiri each day = d

The amount of drug Kiri received on the seventh day = 21 mg

The amount of drug she received on the eleventh day = 29 mg

Therefore, we have an arithmetic progression with the formula for the nth term given as follows;

aₙ = a₁ + (n - 1)·d

Where;

a₁ = u1

n = The number of terms

Therefore, for the 7th day, the amount of drugs she receives, which is 21 milligrams, is given as follows;

a₇ = u1 + (7 - 1)·d = u1 + 6·d = 21

The equation for the amount of drugs she receives in terms of u1 and d on the seventh day is given as follows;

21 = u1 + 6·d

b. For the eleventh day, the amount of drugs she receives, which is 29 milligrams, is given as follows;

a₁₁ = u1 + (11 - 1)·d = u1 + 10·d = 29

Therefore, the equation for the amount of drugs she receives in terms of u1 and d on the eleventh day is given as follows;

29 = u1 + 10·d

c. Therefore, we have two equations which are given as follows;

21 = u1 + 6·d................(1)

29 = u1 + 10·d..............(2)

Subtracting equation (1) from equation (2) gives;

29 - 21 = (u1 + 10·d) - (u1 + 6·d)

8 = 4·d

d = 8/4 = 2

d = 2

From equation (1), we have;

21 = u1 + 6·d = u1 + 6×2 = u1 + 12

21 = u1 + 12

21 - 12 = u1

∴ u1 = 9

d = 2, u1 = 9.

4 0

1 year ago

The Apollo's Chariot is a rollercoaster at Busch Gardens with a top speed of 117 km/hr. Its initial (and tallest) descent can be

nikklg [1K]

Hhhhyhhhuuuuhrerrddddddd

4 0

2 years ago

FreezeFrame, a social networking site, sent out a survey in which the users simply click a response to join the sample. One of t

kramer

Answer:

Here, Convenience sampling is used by the poll

Explanation:

Convenience sampling also called as opportunity, grab or accidental sampling is a kind of non-likelihood testing that includes the sample being drawn from that piece of the populace that is near hand. This sort of sampling is most valuable for pilot testing.

Convenience sampling is a procedure used to make sample according to straightforward entry, readiness to be a sample's part , accessibility at a given scheduled slot.

So, this method of sampling is biased and don't results in desired outcomes.

7 0

2 years ago

F⃗ (x,y)=−yi⃗ +xj⃗ f→(x,y)=−yi→+xj→ and cc is the line segment from point p=(5,0)p=(5,0) to q=(0,2)q=(0,2). (a) find a vector pa

DerKrebs [107]

a. Parameterize $C$ by

$\vec r(t)=(1-t)(5\,\vec\imath)+t(2\,\vec\jmath)=(5-5t)\,\vec\imath+2t\,\vec\jmath$

with $0\le t\le1$ .

b/c. The line integral of $\vec F(x,y)=-y\,\vec\imath+x\,\vec\jmath$ over $C$ is

$\displaystyle\int_C\vec F(x,y)\cdot\mathrm d\vec r=\int_0^1\vec F(x(t),y(t))\cdot\frac{\mathrm d\vec r(t)}{\mathrm dt}\,\mathrm dt$

$=\displaystyle\int_0^1(-2t\,\vec\imath+(5-5t)\,\vec\jmath)\cdot(-5\,\vec\imath+2\,\vec\jmath)\,\mathrm dt$

$=\displaystyle\int_0^1(10t+(10-10t))\,\mathrm dt$

$=\displaystyle10\int_0^1\mathrm dt=\boxed{10}$

d. Notice that we can write the line integral as

$\displaystyle\int_C\vecF\cdot\mathrm d\vec r=\int_C(-y\,\mathrm dx+x\,\mathrm dy)$

By Green's theorem, the line integral is equivalent to

$\displaystyle\iint_D\left(\frac{\partial x}{\partial x}-\frac{\partial(-y)}{\partial y}\right)\,\mathrm dx\,\mathrm dy=2\iint_D\mathrm dx\,\mathrm dy$

where $D$ is the triangle bounded by $C$ , and this integral is simply twice the area of $D$ . $D$ is a right triangle with legs 2 and 5, so its area is 5 and the integral's value is 10.

4 0

1 year ago