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tamaranim1 [39]

3 years ago

11

Only number 7 and 8.Any help is appreciated.Thanks!

1 answer:

dmitriy555 [2]3 years ago

3 0

Number seven is d=yd and r=(yd)/2
8 is D=25.08 and r =12.54

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A convenience store manager notices that sales of soft drinks are higher on hotter days, so he assembles the data in a table, sh

Contact [7]

The regression equation to predict the number of cans sold on a day that is 90°F will be 871.

<h3>What is the linear system?</h3>

A linear system is one in which the parameter in the equation has a degree of one. It might have one, two, or even more variables.

A convenience store manager notices that sales of soft drinks are higher on hotter days, so he assembles the data in the table, shown above. Use technology to find the equation of the least-squares regression line.

Then the regression line will be passing through (55, 340) and (84, 780). Then we have

y - 340 = 15.17(x - 55)

Then the regression equation to predict the number of cans sold on a day that is 90°F will be

y - 340 = 15.17 (90 - 55)

y = 871

More about the linear system link is given below.

brainly.com/question/20379472

#SPJ1

3 0

3 years ago

What value of w satisfies the equation<br> w -10 = 0.5w -7?

Maru [420]

Answer:

w = -6

Step-by-step explanation:

w-10= .5w -7

w = .5w +3

-0.5w = 3

w = -6

8 0

3 years ago

Find e^cos(2+3i) as a complex number expressed in Cartesian form.

ozzi

Answer:

The complex number $e^{\cos(2+31)} = \exp(\cos(2+3i))$ has Cartesian form

$\exp\left(\cosh 3\cos 2\right)\cos(\sinh 3\sin 2)-i\exp\left(\cosh 3\cos 2\right)\sin(\sinh 3\sin 2)$ .

Step-by-step explanation:

First, we need to recall the definition of $\cos z$ when $z$ is a complex number:

$\cos z = \cos(x+iy) = \frac{e^{iz}+e^{-iz}}{2}$ .

Then,

$\cos(2+3i) = \frac{e^{i(2+31)} + e^{-i(2+31)}}{2} = \frac{e^{2i-3}+e^{-2i+3}}{2}$ . (I)

Now, recall the definition of the complex exponential:

$e^{z}=e^{x+iy} = e^x(\cos y +i\sin y)$ .

So,

$e^{2i-3} = e^{-3}(\cos 2+i\sin 2)$

$e^{-2i+3} = e^{3}(\cos 2-i\sin 2)$ (we use that $\sin(-y)=-\sin y)$ .

Thus,

$e^{2i-3}+e^{-2i+3} = e^{-3}\cos 2+ie^{-3}\sin 2 + e^{3}\cos 2-ie^{3}\sin 2)$

Now we group conveniently in the above expression:

$e^{2i-3}+e^{-2i+3} = (e^{-3}+e^{3})\cos 2 + i(e^{-3}-e^{3})\sin 2$ .

Now, substituting this equality in (I) we get

$\cos(2+3i) = \frac{e^{-3}+e^{3}}{2}\cos 2 -i\frac{e^{3}-e^{-3}}{2}\sin 2 = \cosh 3\cos 2-i\sinh 3\sin 2$ .

Thus,

$\exp\left(\cos(2+3i)\right) = \exp\left(\cosh 3\cos 2-i\sinh 3\sin 2\right)$

$\exp\left(\cos(2+3i)\right) = \exp\left(\cosh 3\cos 2\right)\left[ \cos(\sinh 3\sin 2)-i\sin(\sinh 3\sin 2)\right]$ .

5 0

3 years ago

Simplify the expression (√3 + i)(√3- i) and write the result in the form a+bi.

goblinko [34]

Answer: D) 4

Step-by-step explanation:

8 0

3 years ago

[Help asap, will mark brainliest] Find the area of the right triangle.

FrozenT [24]

Answer:

45m

Step-by-step explanation:

1. Multiply the two sides.

2. Divide it by 3 (or multiply by 1/3)

3. The answer is your area.

6 0

3 years ago