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3 years ago

7

the temperature in st paul minnesota was -19F at sunrise. by noon the temperature has risen 25 F. What was the temperature (in F

) at noon?

1 answer:

Ann [662]3 years ago

4 0

Answer: -19 + 25= 6 F

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What is the value of the expression below when y 8 and z = 8? 9y + 3z

aleksley [76]

Answer:

96

Step-by-step explanation:

8y+3z

9(8)+3(8)

72+24

96

6 0

3 years ago

Find two complex numbers that have a sum of i10 a different of -4and a product of -29

lianna [129]

<h2>-2+5i and 2+5i</h2>

Step-by-step explanation:

Let the complex numbers be $a+ib\textrm{ and }c+id$ .

Given, sum is $10i$ , difference is $-4$ and product is $-29$ .

$(a+c)+i(b+d)=10i$ ⇒ $a+c=0,b+d=10$

$(a-c)+i(b-d)=-4$ ⇒ $a-c=-4,b-d=0$

$a=-2,c=2,b=5,d=5$

$(a+ib)(c+id)=(-2+5i)(2+5i)=-4-25=-29$

Hence, all three equations are consistent yielding the complex numbers $-2+5i\textrm{ and }2+5i$ .

8 0

3 years ago

Can I get help with finding the Fourier cosine series of F(x) = x - x^2

trapecia [35]

Assuming you want the cosine series expansion over an arbitrary symmetric interval $[-L,L]$ , $L\neq0$ , the cosine series is given by

$f_C(x)=\dfrac{a_0}2+\displaystyle\sum_{n\ge1}a_n\cos nx$

You have

$a_0=\displaystyle\frac1L\int_{-L}^Lf(x)\,\mathrm dx$
$a_0=\dfrac1L\left(\dfrac{x^2}2-\dfrac{x^3}3\right)\bigg|_{x=-L}^{x=L}$
$a_0=\dfrac1L\left(\left(\dfrac{L^2}2-\dfrac{L^3}3\right)-\left(\dfrac{(-L)^2}2-\dfrac{(-L)^3}3\right)\right)$
$a_0=-\dfrac{2L^2}3$

$a_n=\displaystyle\frac1L\int_{-L}^Lf(x)\cos nx\,\mathrm dx$

Two successive rounds of integration by parts (I leave the details to you) gives an antiderivative of

$\displaystyle\int(x-x^2)\cos nx\,\mathrm dx=\frac{(1-2x)\cos nx}{n^2}-\dfrac{(2+n^2x-n^2x^2)\sin nx}{n^3}$

and so

$a_n=-\dfrac{4L\cos nL}{n^2}+\dfrac{(4-2n^2L^2)\sin nL}{n^3}$

So the cosine series for $f(x)$ periodic over an interval $[-L,L]$ is

$f_C(x)=-\dfrac{L^2}3+\displaystyle\sum_{n\ge1}\left(-\dfrac{4L\cos nL}{n^2L}+\dfrac{(4-2n^2L^2)\sin nL}{n^3L}\right)\cos nx$

4 0

3 years ago

Based on the function h(t), what will be the account value five years after the account is opened?

bija089 [108]

Answer:

Option B

Step-by-step explanation:

f(t) = 5000

g(t) = 250t

h(t) = f(t) + g(t) = 5000 + 250t

After 5 years, the amount of money in the account is:

h(t = 5) = 5000 + 250(5) = 5000 + 1250 = 6250$

5 0

3 years ago

2a2b3 and -4a2b3 are like terms.<br> a. True<br> b. False

Trava [24]

2a²b³ and -4a²b³ are like terms as they both have the same variables with the same degrees.

Your final answer is a. True.

4 0

3 years ago