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4 years ago

8

Subtract 28.9 − 9.25 =_____

2 answers:

yaroslaw [1]4 years ago

8 0

19.65 hope this helps :)

Julli [10]4 years ago

3 0

Hey There!

Here is your answer:

Your answer is: =19.65

How:

28.9
-
9.25
_____
19.65

If you need anymore help just ask me:

Hope this helps!

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Please help ASAP I will give brainliest!

GaryK [48]

Answer:

6 minutes

Step-by-step explanation:

2 degrees per minute

so you need 12 degrees so 12/2 = 6 degrees

3 0

3 years ago

State whether the equation is true, false, or an open sentence. 3(5 – 5) = –6(–2 + 2)

jonny [76]

True.

Remember to follow PEMDAS. First, solve the parenthesis, then multiply

5 - 5 = 0

3(0) = 0

-2 + 2 = 0

-6(0) = 0

0 = 0 ∴ true is your answer

hope this helps

8 0

4 years ago

Ellis multiplies the dimensions of a rug on a drawing by a scale factor to find the dimensions of the actual rug. The dimensions

Digiron [165]

2 * 18 = 36
3 * 18 = 54
so ur scale factor is 18

5 0

4 years ago

Mia can purchase a 40 ounce jar of peanut butter for $5.20 or a 15 once jar for 2.70. how much does mia save per ounce by buying

melomori [17]

Cost per ounce for the 40-ounce peanut butter is
$5.20/40 = $0.13
cost per ounce for the 15-ounce peanut butter is
$2.70 / 15 = $ 0.18
Mia saves 5 cents or $0.05

7 0

4 years ago

Use De Moivre's theorem to write the complex number in trigonometric form: (cos(3pi/5) + i sin (3pi/5))^3

docker41 [41]

By De Moivre's theorem,

$\left(\cos\dfrac{3\pi}5+i\sin\dfrac{3\pi}5\right)^3=\boxed{\cos\dfrac{9\pi}5+i\sin\dfrac{9\pi}5}$

We can stop here ...

# # #

... but we can also express these trig ratios in terms of square roots. Let $x=\dfrac\pi5$ and let $c=\cos x$ . Then recall that

$\cos5x=c^5-10c^3\sin^2x+5c\sin^4x$

$\cos5x=c^5-10c^3(1-c^2)+5c(1-2c^2+c^4)$

$\cos5x=16c^5-20c^3+5c$

On the left, $5x=\pi$ so that

$16c^5-20c^3+5c+1=(c+1)(4c^2-2c-1)^2=0$

Since $c=\cos\dfrac\pi5\neq1$ , we're left with

$4c^2-2c-1=0\implies c=\dfrac{1+\sqrt5}4$

because we know to expect $\cos x>0$ . Then from the Pythagorean identity, and knowing to expect $\sin x>0$ , we get

$\sin x=\sqrt{1-c^2}=\sqrt{\dfrac58-\dfrac{\sqrt5}8}$

Both $\cos$ and $\sin$ are $2\pi$ -periodic, so that

$\cos\dfrac{9\pi}5=\cos\left(\dfrac{9\pi}5-2\pi\right)=\cos\left(-\dfrac\pi5\right)=\cos\dfrac\pi5$

and

$\sin\dfrac{9\pi}5=\sin\left(-\dfrac\pi5\right)=-\sin\dfrac\pi5$

so that the answer we left in trigonometric form above is equal to

$\cos\dfrac\pi5-i\sin\dfrac\pi5=\boxed{\dfrac{1+\sqrt5}4+i\sqrt{\dfrac58+\dfrac{\sqrt5}8}}$

7 0

3 years ago