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maxonik [38]
2 years ago
5

I need help with 7 and 4 and 9 also 10 I will give the most higher points I got

Mathematics
1 answer:
Serga [27]2 years ago
8 0
7. 9/5 = 1 4/5
*When multiplying with a whole number u can turn it into a fraction by putting a one under it
So 3/1 times 3/5 = 9/5
*just multiply straight across

4. 2/1 times 1/4 = 2/4 = THEN SIMPLIFY = 1/2

9. 3/4 times 3/1 (cups)
*multiply straight across
9/4 = 2 1/4 (cups of sugar)
So she uses less than 3 cups to make all 3 loaves

* 10. Sorry I can’t tell if it says 5/6 on both of them
If it says 5/6 on both of them then the answer is obvious and she worked more than that over the course of 2 days

* 10. It must be a 5/9 on the second one
So if so
5/6 times 2/1 (days)
10/6 then we need to multiply the 6 denominator (bottom of the fraction) with something to = 9 denominator
9 divided by 6 = 1.5
10/6 times 1.5 for both the numerator (top) and denominator (bottom)
= 15/9 (simplified 1 6/9)
This one she exercised more than 5/9
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y_1=e^x\cos x
y_2=e^x\sin x

The particular solution is easiest to obtain via variation of parameters. We're looking for a solution of the form

y_p=u_1y_1+u_2y_2

where

u_1=-\displaystyle\frac13\int\frac{y_2e^x\sec x}{W(y_1,y_2)}\,\mathrm dx
u_2=\displaystyle\frac13\int\frac{y_1e^x\sec x}{W(y_1,y_2)}\,\mathrm dx

and W(y_1,y_2) is the Wronskian of the fundamental solutions. We have

W(e^x\cos x,e^x\sin x)=\begin{vmatrix}e^x\cos x&e^x\sin x\\e^x(\cos x-\sin x)&e^x(\cos x+\sin x)\end{vmatrix}=e^{2x}

and so

u_1=-\displaystyle\frac13\int\frac{e^{2x}\sin x\sec x}{e^{2x}}\,\mathrm dx=-\int\tan x\,\mathrm dx
u_1=\dfrac13\ln|\cos x|

u_2=\displaystyle\frac13\int\frac{e^{2x}\cos x\sec x}{e^{2x}}\,\mathrm dx=\int\mathrm dx
u_2=\dfrac13x

Therefore the particular solution is

y_p=\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x

so that the general solution to the ODE is

y=C_1e^x\cos x+C_2e^x\sin x+\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x
7 0
3 years ago
Alan found the value of the expression (15 - 6)2 to be 189. Is he right?
tatuchka [14]
Answer would be C

Because I always use the rule of :
B - Brackets
I - indices
D - division
M - multiplication
A - addition
S - subtraction
3 0
3 years ago
Inferential statistics allow you to decide whether a difference between the experimental and the control group is due to:_______
den301095 [7]

Answer:

Option (A)

Step-by-step explanation:

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1. Descriptive Statistics

2. Inferential Statistics

Descriptive statistics describes the characteristics of observed subjects or items while Inferential statistics makes inferences, based on given or derived data.

Inferential Statistics allow you to decide whether a difference between the experimental group and control group is due to <u>manipulation or chance.</u>

<u />

The Experimental group is the group that an effect is tested on while the Control group is the group that is left untested or uninfluenced. Inferential statistics allow you to decide whether a difference between these 2 groups is due to

- manipulation or interference by any force (which may be the experimenter/researcher)

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5 0
4 years ago
Jake thinks that a rhombus is not always a parallelogram.Is jake correct?
OverLord2011 [107]

Answer:

No, a rhombus is a quadrilateral, with four equal-length sides and opposite sides parallel to each other. All rhombuses are parallelograms, but not all parallelograms are rhombuses. The opposite interior angles of rhombuses are always congruent. A parallelogram is just a four sided closed shape where opposite sides are parallel.

I hope this helps!

8 0
3 years ago
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