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2 years ago

Please help explain and solve angels homework

Mathematics

2 answers:

Nat2105 [25]2 years ago

8 0

Answer:

1. it's a right angle. therefore it has a total of 90°

to get the value of x we start with:

90° - 54° = 36°

(10X+16)°=36°

10X = 36 -16

10X = 20

X = 2

kipiarov [429]2 years ago

6 0

Answer:

Step-by-step explanation:

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Evaluate 2/3 (10-6) 1/2 <br>

Studentka2010 [4]

Answer:

$= \frac{2}{3} (10 - 6) \frac{1}{2}$

$= \frac{2}{3} (4) \frac{1}{2}$

$= \frac{8}{3} \times \frac{1}{2}$

$= \frac{4}{3}$

4 0

2 years ago

jake bought 5 pounds of candy he gave 3 pounds of candy away and he ate 10 ounces candy how many ounces of candy doues jake have

Novosadov [1.4K]

Answer:

22 onces

Step-by-step explanation:

16 ounces in a pound

5-3=2

2x16=32

32-10=22

5 0

2 years ago

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Aveeno and her boyfriend you know who wrote the bus six times in a month it cost $40 for the two of them each time if Avena alwa

MaRussiya [10]

Answer: The cost for the month is $240.

Step-by-step explanation:

Since we have given that

Cost for the two i.e Aveeno and her boyfriend = $40

They were charged 6 times in a month.

As we know that times stands for multiplication.

So, we need to find the cost for the month, where we multiply 6 with $40 to get the amount for the whole month.

So, the amount for the month would be

$40\times 6=\$240$

Hence, the cost for the month is $240.

4 0

2 years ago

can someone show me how to find the general solution of the differential equations? really need to know how to do it for the upc

mariarad [96]

The first equation is linear:

$x\dfrac{\mathrm dy}{\mathrm dx}-y=x^2\sin x$

Divide through by $x^2$ to get

$\dfrac1x\dfrac{\mathrm dy}{\mathrm dx}-\dfrac1{x^2}y=\sin x$

and notice that the left hand side can be consolidated as a derivative of a product. After doing so, you can integrate both sides and solve for $y$ .

$\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac1xy\right]=\sin x$
$\implies\dfrac1xy=\displaystyle\int\sin x\,\mathrm dx=-\cos x+C$
$\implies y=-x\cos x+Cx$

- - -

The second equation is also linear:

$x^2y'+x(x+2)y=e^x$

Multiply both sides by $e^x$ to get

$x^2e^xy'+x(x+2)e^xy=e^{2x}$

and recall that $(x^2e^x)'=2xe^x+x^2e^x=x(x+2)e^x$ , so we can write

$(x^2e^xy)'=e^{2x}$
$\implies x^2e^xy=\displaystyle\int e^{2x}\,\mathrm dx=\frac12e^{2x}+C$
$\implies y=\dfrac{e^x}{2x^2}+\dfrac C{x^2e^x}$

- - -

Yet another linear ODE:

$\cos x\dfrac{\mathrm dy}{\mathrm dx}+\sin x\,y=1$

Divide through by $\cos^2x$ , giving

$\dfrac1{\cos x}\dfrac{\mathrm dy}{\mathrm dx}+\dfrac{\sin x}{\cos^2x}y=\dfrac1{\cos^2x}$
$\sec x\dfrac{\mathrm dy}{\mathrm dx}+\sec x\tan x\,y=\sec^2x$
$\dfrac{\mathrm d}{\mathrm dx}[\sec x\,y]=\sec^2x$
$\implies\sec x\,y=\displaystyle\int\sec^2x\,\mathrm dx=\tan x+C$
$\implies y=\cos x\tan x+C\cos x$
$y=\sin x+C\cos x$

- - -

In case the steps where we multiply or divide through by a certain factor weren't clear enough, those steps follow from the procedure for finding an integrating factor. We start with the linear equation

$a(x)y'(x)+b(x)y(x)=c(x)$

then rewrite it as

$y'(x)=\dfrac{b(x)}{a(x)}y(x)=\dfrac{c(x)}{a(x)}\iff y'(x)+P(x)y(x)=Q(x)$

The integrating factor is a function $\mu(x)$ such that

$\mu(x)y'(x)+\mu(x)P(x)y(x)=(\mu(x)y(x))'$

which requires that

$\mu(x)P(x)=\mu'(x)$

This is a separable ODE, so solving for $\mu$ we have

$\mu(x)P(x)=\dfrac{\mathrm d\mu(x)}{\mathrm dx}\iff\dfrac{\mathrm d\mu(x)}{\mu(x)}=P(x)\,\mathrm dx$
$\implies\ln|\mu(x)|=\displaystyle\int P(x)\,\mathrm dx$
$\implies\mu(x)=\exp\left(\displaystyle\int P(x)\,\mathrm dx\right)$

and so on.

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3 years ago

a salt truck drops 39 kilograms of salt per minute. how many grams of salt does the truck drop per second

PilotLPTM [1.2K]

Multiply by 1 minute/60 seconds

the minutes cancel out and youre left with 0.65 kilograms per second

you then multiply by 1000 grams/1 kilogram

the kilograms cancel out and youre left with 650 grams per second

5 0

2 years ago

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