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3 years ago

What is the gcf of 19 and 22

1 answer:

3 0

The GCF would simply be 1 because 19 is a prime number. This means that the only factors of 19 are 1 and 19, while the factors of 22 would be 1, 2, 11, and 22. The only factor they share is 1, so their greatest common factor is 1.

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Verify the identity. Show your work. (1 + tan^2u)(1 - sin^2u) = 1

yuradex [85]

$\bf \textit{Pythagorean Identities}
\\\\
sin^2(\theta)+cos^2(\theta)=1\implies cos^2(\theta)=1-sin^2(\theta)
\\\\ 1+tan^2(\theta)=sec^2(\theta)\\\\
-------------------------------\\\\\
[1+tan^2(u)][1-sin^2(u)]=1
\\\\\\\
[sec^2(u)][cos^2(u)]\implies \cfrac{1}{cos^2(u)}\cdot cos^2(u)\implies \cfrac{cos^2(u)}{cos^2(u)}\implies 1$

7 0

3 years ago

Find the product. the coefficient of the product of (-5xy^2) and (-4x^2y) is

Art [367]

$(-5xy^2)(-4x^2y)=5xy^2\times4x^2y=20x^1y^2x^2y^1=20x^{1+2}y^{2+1}=20x^3y^3$

8 0

3 years ago

Read 2 more answers

Please help ASAP!!!!!

Viefleur [7K]

Use pythagorean’s theorem
as you can see it’s a right isosceles triangle

so your equation is adapted to this:
p^2 + p^2=44^2
add like terms and simplify
2p^2=1936
divide by 2
p^2=968
square root
p= √968
simplify
p=22√2

the answer would be your second option

3 0

3 years ago

3. (6 Points). Solve the initial value problem y'-y.cosx=0, y(pi/2)=2e

Stells [14]

Answer:

$y=2e^{sin(x)}$

Step-by-step explanation:

Given equation can be re written as

$\frac{\mathrm{d} y}{\mathrm{d} x}-ycos(x)=0\\\frac{\mathrm{d} y}{\mathrm{d} x}=ycos(x)\\\\=> \frac{dy}{y}=cox(x)dx\\\\Integrating \\ \int \frac{dy}{y}=\int cos(x)dx \\\\ln(y)=sin(x)+c$ ............(i)

Now it is given that y(π/2) = 2e

Applying value in (i) we get

ln(2e) = sin(π/2) + c

=> ln(2) + ln(e) = 1+c

=> ln(2) + 1 = 1 + c

=> c = ln(2)

Thus equation (i) becomes

ln(y) = sin(x) + ln(2)

ln(y) - ln(2) = sin(x)

ln(y/2) = sin(x)

$y= 2e^{sinx}$

7 0

2 years ago

insurance company checks police records on 559 accidents selected at random and notes that teenagers were at the wheel in 91 of

ss7ja [257]

Answer: $(13.22\%,\ 19.34\%)$

Step-by-step explanation:

Given : Sample space : n= 559

Sample proportion : $\hat{p}=\dfrac{91}{559}=0.162790697674\approx0.1628$

$=16.28\%$

Significance level : $\alpha= 1-0.95=0.05$

Critical value : $z_{\alpha/2}=1.96$

Confidence level for population proportion:-

$\hat{p}\pm z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}\\\\=0.1628\pm (1.96)\sqrt{\dfrac{0.1628(1-0.1628)}{559}}\\\\=0.1628\pm0.030604971367\\\\\approx 0.1628\pm0.0306=(0.1322,\ 0.1934)=(13.22\%,\ 19.34\%)$

Hence, 95% confidence interval for the percentage of all auto accidents that involve teenage drivers.= $(13.22\%,\ 19.34\%)$

7 0

3 years ago