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3 years ago

9

What is the answer please

1 answer:

Marina86 [1]3 years ago

8 0

Hey there!

Use The Pythagorean theorem. It’s a^2+b^2=c^2. Plug in 11 and 60. The equation would be 11^2+60^2=c^2. Simplify to get 121+3600=c^2. Simplify again to get 3721=c^2. Sqaure root each side to get c=61. The answer is the last one, c=61.

I hope this helps!

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100th term for 5, 2, -1

andrezito [222]

Answer:

100th term = -1485

Step-by-step explanation:

<h3>Arithmetic Progressionthe </h3><h3>nth term, Tn:</h3>

$t_{n} = a + (n - 1)d$

n = number of terms

= 100

a = first term

= 5

d = difference

= second term - first term

= 2 - 5

= -3

substitute the values into the formula

$t_{100} = 5(100 - 1)( - 3)$

$t_{100} = 5(99)( - 3)$

$t_{100} = 495( - 3)$

$t_{100} = - 1485$

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3 years ago

Which statement is true?

Maslowich

Answer:

A

Step-by-step explanation:

on the left of the equation, (6^4)^-5 = 6^-20

the right side you add the exponents so it would be (-7-3) =-10

so it would be 6^-10, which is greater than 6^-20

5 0

2 years ago

A tank contains 1080 L of pure water. Solution that contains 0.07 kg of sugar per liter enters the tank at the rate 7 L/min, and

allsm [11]

(a) Let $A(t)$ denote the amount of sugar in the tank at time $t$ . The tank starts with only pure water, so $\boxed{A(0)=0}$ .

(b) Sugar flows in at a rate of

(0.07 kg/L) * (7 L/min) = 0.49 kg/min = 49/100 kg/min

and flows out at a rate of

(<em>A(t)</em>/1080 kg/L) * (7 L/min) = 7<em>A(t)</em>/1080 kg/min

so that the net rate of change of $A(t)$ is governed by the ODE,

$\dfrac{\mathrm dA(t)}[\mathrm dt}=\dfrac{49}{100}-\dfrac{7A(t)}{1080}$

or

$A'(t)+\dfrac7{1080}A(t)=\dfrac{49}{100}$

Multiply both sides by the integrating factor $e^{7t/1080}$ to condense the left side into the derivative of a product:

$e^{\frac{7t}{1080}}A'(t)+\dfrac7{1080}e^{\frac{7t}{1080}}A(t)=\dfrac{49}{100}e^{\frac{7t}{1080}}$

$\left(e^{\frac{7t}{1080}}A(t)\right)'=\dfrac{49}{100}e^{\frac{7t}{1080}}$

Integrate both sides:

$e^{\frac{7t}{1080}}A(t)=\displaystyle\frac{49}{100}\int e^{\frac{7t}{1080}}\,\mathrm dt$

$e^{\frac{7t}{1080}}A(t)=\dfrac{378}5e^{\frac{7t}{1080}}+C$

Solve for $A(t)$ :

$A(t)=\dfrac{378}5+Ce^{-\frac{7t}{1080}}$

Given that $A(0)=0$ , we find

$0=\dfrac{378}5+C\implies C=-\dfrac{378}5$

so that the amount of sugar at any time $t$ is

$\boxed{A(t)=\dfrac{378}5\left(1-e^{-\frac{7t}{1080}}\right)}$

(c) As $t\to\infty$ , the exponential term converges to 0 and we're left with

$\displaystyle\lim_{t\to\infty}A(t)=\frac{378}5$

or 75.6 kg of sugar.

7 0

3 years ago

The equatio y =3x/2+8 defines y in terms of x. What is x in terms of y.

WITCHER [35]

$y = \frac{3x}{2} + 8$

the L.C.M is 2

since it is the highest denominator.

so multiply through by the L.C.M

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$2y = 3x + 16$

making x the subject.

$3x = 2y - 16$

dividing through by 3

$\frac{3x}{3} = \frac{2y - 16}{3}$

$x = \frac{2y - 16}{3}$

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leonid [27]

Answer:

3.4

Step-by-step explanation:

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