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Mathematics

2 answers:

SSSSS [86.1K]2 years ago

7 0

Answer:

Third one down

Your answer is y= a(x - 3)2 + 5

Anna71 [15]2 years ago

5 0

Mommy is the answer your welcome

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If y is inversely proportional to the square root of x and y=−70 when x=49, find y if x=2401.

Dahasolnce [82]

The value of y when x= 2401 is -10

<h3>How to determine the value of x?</h3>

An inverse variation is represented as:

$y = \frac{k}{\sqrt x}$

Where k is the proportionality constant

When y = -70, x = 49

So, we have:

$-70 = \frac{k}{\sqrt {49}}$

Take the square root of 49

$-70 = \frac{k}{7}$

Multiply through by 7

k = -490

Substitute k = -490 in $y = \frac{k}{\sqrt x}$

$y = -\frac{490}{\sqrt x}$

When x =2401, we have

$y = -\frac{490}{\sqrt {2401}}$

Evaluate the square root

$y = -\frac{490}{49}$

Divide

y = -10

Hence, the value of y when x= 2401 is -10

Read more about variation at:

brainly.com/question/14254277

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4 0

2 years ago

Evaluate the line integral in Stokes? Theorem to evaluate the surface integral integrate S ( x F) n dS. Assume that n is in the

gtnhenbr [62]

Answer:

Step-by-step explanation:

Consider that F(x,y,z) = (x + y, y + z, z + z)

$r(t)=(cos t,4sint,\sqt{5}cost)\\r'(t)=(-sint,4cost,-\sqrt{5}sint)\\F(r(t))=[tex]\int\limits^{2\pi}_0 {F(r(t))r'(t)} \, dt \\\\=\int\limits^{2\pi}_0(cost+4sint,4sint+\sqrt{5}cost+cost)(-sint,4cost,-\sqt{5}sint)dt\\\\=\int\limits^{2\pi}_0(-sintcost-4sin^2t+16costsint+4\sqrt{5}cos^2t-5sintcost-\sqrt{5}sintcost)dt\\\\=\int\limits^{2\pi}_0(10sintcost-4sin^2t+4\sqrt{5}cos^2t-\sqrt{5}sintcost)dt\\\\=\int\limits^{2\pi}_0((10-\sqrt{5})sintcost-(4+4+\sqrt{5})sin^2t+4\sqrt{5})dt\\\\(10-\qrt{5})(0)-(4+4+\sqrt{5}(\pi)+4\sqrt{5}(2\pi)\\\\(\sqrt{5}-1)4\pi)$ [/tex]