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3 years ago

Find the equation of the circle whose center and radius are given.

2 answers:

7 0

As i said before put the center instead of (h,k) in the general formula and put r=1
so (x-(-2))^2 + (y-(-5)^2 = 1
(x+2)^2 + (y+5)^2 = 1

Levart [38]3 years ago

4 0

Answer: The required equation of the circle is $x^2+y^2+4x+10y+28=0.$

Step-by-step explanation: We are given to find the equation of the circle with center (-2, -5) and radius of length 1 unit.

We know that

the standard equation of a circle with center (h, k) and radius of length r units is given by

$(x-h)^2+(y-k)^2=r^2~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)$

For the given circle, we have

center, (h, k) = (-2, -5) and radius, r = 1 units.

Therefore, from equation (i), we get

$(x-(-2))^2+(y-(-5))^2=1^2\\\\\Rightarrow (x+2)^2+(y+5)^2=1\\\\\Rightarrow x^2+4x+4+y^2+10x+25=1\\\\\Rightarrow x^2+y^2+4x+10y+29-1=0\\\\\Rightarrow x^2+y^2+4x+10y+28=0.$

Thus, the required equation of the circle is $x^2+y^2+4x+10y+28=0.$

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3 years ago

If f(x) is an exponential function where f(1.5) = 7 and f(8.5) = 46, then find the value of f(14), to the nearest hundredth.

joja [24]

Answer:

$f(14)=201.932$

Step-by-step explanation:

Set up a system of equations and solve for "b"

$\left \{ {{46=ab^{8.5}} \atop {7=ab^{1.5}}} \right\\ \\\left \{ {{\frac{46}{b^{8.5}}=a } \atop {\frac{7}{b^{1.5}}=a} \right\\\\\frac{46}{b^{8.5}}=\frac{7}{b^{1.5}}\\ \\46b^{1.5}=7b^{8.5}\\\\46=7b^7\\\\\frac{46}{7}=b^7\\ \\b=1.308604899$

Determine the value of "a" using "b"

$y=ab^x\\\\46=a(1.308604899)^{8.5}\\\\46=9.837224985a\\\\a=4.676115477$

Find f(14) using the new function

$f(x)=4.676115477(1.308604899)^x\\\\f(14)=4.676115477(1.308604899)^{14}\\\\f(14)\approx201.932$

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3 years ago

Which is greater 11/16 or 3/4

Anettt [7]

To compare fractions, we need to have the same denominator
The least common multiple is 16
$\frac{11}{16}$
to change $\frac{3}{4}$ , we should multiply the numerator and denominator by 4
$\frac{3*4}{4*4}$ = $\frac{12}{16}$

$\frac{11}{16}$ < $\frac{12}{16}$

so basically, $\frac{12}{16}$ is greater than $\frac{11}{16}$
which means $\frac{3}{4}$ is greater

$\frac{3}{4}$ is greater

6 0

3 years ago