How do you write 39/10 as a percentage?

Answers

Answer 1

Answer: Answer:
39/10

39 ÷ 10= 3.9

3.9 x 100 = 390%

Answer 2

Answer: 39/10
= (39*10) / (10*10)
= 390/ 100
= 390%

The final answer is 390%~

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1. For which value of x is the expression log base 4 of (x+3) not defined?a. 0b. -2c. 3d. -42. If log 7= a, what is the value of log 490 in terms of a?3. For what value of k will the graph of y=log base 7 of x contain the point (1,k)?

Major axis 12 units long and parallel to the y-axis, minor axis 8 units long, center at (-2,5)

Answers

center is located at (-2,5) is:(x+2)^2 / 36 + (y-5)^2 / 64 = 1

The dimensions of the given ellipse are major axis 12 units long and parallel to the y-axis, minor axis 8 units long, and the center is located at (-2,5).Let us find the standard form equation of the ellipse. The standard form equation of an ellipse is given by:(x-h)^2 / a^2 + (y-k)^2 / b^2 = 1Where (h, k) is the center of the ellipse, a is the distance from the center to either the x-axis or the y-axis, and b is the distance from the center to the other axis. Therefore, for the given ellipse, the equation of the ellipse in standard form is:(x+2)^2 / 36 + (y-5)^2 / 64 = 1Thus, the standard form equation of the ellipse whose major axis is 12 units long and parallel to the y-axis, minor axis 8 units long, and center is located at (-2,5) is:(x+2)^2 / 36 + (y-5)^2 / 64 = 1.

Learn more about ellipse

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How do I solve a quadratic equation like this: x squared minus 9x = -14?

Answers

$x^2-9x=-14\n\nx^2-9x+14=0\n--------------------\n-9=-7+(-2)\n14=-7\cdot(-2)\n--------------------\n\nx^2-9x+14=(x-7)(x-2)\n\n(x-7)(x-2)=0\iff x-7=0\ or\ x-2=0\n\nx=7\ or\ x=2$

$x^2-9x+14=0\n\na=1;\ b=-9;\ c=14\n\n\Delta=b^2-4ac\n\n\Delta=(-9)^2-4\cdot1\cdot14=81-56=25\n\nx_1=(-b-\sqrt\Delta)/(2a)\to x_1=(9-√(25))/(2\cdot1)=(9-5)/(2)=(4)/(2)=2\n\nx_2=(-b+\sqrt\Delta)/(2a)\to x_2=(9+√(25))/(2\cdot1)=(9+5)/(2)=(14)/(2)=7$

Bring everything over to one side.
x² -9x=-14
x²-9x+14=0
Now it is an easy trinomial so you need to find two numbers that add up to -9 and multiply to equal 14. The two numbers are -7 and -2.
x²-9x+14=0
(x-7)(x-2)=0
x= 7 and 2

What’s the distance between (1, 7) and (-2, -4) please help

Answers

Answer:

the slope is 11/3

so distance is 130 or 11.4 in decimal form

Answer:

Exact distance = $√(130)$

Approximate distance = 11.40175

Explanation:

We'll use the distance formula

$d = √((x_1-x_2)^2+(y_1-y_2)^2)\n\nd = √((1-(-2))^2+(7-(-4))^2)\n\nd = √((1+2)^2+(7+4)^2)\n\nd = √((3)^2+(11)^2)\n\nd = √(9+121)\n\nd = √(130) \ \ \text{ ... exact distance}\n\nd \approx 11.40175 \ \ \text{ ... approximate distance}\n\n$

What multiplies to negative 4 and adds to 7

Answers

1. multiply to -18 add to -17 = Multiplying -18 and 1 will give -18 as the result and then on adding -18 and + 1 the result comes to -17

Arrange the circles (represented by their equations in general form) in ascending order of their radius lengths.x2 + y2 − 2x + 2y − 1 = 0
x2 + y2 − 4x + 4y − 10 = 0
x2 + y2 − 8x − 6y − 20 = 0
4x2 + 4y2 + 16x + 24y − 40 = 0
5x2 + 5y2 − 20x + 30y + 40 = 0
2x2 + 2y2 − 28x − 32y − 8 = 0
x2 + y2 + 12x − 2y − 9 = 0

Answers

The correct answer is:

x²+y²-2x+2y-1 = 0;
x²+y²-4x+4y-10 = 0;
5x²+5y²-20x+30y+40 = 0;
x²+y²-8x-6y-20 = 0;
x²+y²+12x-2y-9 = 0;
4x²+4y²+16x+24y-40 = 0; and
2x²+2y²-28x-32y-8 = 0

Explanation:

For each of these, we want to write the equation in the form
(x+h)²+(y+k)² = r².

To do this, we evaluate the terms 2hx and 2ky in each equation. We will take half of this; this will tell us what h and k are for each equation.

For the first equation:
2hx = -2x and 2ky = 2y.

Half of -2x = -1x and half of 2y = 1y; this means h = -1 and k = 1:
(x-1)² + (y+1)² + ___ - 1 = 0

When we multiply (x-1)², we get
x²-2x+1.
When we multiply (y+1)², we get
y²+2y+1.

This gives us 1+1 = 2 for the constant. We know we must add something to 2 to get -1; 2 + ___ = -1; the missing term is -3. Add that to each side (to have r² on the right side of the equals) and we have
(x-1)² + (y+1)² = 3
This means that r² = 3, and r = √3 = 1.732.

For the second equation, 2hx = -4x and 2ky = 4y; this means h = -4/2 = -2 and k = 4/2 = 2. This gives us
(x-2)² + (y+2)² -10 + ___ = 0.

Multiplying (x-2)² gives us
x²-4x+4.
Multiplying (y+2)² gives us
y²+4x+4.
This gives us 4+4= 8 for our constant so far.

We know 8 + ___ = -10; this means the missing term is -18. Add this to each side of the equation to have
(x-2)²+(y+2)² = 18; r² = 18; r = √18 = 3√2 = 4.243.

For the third equation, 2hx = -8x and 2ky = -6y. This means h = -8/2 = -4 and k = -6/2 = -3. This gives us:
(x-4)²+(y-3)²-20 = 0

Multiplying (x-4)² gives us
x²-8x+16.
Multiplying (y-3)² gives us
y²-6y+9.

This gives us 16+9 = 25 for the constant. We know that 25+___ = -20; the missing term is -45. Add this to each side for r², and we have that
r²=45; r = √45 = 3√5 = 6.708.

For the next equation, we factor 4 out of the entire equation:
4(x²+y²+4x+6y-10)=0.
This means 2hx = 4x and 2ky = 6y; this gives us h = 4/2 = 2 and k = 6/2 = 3. This gives us
4((x+2)²+(y+3)² - 10) = 0.

Multiplying (x+2)² gives us
x²+4x+4.
Multiplying (y+3)² gives us
y²+6y+9.

This gives us a constant of 4+9 = 13. We know 13+__ = -10; this missing value is -23. Since we had factored out a 4, that means we have 4(-23) = -92. Adding this to each side for r², we have
r²=92; r = √92 = 2√23 = 9.59.

For the next equation, we factor out a 5 first:
5(x²+y²-4x+6y+8) = 0. This means that 2hx = -4x and 2ky = 6y; this gives us h = -4/2 = -2 and k = 6/2 = 3:

5((x-2)²+(y+3)²+8) = 0.

Multiplying (x-2)² gives us
x²-4x+4.
Multiplying (y+3)² gives us
y²+6y+9.

This gives us a constant of 4+9 = 13. We know that 13+__ = 8; the missing value is -5. Since we factored a 5 out, we have 5(-5) = -25. Adding this to each side for r² gives us
r²=25; r = √25 = 5.

For the next equation, we first factor a 2 out:
2(x²+y²-14x-16y-4) = 0. This means 2hx = -14x and 2ky = -16y; this gives us h = -14/2 = -7 and k = -16/2 = -8:

2((x-7)²+(y-8)²-4) = 0.

Multiplying (x-7)² gives us
x²-14x+49.
Multiplying (y-8)² gives us
y²-16x+64.

This gives us a constant of 49+64=113. We know that 113+__ = -4; the missing value is -117. Since we first factored out a 2, this gives us 2(-117) = -234. Adding this to each side for r² gives us
r²=234; r = √234 = 3√26 = 15.297.

For the last equation, 2hx = 12x and 2ky = -2; this means h = 12/2 = 6 and k = -2/2 = -1:
(x+6)²+(y-1)²-9 = 0

Multiplying (x+6)² gives us
x²+12x+36.
Multiplying (y-1)² gives us
y²-2y+1.

This gives us a constant of 36+1 = 37. We know that 37+__ = -9; the missing value is -46. Adding this to each side for r² gives us
r² = 46; r=√46 = 6.78.

Find the radius of each equation:

1.
$x^2 + y^2-2x+2y-1 = 0, \n x^2-2x+1-1 + y^2+2y+1-1-1 = 0, \n (x-1)^2+(y+1)^2=3$ , then $r_1= √(3)$ .

2.
$x^2 + y^2-4x + 4y- 10 = 0, \n x^2 -4x+4-4+ y^2 + 4y+4-4- 10 = 0, \n (x-2)^2+(y+2)^2=18$ , then $r_2= √(18)=3 √(2)$ .

3.
$x^2 + y^2-8x- 6y- 20 = 0, \n x^2-8x+16-16+ y^2- 6y+9-9- 20 = 0, \n (x-4)^2+(y-3)^2=45$ , then $r_3= √(45) =3 √(5)$ .

4.
$4x^2 + 4y^2+16x+24y- 40 = 0, \n 4x^2+16x+16-16+ 4y^2+24y+36-36- 40 = 0, \n 4(x+2)^2+4(y+3)^2=92,\n (x+2)^2+(y+3)^2=23$ , then $r_4= √(23)$ .

5.
$5x^2 + 5y^2-20x+30y+ 40 = 0, \n 5x^2-20x+20-20+ 5y^2+30y+45-45- 40 = 0, \n 5(x-2)^2+5(y+3)^2=105,\n (x-2)^2+(y+3)^2=21$ , then $r_5= √(21)$ .

6.
$2x^2 + 2y^2-28x-32y- 8= 0, \n 2x^2-28x+98-98+ 2y^2-32y+128-128- 8= 0, \n 2(x-7)^2+2(y-8)^2=234,\n (x+2)^2+(y+3)^2=117$ , then $r_6= √(117)=3√(13)$ .

7.
$x^2 + y^2+12x-2y-9 = 0, \n x^2+12x+36-36+ y^2-2y+1-1- 9 = 0, \n (x+6)^2+(y-1)^2=46$ , then $r_7= √(46)$ .

Hence
$r_1= √(3)$ , $r_2=3 √(2)$ , $r_3=3 √(5)$ , $r_4= √(23)$ , $r_5= √(21)$ , $r_6= 3√(13)$ , $r_7= √(46)$ and $r_1\ \textless \ r_2\ \textless \ r_5\ \textless \ r_4\ \textless \ r_3\ \textless \ r_7\ \textless \ r_6$ .

Find the domain and range of the INVERSE of the given function.
Square root x-3

Answers

$f(x)=√(x-3)\n y=√(x-3)\n y^2=x-3\n x=y^2+3\n f^(-1)(x)=x^2+3\n D:x\in\matxbb{R}\n R:y\in\langle3,\infty)$

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