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3 years ago

13

Solve the equation.

1 answer:

Readme [11.4K]3 years ago

8 0

Answer:

= 14/3

Step-by-step explanation:

add 3 to both sides of the equation

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Which expression is equivalent to 52.2

murzikaleks [220]

Answer:

Anwer is (z.50)+(z.2) since expanding it is z.52

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2 years ago

Simplify. 5(22 - 9 · 2) - 2( 3 · 2 4) 7 <br> a.-99 <br> b.-83 <br> c.-75 <br> d.-71

Pachacha [2.7K]

Not completely sure on the symbols, but I think it is 5 x ( 22 - 9 x 2 ) - 2 ( 3 x 2 x 4 ) x 7 = 136. It doesn't really match the answers, so my advice is that you use a calculator for this one, or as the person/teacher who gave you the problem how to solve it.

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3 years ago

If apples cost £1.25 per killogram. How much will 6 kilogram cost?

pochemuha

Your answer will be 7.50. What I did was take 1.25 and times it by 6.

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3 years ago

Find+the+positive+value+for+α+if+the+radius+of+the+circle+3x^2+3y-6αx+12y-3α=0+is+4

Alik [6]

Answer:

$\alpha=3$

Step-by-step explanation:

<u>Equation of a Circle</u>

A circle of radius r and centered on the point (h,k) can be expressed by the equation

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

We are given the equation of a circle as

$3x^2+3y^2-6\alpha x+12y-3\alpha=0$

Note we have corrected it by adding the square to the y. Simplify by 3

$x^2+y^2-2\alpha x+4y-\alpha=0$

Complete squares and rearrange:

$x^2-2\alpha x+y^2+4y=\alpha$

$x^2-2\alpha x+\alpha^2+y^2+4y+4=\alpha+\alpha^2+4$

$(x-\alpha)^2+(y+2)^2=r^2$

We can see that, if r=4, then

$\alpha+\alpha^2+4=16$

Or, equivalently

$\alpha^2+\alpha-12=0$

There are two solutions for $\alpha$ :

$\alpha=-4,\ \alpha=3$

Keeping the positive solution, as required:

$\boxed{\alpha=3}$

8 0

3 years ago

Write an equation for an ellipse centered at the origin, which has foci at (\pm\sqrt{8},0)(± 8 ,0)(, plus minus, square root o

Alchen [17]

Answer:

The equation of ellipse centered at the origin

$\frac{x^2}{18} +\frac{y^2}{10} =1$

Step-by-step explanation:

given the foci of ellipse (±√8,0) and c0-vertices are (0,±√10)

The foci are (-C,0) and (C ,0)

Given data (±√8,0)

the focus has x-coordinates so the focus is lie on x- axis.

The major axis also lie on x-axis

The minor axis lies on y-axis so c0-vertices are (0,±√10)

given focus C = ae = √8

Given co-vertices ( minor axis) (0,±b) = (0,±√10)

b= √10

The relation between the focus and semi major axes and semi minor axes are $c^2=a^2-b^2$

$a^{2} = c^{2} +b^{2}$

$a^{2} = (\sqrt{8} )^{2} +(\sqrt{10} )^{2}$

$a^{2} =18$

$a=\sqrt{18}$

The equation of ellipse formula

$\frac{x^2}{a^2} +\frac{y^2}{b^2} =1$

we know that $a=\sqrt{18} and b=\sqrt{10}$

<u>Final answer:</u>-

<u>The equation of ellipse centered at the origin</u>

<u /> $\frac{x^2}{18} +\frac{y^2}{10} =1$ <u />

8 0

3 years ago

Read 2 more answers