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3 years ago

Can anyone answer these for me pls

Mathematics

1 answer:

wel3 years ago

6 0

Answer:

#7 can be simplified to -10 √15x -40 √5

#8 = -20 √6x +30x√2

#9 = 10x√6+3 √15xx

#10 = -20√6a -15 √2

Step-by-step explanation:

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The length of a rectangle is the width minus 8 units. The area of the rectangle is 9 units. What is the length, in units, of the

frez [133]

Length of rectangle is 1 unit

Solution: 

Given that

Length of a rectangle is width minus 8 units.

Area of rectangle = 9 square units

Need to calculate the length of rectangle.

Let us assume width of rectangle = x

As Length is width minus 8,

Length of the rectangle = x – 8

$\text{ Area of rectangle }=\text{ width of rectangle }\times \text{ Length of the rectangle }$

$\text{ Area of rectangle }= x\times (x-8)$

$\text { Area of rectangle }=\left(x^{2}-8 x\right)$

As given that area of rectangle = 9 square units

$\begin{array}{l}{\Rightarrow x^{2}-8 x=9} \\\\ {\Rightarrow x^{2}-8 x-9=0}\end{array}$

On solving above quadratic equation for x, we get

$\begin{array}{l}{\Rightarrow x^{2}-9 x+x-9=0} \\\\ {\Rightarrow x(x-9)+1(x-9)=0} \\\\ {\Rightarrow (x-9)(x+1)=0}\end{array}$

When $x-9 =9, x = 9$

When $x + 1 = 0, x = -1$

As width cannot be negative, so considering x = 9

$\text{ Length of rectangle }= x-8 = 9-8 = 1 \text{ unit }$

Hence, Length of rectangle is 1 unit.

5 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 $a=\sqrt{18} and b=\sqrt{10}$

Final answer:-

The equation of ellipse centered at the origin

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

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