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3 years ago

Use division to express the fraction in the form quotient + remainder / divisor. y^2-y-11/y-3

1 answer:

3 0

use long division or synthetic division
I will use long

y-3 times y=y^3-3y
subtract that from y^2-y-11
we are left with
2y-11
multiply y-3 by 2 and get 2y-6
subtract that from 2y-11
get 5 as a remainder

answer is y+2+5/(y-3)

You might be interested in

Find the equation of a line that is perpendicular to 4x+6y=1 that has a greater y intercept compared to 2x+3y=18

Marina86 [1]

Answer:

4y = 6x + 40

Step-by-step explanation:

The general equation of a straight line is y = mx + b

m is the slope and b is the y-intercept

let us write both equations in this form;

we have this as;

6y = -4x + 1

y = -4x/6 + 1/6

and;

2x + 3y = 18

3y = -2x + 18

y = -2x/3 + 6

So firstly we want to find an equation that is perpendicular to the first

When two lines are perpendicular, their slopes has a product of -1

The slope of the first line is -4/6

let the slope of the line we want be m

As per they are perpendicular;

-4/6 * m = -1

-4m/6 = -1

-4m = -6

m = 6/4

So now, we want the y-intercept greater than that of the second equation which is a y-intercept of 6

we can choose 10

and we have the equation as:

y = 6x/4 + 10

multiply through by 4

4y = 6x + 40

7 0

3 years ago

The sum of the diagonals of a rhombus is 5√2.

Alex

Greetings!
Let ABCD be a rhombus
AC + BD = 5√2 cm

Area of ABCD = Ar△ABD + Ar △BCD
$= \frac{1}{2} \times BD \times AO + \frac{1}{2} \times BD \times OC$
$= \frac{1}{2} BD (AO + OC)$
$\frac{1}{2} BD \times AC$

$So, \frac{ BD \times AC}{2} = 4 \: cm {}^{2}$
$BD \times AC = 8$
$Now, AC + \: BD = 5 \sqrt{2}$

Squaring both sides, we get
$AC {}^{2} + BD {}^{2} + 2 AC.BD =50$
$AC {}^{2} + BD {}^{2} + 2 \times 8 = 50$
$AC {}^{2} + BD {}^{2} = 50 - 16$
$AC {}^{2} + BD {}^{2} = 34$

In △AOB, we have
$OA {}^{2} + OB {}^{2} =AB {}^{2}$
$( \frac{ AC }{2} ) {}^{2} + ( \frac{ BD}{2} ) {}^{2} = AB {}^{2}$
$\frac{ AC {}^{2} } {4} + \frac{ BD {}^{2} }{4} = AB {}^{2}$
$AC {}^{2} + BD {}^{2} = 4 AB {}^{2}$
$34 = 4 AB {}^{2}$

Square rooting both sides
$\sqrt{34} = 2 AB$
$Perimeter = 4 \: AB \\ = 2 \times 2 \: AB \\ = 2 \: \times \sqrt{34 } \\ = 2 \sqrt{34 \: } units$ .

Hope it helps!

7 0

3 years ago

The range of 16,23,55,64,49

miskamm [114]

Answer: 48

Step-by-step explanation: To find the range of the data set shown here, remember that the range is the difference between the greatest number in the data set and the least number in the data set.

Greatest number → 64

Least number → 16

Now, we need to subtract 16 from 64.

64 - 16 = 48

Therefore, the range of the data set is 48.

7 0

3 years ago

Simplify the expression. ( I got 144, but i'm not sure)

g100num [7]

Answer:

144 is correct

Step-by-step explanation:

13 - 1 is 12

12 times 12 = 144

144 divided by 2 is 72

72 times 2 is 144

8 0

3 years ago

Read 2 more answers

If someone could help me that would be great

sasho [114]

Answer:

please mark me as brainlist please

Step-by-step explanation:

Correct option is B)

The circumference of the circle is 96 units.

Therefore, 2π×r=96

⇒r=

2×22

96×7

=15.27 units

Now, in the quadrilateral OLMN,∠OLM+∠LMN+∠MNO+∠NOL=360

Given that LM and MN are tangents to the circle, ∠OLM=∠ONM=90

∴∠NOL=360−90−90−60=120

The length of arc is given by rθ where θ is the angle made by the arc at the centre in radian.

Thus, arc length LN=15.27×

180

120π

=31.98

8 0

2 years ago