All categories

3 years ago

Plz help me get the right answer to these

Mathematics

2 answers:

-Dominant- [34]3 years ago

8 0

Step-by-step explanation:

$(-7b^5+42b^4)\dib7b=\dfrac{-7b^5}{7b}+\dfrac{42b^4}{7b}=-b^4+6b^3$

Citrus2011 [14]3 years ago

4 0

- $b^{4}$ + 6 $b^{3}$

divide each term on the numerator by 7b

= - 7 $b^{5}$ / 7b + 42 $b^{4}$ / 7b

= - $b^{4}$ + 6 $b^{3}$

You might be interested in

1. Quadrilateral ABCD has vertices A(-1, 1), B(2, 3), C(6, 0) and D(3, -2). Determine using coordinate geometry whether or not t

deff fn [24]

Answer:

The Conclusion is

Diagonals AC and BD,

a. Bisect each other

b. Not Congruent

c. Not Perpendicular

Step-by-step explanation:

Given:

[]ABCD is Quadrilateral having Vertices as

A(-1, 1),

B(2, 3),

C(6, 0) and

D(3, -2).

So the Diagonal are AC and BD

To Check

The diagonals AC and BD

a. Bisect each other. B. Are congruent. C. Are perpendicular.

Solution:

For a. Bisect each other

We will use Mid Point Formula,

If The mid point of diagonals AC and BD are Same Then

Diagonal, Bisect each other,

For mid point of AC

$Mid\ point(AC)=(\dfrac{x_{1}+x_{2} }{2},\dfrac{y_{1}+y_{2} }{2})$

Substituting the coordinates of A and C we get

$Mid\ point(AC)=(\dfrac{-1+6}{2},\dfrac{1+0}{2})=(\dfrac{5}{2},\dfrac{1}{2})$

Similarly, For mid point of BD

Substituting the coordinates of B and D we get

$Mid\ point(BD)=(\dfrac{2+3}{2},\dfrac{3-2}{2})=(\dfrac{5}{2},\dfrac{1}{2})$

Therefore The Mid point of diagonals AC and BD are Same

Hence Diagonals,

a. Bisect each other

B. Are congruent

For Diagonals to be Congruent We use Distance Formula

For Diagonal AC

$l(AC) = \sqrt{((x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2} )}$

Substituting A and C we get

$l(AC) = \sqrt{((6-(-1))^{2}+(0-1)^{2} )}=\sqrt{(49+1)}=\sqrt{50}$

Similarly ,For Diagonal BD

Substituting Band D we get

$l(BD) = \sqrt{((3-2))^{2}+(-2-3)^{2} )}=\sqrt{(1+25)}=\sqrt{26}$

Therefore Diagonals Not Congruent

For C. Are perpendicular.

For Diagonals to be perpendicular we need to have the Product of slopes must be - 1

For Slope we have

$Slope(AC)=\dfrac{y_{2}-y_{1} }{x_{2}-x_{1} }$

Substituting A and C we get

$Slope(AC)=\dfrac{0-1}{6--1}\\\\Slope(AC)=\dfrac{-1}{7}$

Similarly, for BD we have

$Slope(BD)=\dfrac{-2-3}{3-2}\\\\Slope(BD)=\dfrac{-5}{1}$

The Product of slope is not -1

Hence Diagonals are Not Perpendicular.

6 0

4 years ago

Verifique se os numeros 15,20 e 30 sao diretamente proporcionais aos numeros 24,32 e 48

Vikki [24]

We are given to check

whether these numbers are proportional

we know that any set of numbers are proportional when their ratios are equal

so, firstly we will ratios

First set of numbers:

We are given first set of numbers as 15 , 20 and 30

so, it's ratio is

$15:20:30$

now, we can simplify it

$5\times 3:5\times 4:5\times 6$

Since, common factor is 5

so, we can cancel 5

and we get

$3:4:6$

Second set of numbers:

We are given first set of numbers as 24 , 32 and 48

so, it's ratio is

$24:32:48$

now, we can simplify it

$8\times 3:8\times 4:8\times 6$

Since, common factor is 8

so, we can cancel 8

and we get

$3:4:6$

We can see that both sets of numbers are having same ratios

so, they are proportional.........Answer

4 0

3 years ago

Complete the table for the given rule. Rulet y =

siniylev [52]

At x = 8, y = 4

At x = 14, y = 7

At y = 3, x = 6

Solution:

Given rule:

$$y=\frac{x}{2}$

At x = 8,

$$y=\frac{8}{2}=4$

Hence at x = 8, y = 4.

At x = 14.

$$y=\frac{14}{2}=7$

Hence at x = 14, y = 7.

At y = 3

$$3=\frac{x}{2}$

Multiply by 2 on both sides.

$$3\times2=\frac{x}{2} \times2$

$6 =x$

Hence at y = 3, x = 6.

The image of the table is attached below.

3 0

4 years ago

The question is the picture

BabaBlast [244]

Tha answer is 80+45+1.00

7 0

3 years ago

Evaluate -|-8 - 2| -10 -6 10

Natasha_Volkova [10]

Answer:

-10

Step-by-step explanation:

I used a calculator.

5 0

4 years ago

Read 2 more answers