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2 years ago

What is the quotient of (-6) divided by (-7)

Mathematics

1 answer:

stepladder [879]2 years ago

5 0

Answer:

6/7 or .857

Step-by-step explanation:

Since they are both negative, it cancels each other and becomes positive

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PLEASE HELP ME ASAP!

romanna [79]

Answer:

x =- 3 and y = -7

Step-by-step explanation:

2X-2Y=-8

X=2y + 11

We need to isolate both X and Y in both equations

2x-2y=-8

(add 2y to both sides)

2x=-8+ 2y

(divide both sides by 2)

x=-4+y and x=2y+11

because both of these equations are the same we can put them together

4+y=2y+11

(subtract y)

4=y+11

(subtract 11)

-7=y

so y = -7

then to find x you just need to plug in y to one of the equations

x=2(-7) + 11

x= -14 +11

x = -3

6 0

3 years ago

Read 2 more answers

Point R divides in the ratio 1 : 3. If the x-coordinate of R is -1 and the x-coordinate of P is -3, what is the x-coordinate of

ollegr [7]

Question:

Point R divides in the ratio 1 : 3. If the x-coordinate of R is -1 and the x-coordinate of P is -3, what is the x-coordinate of Q?

Answer:

x-coordinate of Q is 5

Solution:

Given that,

Point R divides in the ratio 1 : 3

Which means,

Point R divides the line segment PQ internally

The x-coordinate of the point which divides the line segment in ration m:n internally is given as:

$x=(\frac{mx_{2}+nx_{1}}{m+n} )$

Where,

$x$ = x-coordinate of point dividing the segment R = -1

$x_1$ = x-coordinate of P = -3

$x_2$ = x-coordinate of Q = ?

m : n = 1 : 3

m = 1

n = 3

Therefore,

$-1 = \frac{1 \times x_2 + 3 \times -3}{1+3}\\\\-1 = \frac{x_2 -9}{4}\\\\x_2 - 9 = -4\\\\x_2 = 9 - 4\\\\x_2 = 5$

Thus x-coordinate of Q is 5

5 0

2 years ago

Does anybody know this? 1/2 divided by 4/7

Stells [14]

The answer to the question is 7/8

7 0

2 years ago

Read 2 more answers

Find the value of c so that (x-5) is a factor of the polynomial p(x)

liubo4ka [24]

I think the question is

Find the value of c so that (x-5) is a factor of the polynomial

$p(x) = x^3 + 2x^2 + cx + 10$

The other factor is going to be some quadratic. We can say a few things about its coefficients but let's start by saying in general it's

$q(x)= ax^2 + bx + k$

$p(x) = (x-5)q(x)$

$x^3 + 2x^2 + cx + 10 = (x-5)(ax^2 + bx+k) = ax^3 + (b-5a)x^2 + (k-5b)x - 5k$

Equating respective coefficients,

$a=1$

$b-5a = 2$

$k - 5b = c$

$-5k = 10$

so we get

$b = 2 + 5 = 7$

$k = 10/-5 = -2$

$c = k - 5b = 2 - 5(7)= -37$

Answer: -37

Check:

$(x^2 + 7x - 2)(x - 5) = x^3 + 2 x^2 - 37 x + 10\quad\checkmark$

7 0

2 years ago

What are the zeros of the function y = x2 + 10x – 171, and why?

aleksandr82 [10.1K]

Y = x^2 + 10x - 171
y = (x - 9)(x + 19)

x - 9= 0 x + 19 = 0
x = 9 x = -19

Answer B covers all requirements... the factored form is
y= (x + 19)(x - 9) 
and the zeros are -19 and 9

4 0

3 years ago

Read 2 more answers