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3 years ago

What is the quotient of - 48 ÷ -6 , answer and show how to solve

Mathematics

2 answers:

KonstantinChe [14]3 years ago

7 0

The quotient is -8. This being because when a negative number is in parentheses, the negative will be the opposite: Positive. And we know 48 divided by 6 is 8. So the answer is B.

Brums [2.3K]3 years ago

3 0

The best answer is B)-8 because if you do a negative times a positive it will give you a Negative answer and if you do a negative times a negative then you will get a positive answer so I hope this helps

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What is quivalent to 5/8?

coldgirl [10]

Answer:

10/16 is also equivalent

7 0

2 years ago

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Line segment AB has a length of 2 units. It is translated 6 units to the right on a coordinate plane to obtain line segment A'B'

NeTakaya

2 units, because translations do not change the length of segments. Hope this helps!

6 0

3 years ago

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The equation for line r can be written as y=-1/4x+3. Line s which is parallel to lone r includes the point (-4,3). What is the e

MaRussiya [10]

Answer:

$y=-\frac{1}{4}x+2$

Step-by-step explanation:

Hi there!

What we need to know:

Linear equations are typically organized in slope-intercept form: $y=mx+b$ where m is the slope and b is the y-intercept (the value of y when x is 0)
Parallel lines always have the same slope

1) Determine the slope of line S using line R (m)

$y=-\frac{1}{4} x+3$

We can identify clearly that the slope of the line is $-\frac{1}{4}$ , as it is in the place of m. Because parallel lines always have the same slope, the slope of line S would also be $-\frac{1}{4}$ . Plug this into $y=mx+b$ :

$y=-\frac{1}{4}x+b$

2) Determine the y-intercept of line S (b)

$y=-\frac{1}{4}x+b$

Plug in the given point (-4,3) and solve for b

$3=-\frac{1}{4}(-4)+b\\3=1+b$

Subtract 1 from both sides to isolate b

$3-1=1+b-1\\2=b$

Therefore, the y-intercept is 2. Plug this back into $y=-\frac{1}{4}x+b$ :

$y=-\frac{1}{4}x+2$

I hope this helps!

6 0

2 years ago

Solve y'' + 10y' + 25y = 0, y(0) = -2, y'(0) = 11 y(t) = Preview

svetlana [45]

Answer: The required solution is

$y=(-2+t)e^{-5t}.$

Step-by-step explanation: We are given to solve the following differential equation :

$y^{\prime\prime}+10y^\prime+25y=0,~~~~~~~y(0)=-2,~~y^\prime(0)=11~~~~~~~~~~~~~~~~~~~~~~~~(i)$

Let us consider that

$y=e^{mt}$ be an auxiliary solution of equation (i).

Then, we have

$y^prime=me^{mt},~~~~~y^{\prime\prime}=m^2e^{mt}.$

Substituting these values in equation (i), we get

$m^2e^{mt}+10me^{mt}+25e^{mt}=0\\\\\Rightarrow (m^2+10y+25)e^{mt}=0\\\\\Rightarrow m^2+10m+25=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow m^2+2\times m\times5+5^2=0\\\\\Rightarrow (m+5)^2=0\\\\\Rightarrow m=-5,-5.$

So, the general solution of the given equation is

$y(t)=(A+Bt)e^{-5t}.$

Differentiating with respect to t, we get

$y^\prime(t)=-5e^{-5t}(A+Bt)+Be^{-5t}.$

According to the given conditions, we have

$y(0)=-2\\\\\Rightarrow A=-2$

and

$y^\prime(0)=11\\\\\Rightarrow -5(A+B\times0)+B=11\\\\\Rightarrow -5A+B=11\\\\\Rightarrow (-5)\times(-2)+B=11\\\\\Rightarrow 10+B=11\\\\\Rightarrow B=11-10\\\\\Rightarrow B=1.$

Thus, the required solution is

$y(t)=(-2+1\times t)e^{-5t}\\\\\Rightarrow y(t)=(-2+t)e^{-5t}.$

6 0

3 years ago

Select the following numbers that cannot represent the area of a polygon.

Alborosie

-5,0 and-200 cannot represent the area of polygon
As 0 has no value and there are impossible to make a polygon with negative value

3 0

3 years ago