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4 years ago

Which inequality represents the graph below

Mathematics

2 answers:

givi [52]4 years ago

6 0

q<5

The circle is opened so you don't choose the answers that have a line under the inequality because that would be a closed circle. That already gets rid of 2 answers. Then you choose the answer that has the direction that the arrow is pointing towards.

Basile [38]4 years ago

3 0

Answer:

q<5

Step-by-step explanation:

The graph of number is given

In the number line we have shaded part at starts at 5

we have open circle at 5

for open circle we use < or > symbol

The number line is shaded left

For shading left we use < symbol and for shading right we use > symbol

So inequality is q< 5

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-48 + 37 - 28 - 1 please help ! please include steps too

Svetach [21]

Answer:

-40

Step-by-step explanation:

-48 + 37 - 28 - 1

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-40

6 0

3 years ago

Generalize the pattern by finding the nth term. 2, 6, 12, 20...

natulia [17]

this is neither a geometric nor an arithmetic sequence

add 4, add 6, add 8

Since you have options, I would guess and check

n=2

2^2+2 =6

3(2)-1 = 5 no

(2+1) (2+2) =12 no

Choice A

8 0

3 years ago

The students in Mrs. Smith's class placed 26 math textbooks on a scale. The scale indicated a weight of 1,248 ounces. Each math

irina [24]

Answer:

48 ounces

Step-by-step explanation:

1,248 divided by 26 is 48

5 0

3 years ago

Solve 5y'' + 3y' – 2y = 0, y(0) = 0, y'(0) = 2.8 y(t) = 0 Preview

mario62 [17]

Answer: The required solution is

$y(t)=-\dfrac{7}{3}e^{-t}+\dfrac{7}{3}e^{\frac{1}{5}t}.$

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

$5y^{\prime\prime}+3y^\prime-2y=0,~~~~~~~y(0)=0,~~y^\prime(0)=2.8~~~~~~~~~~~~~~~~~~~~~~~~(i)$

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

$5m^2e^{mt}+3me^{mt}-2e^{mt}=0\\\\\Rightarrow (5m^2+3y-2)e^{mt}=0\\\\\Rightarrow 5m^2+3m-2=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow 5m^2+5m-2m-2=0\\\\\Rightarrow 5m(m+1)-2(m+1)=0\\\\\Rightarrow (m+1)(5m-1)=0\\\\\Rightarrow m+1=0,~~~~~5m-1=0\\\\\Rightarrow m=-1,~\dfrac{1}{5}.$

So, the general solution of the given equation is

$y(t)=Ae^{-t}+Be^{\frac{1}{5}t}.$

Differentiating with respect to t, we get

$y^\prime(t)=-Ae^{-t}+\dfrac{B}{5}e^{\frac{1}{5}t}.$

According to the given conditions, we have

$y(0)=0\\\\\Rightarrow A+B=0\\\\\Rightarrow B=-A~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)$

and

$y^\prime(0)=2.8\\\\\Rightarrow -A+\dfrac{B}{5}=2.8\\\\\Rightarrow -5A+B=14\\\\\Rightarrow -5A-A=14~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{Uisng equation (ii)}]\\\\\Rightarrow -6A=14\\\\\Rightarrow A=-\dfrac{14}{6}\\\\\Rightarrow A=-\dfrac{7}{3}.$

From equation (ii), we get

$B=\dfrac{7}{3}.$

Thus, the required solution is

$y(t)=-\dfrac{7}{3}e^{-t}+\dfrac{7}{3}e^{\frac{1}{5}t}.$

7 0

3 years ago

Which classification best describes the following system of equations? 3x+6y-12z=36 x=2y-4z=12 4x+8y-16z=48 inconsistent and dep

Viktor [21]

Answer:

Consistent and dependent

Step-by-step explanation:

We are given the following equation:

1. $3x+6y-12z=36$

2. $x+2y-4z=12$

3. $4x+8y-16z=48$

For equation 1 and 3, if we take out the common factor (3 and 4 respectively) out of it then we are left with $x+2y-4z=12$ which is the same as the equation number 2.

There is at least one set of the values for the unknowns that satisfies every equation in the system and since there is one solution for each of these equations, this system of equations is consistent and dependent.

6 0

3 years ago