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Monica [59]
3 years ago
14

I dont understand this but maybe some one will and they will help me. The quotient of a number and -7 decreased by 2 is ten find

the answer
Mathematics
1 answer:
givi [52]3 years ago
8 0
It might be nineteen because nineteen minus seven is twelve. twelve minus two is ten.
You might be interested in
Consider U = {x|x is a real number}. A = {x|x ∈ U and x + 2 > 10} B = {x|x ∈ U and 2x > 10} Which statements are true? 5 ∉
-Dominant- [34]

x+2 > 10 solves to x > 8 after we subtract 2 from both sides

So set A is the set of real numbers that are larger than 8. The value 8 itself is not in set A. The same can be said about 5 as well.


Set B is the set of values that are larger than 5 since 2x > 10 turns into x > 5 after dividing both sides by 2. The value x = 5 is not in set B since x > 5 would turn into 5 > 5 which is false. The values x = 6, x = 8, and x = 9 are in set B.


----------------


Summarizing everything, we can say...

5 is not in set A. True

5 is in set B. False

6 is in set A. False

6 is not in set B. False

8 is not in set A. True

8 is in set B. True

9 is in set A. True

9 is not in set B. False


5 0
2 years ago
Read 2 more answers
BRAINLIEST AND STARS
IrinaK [193]

Answer:

D

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
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
2 years ago
PLS I need help with this,
Over [174]

Answer:

5

Step-by-step explanation:

2:16-2:18=3

180ft-185ft=5

6 0
3 years ago
Which expression is the factorization of x2 + 10x + 21?
maxonik [38]

Answer:

(x + 3)(x + 7)

Step-by-step explanation:

Find two numbers that when added up to , they ALSO have to multiply up to 21. This is simple because of the fact that there is no leading coefficient greater than 1⃣.

7 0
2 years ago
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