Ask question

Ask question

All categories

3 years ago

11

There are 30 students. 8 students have cats, 15 students have dogs, and 5 students have both a cat and a dog. How many students

don't have a cat or a dog?

1 answer:

Mkey [24]3 years ago

6 0

Answer:

2 students have both cat and dog

Step-by-step explanation:

8+15+5=28

30-28=2

You might be interested in

The most popular items at a bakery are its raspberry scones and its lemon poppy

Leno4ka [110]

The system r+l = 46 and 15r + 9l = 396 can be used to determine the number of raspberries boxes and lemon seed boxes sold.

Step-by-step explanation:

Given,

Cost per box of raspberry = $15

Cost per box of lemon seeds = $9

Total boxes sold = 46

Total revenue generated = $396

Let,

r represent the number of raspberries sold

l represent the number of lemon seeds sold

According to given statement;

r+l = 46 Eqn 1

15r + 9l = 396 Eqn 2

The system r+l = 46 and 15r + 9l = 396 can be used to determine the number of raspberries boxes and lemon seed boxes sold.

7 0

3 years ago

How do I solve: -20<4-2x

Sunny_sXe [5.5K]

Steps to solve:

-20 < 4 - 2x

~Subtract 4 to both sides

-24 < -2x

~Divide -2 to both sides

12 > x

Best of Luck!

6 0

3 years ago

Help needed asap!!!!!!!

Alexxandr [17]

Answer:

8.2

Step-by-step explanation:

w-x+z

-2.1-(-7)+3.3

-2.1+7+3.3

8.2

Hope this helps!

6 0

3 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

16 - 2t = t + 9 + 4t

Inga [223]

Answer:

t = 1

Step-by-step explanation:

16 - 2t = t + 9 + 4t

=> 16 - 9 = 7t

=> 7 = 7t

=> t = 1

$GeniusUser$

6 0

2 years ago