The question is asking you to find out what number is a fraction of something. For example, the problem on the top right is asking 35 is 50% (one half) of a number so what you have to do is multiply 35 times 2 and the answer is 70
Answer:
I think the answer is 9.920 please be careful with your answer I think it 9.920 because it says nine hundred and ninety two thousands so therefor the decimal need to be in between 9 and 9 I hope you get this correct!
step-by-step explanation:
Answer:
(x-3)(x^2-7) = x^3 -3x^2 -7x + 21
just change the sign of the root and put x in front of it, then multiply the factors all together
(x-3)(x-sqr7)(x+sqr7) roots come in conjugate pairs to eliminate irrational coefficients
(x-sqr7)(x+sqr7) = x^2-7. similar to (a-b)(a+b) = a^2-b^2
Step-by-step explanation:
Answer:x=31.4919
Step-by-step explanation:
Step1:isolate a square root on the left hand side√x+3=√2x-1-2
Step2:eliminate the radicals on the left hand side
Raise both sides to the second power
√x+3)^2=(√2x-1-2)^2
After squaring
x+3=2x-1+4-4-4√2x-1
Step3:get the remaining radicals by itself
x+3=2x-1+4-4√2x-1
Isolate radical on the left hand side
4√2x-1=-x-3+2x-1+4
4√2x-1=x
Step4:eliminate the radicals on the left hand side
Raise both side to the second power
(4√2x-1)^2=x^2
After squaring
32x-16=x^2
Step 5:solve the quadratic equation
x^2-32x-16
This equation has two real roots
x1=32+√960/2=31.4919
x2=32-√960/2=0.5081
Step6:check that the first solution is correct
Put in 31.4919 for x
√31.4919+3=√2•31.4919-1-2
√34.492=5.873
x=31.4919
Step7:check that the second solution is correct
√x+3=√2x-1-2
Put in 0.5081 for x
√0.5081+3=√2•0.5081-1-2
√3.508=-1.873
1.873#-1.873
One solution was found
x=31.4919
Answer:

Step-by-step explanation:
We have a separable equation, first let's rewrite the equation as:

But:

So:

Multiplying both sides by dx and dividing both sides by 3a+y:

Integrating both sides:

Evaluating the integrals:

Where C1 is an arbitrary constant.
Solving for y:


So:

Finally, let's evaluate the initial condition in order to find C1:

Solving for C1:

Therefore:
