Sole equation 4×6-24÷4

If 70 mL of solution contains 280 mg of drug, how many milligrams of drug are in 50 mL of solution?

Ierofanga [76]

280 mg = 70 Ml
280/70 = 4
50 mL x 4 = 200 mg

200 mg

8 0

1 year ago

Let h(x)=2x−5 and g(x)=−3x+1.Find h(x)+g(x).? help please

Aleksandr-060686 [28]

This may look a little confusing but all you have to do is plug the equation given for h(x) which is 2x-5 in to the h(x) area in the equation where it says h(x) + g(x).

So far that’s (2x-5) + g(x)

Then,

Do the same with equation given for g(x) which is now,

(2x-5) + (3x+1)

Then solve,

2x - 5 + 3x + 1
2x + 3x- 5 + 1
5x - 4

Therefore the answer is 5x - 4.

8 0

3 years ago

The point (3, −5) is reflected over the x-axis. What are the coordinates of the new point? Group of answer choices

statuscvo [17]

(3,5) is the answer i think

hope it helpss

8 0

2 years ago

Write a linear function f with the values f (-1) = 5 and f (3) = 4.<br> A function is f(x) =

Drupady [299]

Answer:

Step-by-step explanation:

Since we have points (-1,5) and (3,4) we can find the slope m of y=mx+b

m=(y2-y1)/(x2-d1)=(5-4)/(-1-3)=1/-4=-1/4 so we have

y=-x/4+b, using point (3,4) we can solve for the y intercept, b.

4=-3/4+b, b=19/4, so the line is

y=(-x+19)/4

5 0

2 years ago

(x +y)^5<br> Complete the polynomial operation

Vesna [10]

Answer:

Please check the explanation!

Step-by-step explanation:

Given the polynomial

$\left(x+y\right)^5$

$\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i$

$a=x,\:\:b=y$

$=\sum _{i=0}^5\binom{5}{i}x^{\left(5-i\right)}y^i$

so expanding summation

$=\frac{5!}{0!\left(5-0\right)!}x^5y^0+\frac{5!}{1!\left(5-1\right)!}x^4y^1+\frac{5!}{2!\left(5-2\right)!}x^3y^2+\frac{5!}{3!\left(5-3\right)!}x^2y^3+\frac{5!}{4!\left(5-4\right)!}x^1y^4+\frac{5!}{5!\left(5-5\right)!}x^0y^5$

solving

$\frac{5!}{0!\left(5-0\right)!}x^5y^0$

$=1\cdot \frac{5!}{0!\left(5-0\right)!}x^5$

$=1\cdot \:1\cdot \:x^5$

$=x^5$

also solving

$=\frac{5!}{1!\left(5-1\right)!}x^4y$

$=\frac{5}{1!}x^4y$

$=\frac{5x^4y}{1}$

$=5x^4y$

similarly, the result of the remaining terms can be solved such as

$\frac{5!}{2!\left(5-2\right)!}x^3y^2=10x^3y^2$

$\frac{5!}{3!\left(5-3\right)!}x^2y^3=10x^2y^3$

$\frac{5!}{4!\left(5-4\right)!}x^1y^4=5xy^4$

$\frac{5!}{5!\left(5-5\right)!}x^0y^5=y^5$

so substituting all the solved results in the expression

$=\frac{5!}{0!\left(5-0\right)!}x^5y^0+\frac{5!}{1!\left(5-1\right)!}x^4y^1+\frac{5!}{2!\left(5-2\right)!}x^3y^2+\frac{5!}{3!\left(5-3\right)!}x^2y^3+\frac{5!}{4!\left(5-4\right)!}x^1y^4+\frac{5!}{5!\left(5-5\right)!}x^0y^5$

$=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$

Therefore,

$\left(x\:+y\right)^5=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$

6 0

2 years ago