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3 years ago

Does antibiotics help kill the common cold??

2 answers:

5 0

<span>Antibiotics don't work on the common cold. They fight bacteria-related illnesses. Colds are caused by viruses, so antibiotics won't do you any good. They can, however, do you harm.</span><span>Mar 29, 2016</span>

sasho [114]3 years ago

3 0

No it doesn't because colds are made up of viruses. So the antibiotics won't help.

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87.5% of what number is 49?

kozerog [31]

Answer=56

87.5%*y=49
.875y=49
____ ___
.875 .875
divide both sides by .875
y=56

7 0

3 years ago

Six values on the number line above are marked with letters. Which letters represent integers when the value on the number line

r-ruslan [8.4K]

Answer:

A) L, N, O, P

Step-by-step explanation:

The even numbers on the number line are marked by the letters L, N, O, P.

_____

If a number gives an integer result when divided by 2, it is an even number. Zero is an even number, giving the integer zero when divided by 2.

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3 years ago

HEKEKEKKELWMEMEKSMEJERJJEHEHEHEHEHEGRVEBS

faltersainse [42]

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3 years ago

A self own company charges $40 per month +0.05 cents for each text message sent.Select expression is there a percent of cost in

kondaur [170]

<h3>The monthly cost of cell phone is A = 40 + 0.005 m</h3>

Step-by-step explanation:

The monthly fixed charges by mobile company = $40

The cost of sending each text message = 0.05 cents

Now, 100 cents = 1 dollar

⇒ 1 cent =$ $(\frac{1}{100} )$

⇒ 0.05 cent s =$ $(\frac{1}{100} \times 0.05) =0.0005$

Also, the number of text messages send = m

So, the cost of sending m text message = m x ( cost of 1 text )

= m x ( $ 0.0005) = 0.0005 m

Now, Total bill = Fixed rate + cost of m texts

⇒ A =$40 + $ 0.0005 m

Hence, the monthly cost of cell phone is A = 40 + 0.005 m

7 0

3 years ago

In the third degree polynomial function f(x) =x^3 + 4x, why are 2i and -2i zeros?

IgorC [24]

Answer:

We just need to evaluate and get f(2i)=0, f(-2i)=0.

Step-by-step explanation:

Since $i^2=-1$ , then $i^3=i^2i=-i$ , and we can apply this when we evaluate $f(x) =x^3 + 4x$ for 2i and -2i.

First we have:

$f(2i) =(2i)^3 + 4(2i)=2^3i^3+8i=8(-i)+8i=0$

Which shows that 2i is a zero of f(x).

Then we have:

$f(-2i) =(-2i)^3 + 4(-2i)=(-2)^3i^3-8i=-8(-i)-8i=8i-8i=0$

Which shows that -2i is a zero of f(x).

4 0

3 years ago