Ask question

Ask question

All categories

2 years ago

8

Please help me with this problem

1 answer:

melomori [17]2 years ago

7 0

Answer:

the positive and negative square root of 196 is 14

You might be interested in

Find the missing terms in the following sequence 2,4,8,16,30, , , , 186,

ivann1987 [24]

Are you sure it is correct because for each one your just doubling the number. 2+2=4. 4+4=8. 8+8=16. I think its suppose to be 16+16= 32 and so on, but I may be wrong

5 0

3 years ago

What is 936,292 rounded to the nearest hundred thousand

Romashka [77]

The answer is 900,000?

3 0

2 years ago

A scientist writes the equation n(h)100e^0.25h to model the growth of a certain bacteria in a petri dish, where N represents the

Natasha2012 [34]

Answer: Second option is correct.

Step-by-step explanation:

Since we have given that

$n(h)=100e^{0.25h}$

where, N represents the number of bacteria after 'h' hours.

So, we have given that there are 450 bacteria present.

N = 450

According to question,

$450=100e^{0.25h}\\\\\frac{450}{100}=e^{0.25h}\\\\4.5=e^{0.25h}\\\\\text{Taking natural log 'ln' on both sides}\\\\\ln (4.5)=0.25h\\\\1.50=0.25h\\\\h=\frac{1.50}{0.25}\\\\h\approx 6\ hours$

Hence, Second option is correct.

8 0

3 years ago

Read 2 more answers

Verify the identity. cotangent x equals StartFraction 1 plus cosine 2 x Over sine 2 x EndFractioncot x= 1+cos2x sin2x Use the ap

lutik1710 [3]

Answer with Step-by-step explanation:

We are given that

RHS

$\frac{1+Cos2x}{Sin2x}$

We have to verify the identity.

We know that

$1+Cos2x=2Cos^2 x$

$Sin2x=2SinxCos x$

Using the formula

$\frac{2Cos^2x}{2SinxCosx}$

$\frac{Cosx}{Sinx}$

$Cot x$

By using the formula

$Cotx=\frac{Cosx}{Sinx}$

LHS=RHS

Hence, verified

6 0

3 years ago

Use the rules of exponents to simplify the expressions. Match the expression with its equivalent value.

Lelechka [254]

Answer:

1) $\frac{(-2)^{-5}}{(-2)^{-10}}=-32$

2) $2^{-1}.2^{-4} = \frac{1}{32}$

3) $(-\frac{1}{2} )^3.(-\frac{1}{2} )^2=-\frac{1}{32}$

4) $\frac{2}{2^{-4}} = 32$

Step-by-step explanation:

1) $\frac{(-2)^{-5}}{(-2)^{-10}}$

Solving using exponent rule: $a^{-m}=\frac{1}{a^m}$

$\frac{(-2)^{-5}}{(-2)^{-10}}\\=(-2)^{-5+10}\\=(-2)^{5}\\=-32$

So, $\frac{(-2)^{-5}}{(-2)^{-10}}=-32$

2) $2^{-1}.2^{-4}$

Using the exponent rule: $a^m.a^n=a^{m+n}$

We have:

$2^{-1}.2^{-4}\\=2^{-1-4}\\=2^{-5}$

We also know that: $a^{-m}=\frac{1}{a^m}$

Using this rule:

$2^{-5}\\=\frac{1}{2^5}\\=\frac{1}{32}$

So, $2^{-1}.2^{-4} = \frac{1}{32}$

3) $(-\frac{1}{2} )^3.(-\frac{1}{2} )^2$

Solving:

$(-\frac{1}{2} )^3.(-\frac{1}{2} )^2\\=(-\frac{1}{8} ).(\frac{1}{4} )\\=-\frac{1}{32}$

So, $(-\frac{1}{2} )^3.(-\frac{1}{2} )^2=-\frac{1}{32}$

4) $\frac{2}{2^{-4}}$

We know that: $a^{-m}=\frac{1}{a^m}$

$\frac{2}{2^{-4}}\\=2\times 2^4\\=2(16)\\=32$

So, $\frac{2}{2^{-4}} = 32$

3 0

3 years ago