Ask question

Ask question

All categories

2 years ago

10

What is the answer to these questions---

1 answer:

shutvik [7]2 years ago

5 0

1. w= 52
2. x= 5
3. n=1.4
4. x= -28/9 or x=-3.1

You might be interested in

Need help please :( What is the second term of (s+v)^5?

Citrus2011 [14]

First compute the coefficient like this:
$C_5^1=\frac{5!}{1!(5-1)!}=\frac{5!}{4!}=\frac{5\times4!}{4!}$
Simplifying the fraction over 4! we get:
$C_5^1=5$
and the variables are $s^4v$ . So answer $5s^4v$ .

The correct answer is C then.

3 0

3 years ago

If a menu has a choice of 4 appetizers, 5 main courses, and 5 desserts, how many dinners are possible if each includes one appet

emmasim [6.3K]

Only 4 dinners are possible.

3 0

3 years ago

The population in thousands of people of a city is approximated by the function P(t) = 1200(2)0.1038twhere t is the number of ye

asambeis [7]

The population of the group in 2019 is 2,134,000 people

Let P represent the population in thousands of people in the year t after 2011

Given the equation:

$P=1200(2)^{0.1038t}$

a) The population of the group in 2019, that is t = 8(2019 - 2011):

$P=1200(2)^{0.1038*8}=2134$

b) The population of the group in 2029, that is t = 18(2029 - 2011):

$P=1200(2)^{0.1038*18}=4382$

The population of the group in 2019 is 2,134,000 people

Find out more on equation at: brainly.com/question/2972832

7 0

2 years ago

A hog producer is feeding soybean meal and corn to his pigs. He needs to know how many pounds he needs to feed of each in order

Mazyrski [523]

Answer:

Feed 1, Soybean meal required = $\dfrac{2000}{3} \text{ lbs.}$

Feed 2, Corn meal required = $\dfrac{4000}{3} \text{ lbs.}$

Step-by-step explanation:

Total feed is <em>1 ton</em> i.e. <em>2000 lbs</em>.

Let <em>x</em> be the amount of Feed 1 required.

Feed 1 has $45\%$ of protein.

$\text{Protein in Feed 1 = }x \times \dfrac{45}{100} ..... (1)$

Then, amount of Feed 2 required = $(2000 - x) \text{ lbs.}$

Feed 2 has $15\%$ of protein.

$\text{Protein in Feed 2 = }(2000 - x) \times \dfrac{15}{100} ..... (2)$

As per question, total protein required is $25\%$ of 2000 lbs .

Adding (1) and (2) and putting it equal to total protein required.

$\Rightarrow x \times \dfrac{45}{100} + (2000 - x) \times \dfrac{15}{100} = 2000 \times \dfrac{25}{100}\\\Rightarrow 45x + 30000 - 15x = 50000\\\Rightarrow 30x = 20000\\\Rightarrow x = \dfrac{2000}{3}$

Feed 1 required = $\dfrac{2000}{3}\text{ lbs .}$

Feed 2 required = $2000 - \dfrac{2000}{3}\\$

$\Rightarrow \dfrac{4000}{3}\text{ lbs.}$

7 0

3 years ago

Convert 12 m to centimeters.

DiKsa [7]

The answer to your question is "1,200 cm".

5 0

2 years ago

Read 2 more answers