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3 years ago

2. Which statement best describes the relationship between the two given equations?

Mathematics

2 answers:

Hoochie [10]3 years ago

6 0

Answer:

Step-by-step explanation:

2x + 4y = 6

4y = -2x + 6

y = -1/2x + 3/2.....slope = -1/2, y int = 3/2

x - 2y = 3

-2y = -x + 3

y = 1/2x - 3/2....slope = 1/2, y int = -3/2

different slopes, different y int's.....means one solution <==

Morgarella [4.7K]3 years ago

3 0

The answer to your question is B

You might be interested in

Ella has 0.5 lbs of sugar. How much water should she add to make the following concentrations? Tell Ella how much syrup she will

Oksanka [162]

Answer:

Ella should add $32\dfrac{5}{6}$ lbs of water.

Total weight of syrup $33\dfrac{1}{3}$ lbs

Step-by-step explanation:

Weight:

Weight of sugar = 0.5 lbs.

Weight of water added = x lbs

Total weight = 0.5 + x lbs

Percentage:

0.5 + x -- 100%

0.5 -- 1.5%

Write a proportion:

$\dfrac{0.5+x}{0.5}=\dfrac{100}{1.5}$

Cross multiply:

$1.5(0.5+x)=0.5\cdot 100\\ \\0.75+1.5x=50\\ \\1.5x=50-0.75\\ \\1.5x=49.25\\ \\x=\dfrac{49.25}{1.5}=\dfrac{4,925}{150}=\dfrac{197}{6}=32 \dfrac{5}{6}\ lbs$

Ella should add $32\dfrac{5}{6}$ lbs of water.

Total weight of syrup

$32\dfrac{5}{6}+\dfrac{1}{2}=32\dfrac{5}{6}+\dfrac{3}{6}=33\dfrac{1}{3}\ lbs$

3 0

3 years ago

Stephen uses

34kurt

Answer:

13.75 servings or 13 ¾ servings

Step-by-step explanation:

This is the correct question

Stephen uses 2/25 kilograms of tofu in each serving of his famous tofu dish. He has 1 1/10 kilograms of tofu.

To get answer

We will have to divide the number of tofu required with the number of tofu own.

Stephen uses 2/25 kilograms of tofu in each serving and he has 11/10 kilograms tofu.

= (11/10) ÷ (2/25)

11/10 * 25/2 =

275 ÷ 20= 13.75 servings or 13 ¾ servings

Stephen can make 13 ¾ or 13.75 servings

3 0

3 years ago

Read 2 more answers

If n is a positive integer, how many 5-tuples of integers from 1 through n can be formed in which the elements of the 5-tuple ar

erma4kov [3.2K]

Answer:

$\frac{(n+4)*(n+3)*(n+2)*(n+1)*n}{120}$

Step-by-step explanation:

Given

5 tuples implies that:

$n = 5$

$(h,i,j,k,m)$ implies that:

$r = 5$

Required

How many 5-tuples of integers $(h, i, j, k,m)$ are there such that $n\ge h\ge i\ge j\ge k\ge m\ge 1$

From the question, the order of the integers h, i, j, k and m does not matter. This implies that, we make use of combination to solve this problem.

Also considering that repetition is allowed: This implies that, a number can be repeated in more than 1 location

So, there are n + 4 items to make selection from

The selection becomes:

$^{n}C_r => ^{n + 4}C_5$

$^{n + 4}C_5 = \frac{(n+4)!}{(n+4-5)!5!}$

$^{n + 4}C_5 = \frac{(n+4)!}{(n-1)!5!}$

Expand the numerator

$^{n + 4}C_5 = \frac{(n+4)!(n+3)*(n+2)*(n+1)*n*(n-1)!}{(n-1)!5!}$

$^{n + 4}C_5 = \frac{(n+4)*(n+3)*(n+2)*(n+1)*n}{5!}$

$^{n + 4}C_5 = \frac{(n+4)*(n+3)*(n+2)*(n+1)*n}{5*4*3*2*1}$

$^{n + 4}C_5 = \frac{(n+4)*(n+3)*(n+2)*(n+1)*n}{120}$

<u><em>Solved</em></u>

6 0

3 years ago

3.6.PS-8 If the simple interest on $5,000 for 9 years is $2,700, then what is the interest rate? The rate is %. .

Arada [10]

$~~~~~~ \textit{Simple Interest Earned}\\\\I = Prt\qquad\begin{cases}I=\textit{interest earned}\dotfill &\$2700\\P=\textit{original amount deposited}\dotfill & \$5000\\r=rate\to r\%\to \frac{r}{100}\dotfill &0.0r\\t=years\dotfill &9\end{cases}\\\\\\2700=(5000)(0.0r)(9)\implies 2700=45000(0.0r)\implies \cfrac{2700}{45000}=0.0r\\\\\\\cfrac{3}{50} = 0.0r\implies \cfrac{3}{50} = \cfrac{r}{100}\implies \cfrac{3\cdot 100}{50}=r\implies 6 = \stackrel{\%}{r}$

5 0

3 years ago

What should be done to both sides of the equation in order to solve y - 4.5 = 12.2?

maw [93]

Y - 4.5 = 12.2

your solving for y which means your trying to get rid of 4.5 so that y can equal something. The opposite of -4.5 would be +4.5, so you would add 4.5 to both sides to get

y = 16.7

So your answer would be

A. Add 4.5

and also dont' say your stupid because all you need is a little explanation and you'll get it, its very easy you just need to take the time to practice :-)

6 0

3 years ago