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4 years ago

11

Going to highschool next year! Can anyone help me understand Algebra? Math isn't a good subject of mine. My class sort of tried

a bit of Algebra and I just couldn't seem to get it.

1 answer:

AlekseyPX4 years ago

8 0

WHat would you like to know? I took algebra 1 in my 7th grade year. Algebra 2 my 8th, geometry 9th, and rn im in Pre-calc. I bet I can help!

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On the day after my birthday this year, I can truthfully say: "The day after tomoroow is Monday". ON which day is my birthday?

borishaifa [10]

Answer:

Saturday

Step-by-step explanation:

2 days before (starting from Monday) is Saturday

8 0

2 years ago

Please help!!As soon as possible!!

pychu [463]

Answer: Their equations have different y-intercept but the same slope

Step-by-step explanation:

Since, the slope of the line passes through the points (2002,p) and (2011,q) is,

$m_1 = \frac{q-p}{2011-2002} = \frac{q-p}{9}$

Similarly, the slope of the line asses through the points (2,p) and (11,q) is,

$m_2 = \frac{q-p}{11-2} = \frac{q-p}{9}$

Since, $m_1 = m_2$

Hence, both line have the same slope.

Now, the equation of the line one having slope $m_1$ and passes through the point (2002,p) is,

$y - p = \frac{q-p}{9}(x-2002)$

Put x = 0 in the above equation,

We get, $y = \frac{2000(p-q)}{9}+p$

The y-intercept of the line one is $(0, \frac{2000(p-q)}{9}+p)$

Also, the equation of second line having slope $m_2$ and passes through the point (2,p)

$y-p=\frac{q-p}{9}(x-2)$

Put x = 0 in the above equation,

We get, $y = \frac{2(p-q)}{9}+p$

The y-intercept of the line one is $(0, \frac{2(p-q)}{9}+p)$

Thus, both line have the different y-intercepts.

⇒ Third option is correct.

6 0

3 years ago

Solve The Following <br> ( Solving equations)

kondaur [170]

1. 15/2 | 2. 1 … if you need examples just reply here

8 0

2 years ago

Read 2 more answers

player A led a baseball league in runs batted in for the 2006 regular season. player B. who came in second player to A, had 13 f

Alla [95]

A= player A runs batted in
b= player B runs batted in= a-13

250= a + b
substitute b= a-13 for b

250= a + (a-13)
combine like terms
250= 2a -13
add 13 to both sides
263= 2a
divide both sides by 2
131.5= a

Substitute a= 131.5 into equation
250= a + b
250= 131.5 + b
subtract 131.5 from both sides
118.5= b

Hope this helps! :)

7 0

3 years ago

A jeweler wants to make 14 grams of an alloy that is precisely 75% gold.. The jeweler has alloys that are 25% gold, 50% gold, &a

Goryan [66]

Given that the jeweler has alloys that are 25% gold, 50% gold, and 82% gold.

As he wants to make 14 grams of an alloy by adding two different alloys that is precisely 75% gold, so one alloy must have a percentage of gold more than 75%.

One alloy is 82% gold and, the second can be chosen between 25% gold, 50% gold, so there are two cases.

Case 1: 82% gold + 50% gold

Let x grams of 82% gold and y grams of 50% gold added to make x+y=14 grams of 75% gold, so

75% of 14 = 82% of x + 50% of y

$\Rightarrow 75/100 \times 14 = 82/100 \times x + 50/100 \times y \\\\$

$\Rightarrow 75/100 \times 14 = 82/100 \times x + 50/100 \times (14-x)$ [as x+y=14]

$\Rightarrow 75 \times 14 = 82 \times x + 50 \times (14-x) \\\\\Rightarrow 75 \times 14 = 82 \times x + 50 \times14-50\times x \\\\\Rightarrow 75 \times 14 = 32 \times x + 50 \times14 \\\\\Rightarrow 32 \times x =75 \times 14 - 50 \times14 \\\\$

$\Rightarrow x =(25 \times 14)/32=10.9375$ grams

and $y = 14-x= 14-10.9375=3.0625$ grams.

Hence, 10.9375 grams of 82% gold and 3.0625 grams of 50% gold added to make 14 grams of 75% gold.

Case 2: 82% gold + 25% gold

Let x grams of 82% gold and y grams of 25% gold added to make x+y=14 grams of 75% gold, so

75% of 14 = 82% of x + 25% of y

$\Rightarrow 75/100 \times 14 = 82/100 \times x + 25/100 \times y \\\\\Rightarrow 75/100 \times 14 = 82/100 \times x + 25/100 \times (14-x) \\\\ \Rightarrow 75 \times 14 = 82 \times x + 25 \times (14-x) \\\\\Rightarrow 75 \times 14 = 82 \times x + 25 \times14-25\times x \\\\\Rightarrow 75 \times 14 = 57 \times x + 25 \times14 \\\\\Rightarrow 57 \times x =75 \times 14 - 25 \times14 \\\\$

$\Rightarrow x =(50 \times 14)/57=12.28$ grams

and $y = 14-x= 14-12.28=1.72$ grams.

Hence, 12.28 grams of 82% gold and 1.72 grams of 50% gold added to make 14 grams of 75% gold.

3 0

3 years ago