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2 years ago

The points (9, 9) and (5, 1) fall on a particular line. What is its equation in slope-intercept form?

Mathematics

1 answer:

Ilia_Sergeevich [38]2 years ago

3 0

Your equation would be y= 2x-9

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40 green cars 20 silver if same # of green stayed the same how many more silver would need to be added so the ratio of green to

slega [8]

So, there will be 40 green cars. if we want the ratio to be 1:3, we will need there to b three times as many silver cars as green cars: 40*3=120.

we already have 20 silver, so we need to add 120-20, that is 100 silver cars!

4 0

3 years ago

−4p + 9 = −5 <br> what is the answer

12345 [234]

Answer is P=3.5 or p=7/2

They are both correct

7 0

2 years ago

Read 2 more answers

Derive this with respect to x<br><img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B3%7D%7B%281%20%2B%20%20secx%29%20%7D%20" id="Tex

Elenna [48]

$~~~~\dfrac{d}{dx} \left(\dfrac{3}{1 + \sec x} \right)\\\\\\=3 \dfrac{d}{dx} \left( \dfrac 1{ 1+ \sec x} \right)\\\\\\=3 \dfrac{d}{dx} (1+ \sec x)^{-1}\\\\\\=3 (-1) (1 + \sec x )^{-1 -1} \dfrac{d}{dx}( 1 + \sec x)\\\\\\=-3(1 + \sec x)^{-2} ( 0 + \sec x \tan x)\\\\\\=-\dfrac{3\sec x \tan x}{(1 + \sec x)^2}$

8 0

1 year ago

Need help with question 2

Alex777 [14]

The answer is y= -x-1

8 0

3 years ago

Before a bag of flour can be sold, it must be 40 kilograms but can be within 0.5

Serjik [45]

Answer:

The minimum acceptable weight of the flour bag is 39.5 kg,

and its maximum acceptable weight is 40.5 kg

Step-by-step explanation:

Notice that the selling weight of the bag to be sold (x) needs to differ from 40 kg at most 0.5 kg. Therefore, we can write the following inequality:

$|x-40|\leq 0.5$

The inequality can be solve once we remove the absolute value symbol, and solve for the unknown "x". Recall that in order to solve it, we need to consider the two possible cases:

a) that the expression within the absolute value symbol is larger than or equal to zero, and b) that the expression is less than zero.

a) If $x-40\geq 0$ then its absolute value is equal to itself (x-40), and when we remove the absolute value symbols, we get:

$x-40\leq 0.5\\x\leq 0.5+40\\x\leq 40.5\,\,kg$

which means that the weigh "x" of the flour bag should be smaller or equal than 40.5 kg

b) If $x-40$ , then the absolute value of this is its opposite : -x+40, and the inequality becomes:

$-x+40\leq 0.5\\40\leq 0.5+x\\40-0.5\leq x\\39.5\leq x$

which means that the weight of the flour bag must be larger than or equal to 39.5 kg

So, the minimum acceptable weight of the flour bag is 39.5 kg, and the maximum acceptable weight is 40.5 kg

6 0

3 years ago