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3 years ago

This is for Algebra 2 I’m stuck I don’t know if I’m right please help

Mathematics

1 answer:

OLEGan [10]3 years ago

3 0

Answer:

y=a|x-h|+k (absolute value function)

a= 3/5 (a easier way to know is it is somehow basically the slope of the line but it must be positive.)

h= -3 (you are correct)

k= -1 (you are correct)

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Could someone please list all of the basic derivative rules?

olchik [2.2K]

Answer:

The sum rule is f' + g'

The difference rule is f' − g'

The product rule is f g' + f' g

The quotient rule is (f' g − g' f )/g2

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3 years ago

Please help! i promise ill give brainliest, tysm

Mars2501 [29]

Answer:

f(x) = x(x-1)

Step-by-step explanation:

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3 years ago

Read 2 more answers

Please answer this please

motikmotik

Answer:

Step-by-step explanation:

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2 years ago

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Student had 500 milliliters of water in a water bottle. She drank 25% of the water before soccer practice. After practice, she d

sesenic [268]

250 milliters of water remains

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3 years ago

three bags of sweets weigh 27/4 kg. two of them have the same weight and the third bag is heavier than each of the bags of equal

Nana76 [90]

I'm here buddy,

so, let's take the value of the two bags with equal weight as x.

= x + x + (x + $\frac{6}{5}$ ) = $\frac{27}{4}$

= 3x + $\frac{6}{5}$ = $\frac{27}{4}$

= 3x = $\frac{27}{4}$ - $\frac{6}{5}$

( let's take the LCM of 4 and 5 = 20

= 3x = $\frac{135}{20}$ - $\frac{24}{20}$

= 3x = $\frac{111}{20}$

= x = $\frac{111}{20}$ ÷ $\frac{3}{1}$ = $\frac{111}{20}$ × $\frac{1}{3}$ = $\frac{37}{20}$

So, the weight of the equal bags are $\frac{37}{20}$ and the weight of the third bag ( heavy one ) is $\frac{37}{20}$ + $\frac{6}{5}$ = $\frac{37}{20}$ + $\frac{24}{20}$ = $\frac{61}{20}$

1st bag = $\frac{37}{20}$ kg

2nd bag = $\frac{37}{20}$ kg

3rd bag = $\frac{61}{20}$ kg

Hope it helps...

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3 years ago