The swimming pool is open when the high temperature is higher than 20°C. Lainey tried to swim on

Mathematics

1 answer:

lapo4ka [179]1 year ago

5 0

Answer: $16.40

Step-by-step explanation:

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PLS HELP QUICKKKKK! Tap the photo to see the full problem. Answer all 3 questions please and I'll give brainliest if it is right

Ghella [55]

3x + 8 = 3x + []

Subtract 3x from both sides.

[] = 8

If you want no values of x, it can be any positive value other than 8.
If you want all values of x, the value must be 8.
If you want only one value of x, it is impossible because both equations have the same slope.

4 0

2 years ago

Hey! Can someone check my answers?

Andrew [12]

The greatest common factor is the biggest number taken from the values.

Q1. The answer is A. 5y^6

$20 y^{9} +5 y^{6}= 4*5 y^{9}+5 y^{6}$
Since $x^{a}* x^{b} =x^{a+b}$ , then $x^{9}= x^{3}* x^{6}$

Back to our expression:
$4*5 y^{9}+5 y^{6}=4*5 y^{3}*y^{6}+5 y^{6}=4 y^{3}*5y ^{6}+ 5y ^{6}*1=5 y^{6} (4y ^{3} +1)$
The greatest common factor is thus $5 y^{6}$

Q2. The answer is D. 12xy^2

$12x y^{5}+60 x^{4} y^{2} -24 x^{3} y^{3}=12x y^{5}+5*12 x^{4} y^{2} -2*12 x^{3} y^{3}$
We will use the rule $x^{a}* x^{b} =x^{a+b}$ to factorise the exponents:
$12x y^{5}+5*12 x^{4} y^{2} -2*12 x^{3} y^{3}= \\ =12x*y^{2}*y^{3}+5*12*x* x^{3} *y^{2}-2*12x* x^{2} *y*y^{2}= \\ =12xy^{2}*y^{3}+12xy^{2}*5 x^{3} -12xy^{2}*2 x^{2} y= \\ =12xy^{2}(y^{3}+5 x^{3}-2 x^{2} y)$
The greatest common factor is thus $12xy^{2}$

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2 years ago

What does W = too please help

dedylja [7]

$-7w+3(w+6)=30\\\\-7w+3w+18=30\\\\-4w+18=30\ \ \ \ \ |subrtact\ 18\ from\ both\ sides\\\\-4w=12\ \ \ \ \ \ |divide\ both\ sides\ by\ (-4)\\\\\boxed{w=-3}$

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2 years ago

Read 2 more answers

The limit as h approaches 0 of (e^(2+h)-e^2)/h is ?

Whitepunk [10]

If you plug in 0, you get the indeterminate form 0/0. You can, therefore, apply L'Hopital's Rule to get the limit as h approaches 0 of e^(2+h),
which is just e^2.
[e^(2+h) − e^2]/h = [e^2(e^h−1)]/h

so in the limit, as h goes to 0, you'll notice that the numerator and denominator each go to zero (e^h goes to 1, and so e^h-1 goes to zero). This means the form is 'indeterminate' (here, 0/0), so we may use L'Hoptial's rule:

=e^2

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2 years ago

Plz prove the given equation!!!

elixir [45]

Answer:

Please see the attached picture for the full solution.

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2 years ago