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Studentka2010 [4]
3 years ago
15

Can you help me this has to be done

Mathematics
1 answer:
LenaWriter [7]3 years ago
7 0
144 cm3 is what i came up with!
You might be interested in
Q10 Q2.) Find the standard form of the equation of the parabola satisfying the given conditions.
Kipish [7]
Hello!!

Equation forms:

y =  \frac{x^2}{4} +  \frac{x}{2} +  \frac{13}{4}

Therefore, x^2 - 2x + 4y - 13 = 0

And, this is how your graph would look

Good luck :)

8 0
3 years ago
Find a range of th fucunction f(x)=4x-1 for the domain{-1,0,1,2,3}
vlada-n [284]
F(x) = 4x - 1

All you have to do is substitute the values in the domain for x in the function.

f(-1) = 4(-1) - 1
f(-1) = -4 - 1
f(-1) = -5

f(0) = 4(0) - 1
f(0) = -1

f(1) = 4(1) - 1
f(1) = 4 - 1
f(1) = 3

And so on.
7 0
3 years ago
How many liters are in 8 quarts?
sleet_krkn [62]

Answer:

7.57082 liters

Step-by-step explanation:

7 0
3 years ago
Select the correct answer.
katen-ka-za [31]

Answer:

3x - 4/2

Step-by-step explanation:

2 result(s) for "f(x) = 4x³ - 10 g(x)

6 0
2 years ago
Please help i dont understand what its asking
Arisa [49]

Multiply both the numerator and denominator by 2+\sqrt3, which is called the "conjugate" of 2-\sqrt3:

\dfrac{5+\sqrt3}{2-\sqrt3}\cdot\dfrac{2+\sqrt3}{2+\sqrt3}

Why do we pick this number? Recall the difference of squares factorization:

a^2-b^2=(a-b)(a+b)

Now replace a=2 and b=\sqrt3. Then

(2-\sqrt3)(2+\sqrt3)=2^2-(\sqrt3)^2=4-3=1

So in your fraction, you end up with

\dfrac{5+\sqrt3}{2-\sqrt3}=\dfrac{(5+\sqrt3)(2+\sqrt3)}1=(5+\sqrt3)(2+\sqrt3)

Finally, just expand the product.

(5+\sqrt3)(2+\sqrt3)=10+7\sqrt3+(\sqrt3)^2=\boxed{13+7\sqrt3}

7 0
3 years ago
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