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3 years ago

What is the GCF of 54 and 64

Mathematics

1 answer:

Klio2033 [76]3 years ago

7 0

Answer:

1,2,3,6,9,18,27,54 is 54, 1,2,4,8,16,32,64 is 64

Step-by-step explanation:

learn di divide geez

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H is the circumcenter, or point of concurrency, of the perpendicular bisectors of ΔACE.

Radda [10]

Answer:

∠HDC ≅ ∠HDE, AB ≅ BC, HC ≅ HE

Step-by-step explanation:

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2 years ago

A coin will be flipped three times. The result of each flip is either heads (H) or tails (T). The outcome HTH represents the fir

Yuki888 [10]

The answer is 4/7
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2 years ago

Guys, Can you please help me with these questions

MakcuM [25]

Answer:

a) $w^{13} x^{5} y^{6}$

b) $\frac{x}{3y^{6} }$

Step-by-step explanation:

a) $(w^{2} xy^{3} )^{2}(w^{3}x )^{3}$

1. Distribute the second power (2) outside the first pair of parenthesis:

$(w^{2(2)} x^{2} y^{3(2)} )$

= $w^{4} x^{2} y^{6} (w^{3}x )^{3}$

2. Distribute the third power (3) outside the second pair of parenthesis:

$(w^{3(3)} x^{3} )$

= $w^{4} x^{2} y^{6} w^{9} x^{3}$

3. Combine like terms:

$w^{13} x^{5} y^{6}$

--------------------------------------------

b) $\frac{2x^{2} y^{5} }{6xy^{11} }$

1. Factor the number 6 (= 2 · 3):

$\frac{2x^{2} y^{5} }{2(3)xy^{11} }$

2. Cancel the common factor (2):

$\frac{x^{2} y^{5} }{3xy^{11} }$

3. Cancel out $xy^{5}$ in the numerator an denominator:

$\frac{x}{3y^{6} }$

hope this helps!

4 0

1 year ago

6. If the net investment function is given by

Pachacha [2.7K]

The capital formation of the investment function over a given period is the

accumulated capital for the period.

(a) The capital formation from the end of the second year to the end of the fifth year is approximately 298.87.

(b) The number of years before the capital stock exceeds $100,000 is approximately 46.15 years.

Reasons:

(a) The given investment function is presented as follows;

$I(t) = 100 \cdot e^{0.1 \cdot t}$

(a) The capital formation is given as follows;

$\displaystyle Capital = \int\limits {100 \cdot e^{0.1 \cdot t}} \, dt =1000 \cdot e^{0.1 \cdot t}} + C$

From the end of the second year to the end of the fifth year, we have;

The end of the second year can be taken as the beginning of the third year.

Therefore, for the three years; Year 3, year 4, and year 5, we have;

$\displaystyle Capital = \int\limits^5_3 {100 \cdot e^{0.1 \cdot t}} \, dt \approx 298.87$

The capital formation from the end of the second year to the end of the fifth year, C ≈ 298.87

(b) When the capital stock exceeds $100,000, we have;

$\displaystyle \mathbf{\left[1000 \cdot e^{0.1 \cdot t}} + C \right]^t_0} = 100,000$

Which gives;

$\displaystyle 1000 \cdot e^{0.1 \cdot t}} - 1000 = 100,000$

$\displaystyle \mathbf{1000 \cdot e^{0.1 \cdot t}}} = 100,000 + 1000 = 101,000$

$\displaystyle e^{0.1 \cdot t}} = 101$

$\displaystyle t = \frac{ln(101)}{0.1} \approx 46.15$

The number of years before the capital stock exceeds $100,000 ≈ 46.15 years.

Learn more investment function here:

brainly.com/question/25300925

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1 year ago

Hey everybody I have a quick math question but idk what the answer is! It would mean a lot if someone helped me. PS the pic is t

mote1985 [20]

Answer:

it would be 2.33 I divided it I hope I'm right :)

8 0

2 years ago