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Nataly_w [17]
3 years ago
5

If a gold bar has a length of 17.78 cm, a width of 9.208 cm, a height of 4.445 cm, and the density of gold is 19.3 g/cm^3, what

is the
mass of the gold bar. Answer with the correct number of significant figures. Use the rounding rules.

A. 1.40x10^4 grams
B. 14 040 grams
C. 14 045.142 grams
D. 14 045 grams
Mathematics
1 answer:
kozerog [31]3 years ago
4 0

Answer:

17.78×9.208×4.445=727.7275768

D=mass÷volume

mass=density×volume

mass=19.3g/cm3×727,7275768=14045,14223224=14045grams

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Answer:

MULTIPLY!!!!!11!!!!!!!

Step-by-step explanation:

3x2+1/x2-26

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Answer:

8.1

Step-by-step explanation:

Since we dealing with right triangles, we are going to use pythagoern theorem.

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Let plug in the numbers

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5 0
3 years ago
Applying properties of Exponents in exercise,use the properties of exponents to simplify the expression.See example 1.
Paladinen [302]

Answer:  a) 1024, b) 1, c) 64 and d) 0.125.

Step-by-step explanation:

Since we have given that

(a) (4^3)(4^2)

As we know that

a^m\times a^n=a^{m+n}

So, it becomes,

4^3\times 4^2=4^{3+2}=4^5=1024

(b) (\dfrac{1}{4})^2(4^2)

As we know that

a^{-m}=\dfrac{1}{a^m}

(\dfrac{1}{4})^2(4^2)=4^-2\times 4^2=4^{-2+2}=4^0=1

c) (4^6)1/2

As we know that

(4^6)^{\frac{1}{2}}\\\\=4^{\frac{6}{2}}\\\\=4^3\\\\=64

(d) [(8^{-1})(8^{\frac{2}{3}})]^3

As we know that

[(8^{-1})(8^{\frac{2}{3}})]^3\\\\=[8^{-1+\frac{2}{3}}]^3\\\\=[8^{\frac{-1}{3}}]^3\\\\=8^{-1}\\\\=\dfrac{1}{8}=0.125

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5 0
2 years ago
Does anyone know the question?
yan [13]

It definitely is an isosceles triangle, but I'm not certain as to what an acute isosceles triangle is.

7 0
3 years ago
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Leviafan [203]

Answer:

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Step-by-step explanation:

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-30x - 12y + 36x - 12y

Then group together the same variables:

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arguably, you could factor out a 6 and get:

6(x-4y)

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