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3 years ago

The _____ were victims of attempted genocide ?

Mathematics

1 answer:

Marta_Voda [28]3 years ago

7 0

B Jews and Armenians

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Find the area of the shape shown below.

morpeh [17]

Answer:

Step-by-step explanation:

$4\cdot \frac{12 + 4}{2} = 32$

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3 years ago

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The equation f = v + at represents the final velocity of an object, f, with an initial velocity, v, and an acceleration rate, a,

Elenna [48]

Answer:

(f-v)/a = t

Step-by-step explanation:

f = v + at

Subtract v from each side

f-v = v-v + at

f-v = at

Divide each side by a

(f-v)/a = at/a

(f-v)/a = t

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3 years ago

What times what equals 59

sweet-ann [11.9K]

59 is a "prime number". It has no factors except ' 1 ' and itself.

So the only possibility is 1 x 59 .

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3 years ago

Antwan determines the distance between the points –7 and 2 on a numberline. Maggie determines the difference between the numbers

Orlov [11]

Answer:

Their solutions are related because they both are |9| (the absolute value of 9)

Step-by-step explanation:

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3 years ago

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Can someone please help me with my maths question

DIA [1.3K]

Answer:

$a. \ \dfrac{625 \cdot m}{27 \cdot n^{11}}$

$b. \ \dfrac{x^{3 \cdot m - 2}}{y^{ 3 + n}}$

Step-by-step explanation:

The question relates with rules of indices

(a) The give expression is presented as follows;

$\dfrac{m^3 \times \left (n^{-2} \right )^4 \times (5 \cdot m)^4}{\left (3 \cdot m^2 \cdot n \right )^3}$

By expanding the expression, we get;

$\dfrac{m^3 \times n^{-8} \times 5^4 \times m^4}{\left 3^3 \times m^6 \times n^3}$

Collecting like terms gives;

$\dfrac{m^{(3 + 4 - 6)} \times 5^4}{ 3^3 \times n^{3 + 8}} = \dfrac{625 \cdot m}{27 \cdot n^{11}}$

$\dfrac{m^3 \times \left (n^{-2} \right )^4 \times (5 \cdot m)^4}{\left (3 \cdot m^2 \cdot n \right )^3}= \dfrac{625 \cdot m}{27 \cdot n^{11}}$

(b) The given expression is presented as follows;

$x^{3 \cdot m + 2} \times \left (y^{n - 1} \right )^3 \div (x \cdot y^n)^4$

Therefore, we get;

$x^{3 \cdot m + 2} \times \left (y^{n - 1} \right )^3 \times x^{-4} \times y^{-4 \cdot n}$

Collecting like terms gives;

$x^{3 \cdot m + 2 - 4} \times \left (y^{3 \cdot n - 3 -4 \cdot n}} \right ) = x^{3 \cdot m - 2} \times \left (y^{ - 3 -n}} \right ) = x^{3 \cdot m - 2} \div \left (y^{ 3 + n}} \right )$

$x^{3 \cdot m - 2} \div \left (y^{ 3 + n}} \right ) = \dfrac{x^{3 \cdot m - 2}}{y^{ 3 + n}}$

$x^{3 \cdot m + 2} \times \left (y^{n - 1} \right )^3 \times x^{-4} \times y^{-4 \cdot n} =\dfrac{x^{3 \cdot m - 2}}{y^{ 3 + n}}$

4 0

3 years ago