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2 years ago

Whats 2 plus 2?////////////////////////////////////////////////

Mathematics

2 answers:

sergiy2304 [10]2 years ago

6 0

Answer:

Step-by-step explanation:

its like you have 2 fishes and you get two more

Mazyrski [523]2 years ago

5 0

Answer:

Step-by-step explanation:

When you have two question marks and add two more you have four

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1234

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Suppose you are given that p = 2q+1 and that p = 8.

iVinArrow [24]

Answer:

We conclude that the following statements can be proved such as:

$2q+1 = 8$

Step-by-step explanation:

Given the expression

$p = 2q+1$

Given

$p = 8$

IF WE SUBSTITUTE p=2q+1 in p=8

Now, substituting p = 2q+1 in the expression p=8

$p = 8$

$2q+1 = 8$ ∵ p = 2q+1

Therefore, we conclude that the following statements can be proved such as:

$2q+1 = 8$

3 0

3 years ago

Harmonic mean between x and y is

Komok [63]

Harmonic mean between two positive numbers x and y, The simplified formula is $\boxed { \mathfrak{h = \frac{2 \times x \times y}{(x + y)} }}$ I.e. you need to double the product of x and y then divide it by the sum of x and y...~

4 0

1 year ago

Read 2 more answers

Solve the following question

White raven [17]

Answer:

g) $u^{4}\cdot v^{-1}\cdot z^{3}$ , h) $\frac{(x+4)\cdot (x+2)}{3\cdot (x-5)}$

Step-by-step explanation:

We proceed to solve each equation by algebraic means:

g) $\frac{u^{5}\cdot v}{z}\div \frac{u\cdot v^{2}}{z^{4}}$

1) $\frac{u^{5}\cdot v}{z}\div \frac{u\cdot v^{2}}{z^{4}}$ Given

2) $\frac{\frac{u^{5}\cdot v}{z} }{\frac{u\cdot v^{2}}{z^{4}} }$ Definition of division

3) $\frac{u^{5}\cdot v\cdot z^{4}}{u\cdot v^{2}\cdot z}$ $\frac{\frac{a}{b} }{\frac{c}{d} } = \frac{a\cdot d}{b\cdot c}$

4) $\left(\frac{u^{5}}{u} \right)\cdot \left(\frac{v}{v^{2}} \right)\cdot \left(\frac{z^{4}}{z} \right)$ Associative property

5) $u^{4}\cdot v^{-1}\cdot z^{3}$ $\frac{a^{m}}{a^{n}} = a^{m-n}$ /Result

h) $\frac{x^{2}-16}{x^{2}-10\cdot x + 25} \div \frac{3\cdot x - 12}{x^{2}-3\cdot x -10}$

1) $\frac{x^{2}-16}{x^{2}-10\cdot x + 25} \div \frac{3\cdot x - 12}{x^{2}-3\cdot x -10}$ Given

2) $\frac{\frac{x^{2}-16}{x^{2}-10\cdot x+25} }{\frac{3\cdot x - 12}{x^{2}-3\cdot x - 10} }$ Definition of division

3) $\frac{(x^{2}-16)\cdot (x^{2}-3\cdot x -10)}{(x^{2}-10\cdot x + 25)\cdot (3\cdot x - 12)}$ $\frac{\frac{a}{b} }{\frac{c}{d} } = \frac{a\cdot d}{b\cdot c}$

4) $\frac{(x+4)\cdot (x-4)\cdot (x-5)\cdot (x+2)}{3\cdot (x-5)^{2}\cdot (x-4) }$ Factorization/Distributive property

5) $\left(\frac{1}{3} \right)\cdot (x+4)\cdot (x+2)\cdot \left(\frac{x-4}{x-4} \right)\cdot \left[\frac{x-5}{(x-5)^{2}} \right]$ Modulative and commutative properties/Associative property

6) $\frac{(x+4)\cdot (x+2)}{3\cdot (x-5)}$ $\frac{a^{m}}{a^{n}} = a^{m-n}$ / $\frac{a}{b}\times \frac{c}{d} = \frac{a\cdot c}{b\cdot d}$ /Definition of division/Result

3 0

2 years ago

The graph of hix) = |x-10| +6 is shown. On which<br> interval is this graph increasing?

goldfiish [28.3K]

Answer:

3rd answer

Step-by-step explanation:

its the 3rd answer trust me on it

8 0

2 years ago

Which of the following functions f : {0, 1, 2, 3} ! {0, 1, . . . , 7} are one-to-one?

allsm [11]

Answer:

1. No.

$f(0)=0\\1^2=1; \text{then } f(1)=1\\2^2=4\equiv 4 \text{ mod 8}; \text{then } f(2)=4\\3^2=9\equiv 1 \text{mod 8 }; \text{then } f(3)=1$

Since f(1)=f(3) and $1\neq 3$ then f isn't one-to-one.

2. No

$f(0)=0\\1^3=1\equiv 1\text{ mod 8}; \text{then } f(1)=1\\2^3=8\equiv 0\text{ mod 8}; \text{then } f(2)=0\\3^3=27\equiv 3 \text{ mod 8}; \text{then } f(3)=3$

Since f(0)=f(2) and $0\neq 2$ then f isn't one-to-one.

3. No

$0^3-8=-8\equiv 0\text{ mod 8}; \text{then } f(0)=0\\1^3-8=-7\equiv 1\text{ mod 8}; \text{then } f(1)=1\\2^3-8=0\equiv 0 \text{ mod 8}; \text{then } f(2)=0\\3^3-8=27-8=19\equiv 3 \text{ mod 8}; \text{then } f(3)=3\\$

Since f(0)=f(2) and $0\neq 2$ then f isn't one-to-one.

4. Yes

$0^3+2*0=0; \text{then } f(0)=0\\1^3+2*1=3\equiv 3\text{ mod 8}; \text{then } f(1)=3\\2^3+2*2=8+4=12\equiv 4 \text{ mod 8}; \text{then } f(2)=4\\3^3+2*3=27+6=33\equiv 1\text{ mod 8}; \text{then } f(3)=1$

Since $f(0)\neq f(1)\neq f(2) \neq f(3)$ , then f is one-to-one

5. Since f(1)=f(3) and $1\neq 3$ then, f isn't one-to-one

3 0

3 years ago