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4 years ago

Does 10p/2q = 5p/q I don't think so but it doesn't seem right.

Mathematics

2 answers:

Georgia [21]4 years ago

6 0

They are unequal because it’s a 2/1 ratio 10p can not equal 5p

Vladimir [108]4 years ago

4 0

Yes it does equal since they are equivalent!

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Helpppp and explain :)

ZanzabumX [31]

Answer:

Problem 1: 25

Problem 2: 693

Step-by-step explanation:

Problem 1:

f(x) = -5n + 1

g(x) = -6n + 2

(f + g)(-2) = f(-2) + g(-2)

f(-2) = -5(-2) + 1

f(-2) = 10 + 1

f(-2) = 11

g(-2) = -6(-2) + 2

g(-2) = 12 + 2

g(-2) = 14

(f + g)(-2) = 11 + 14

(f + g)(-2) = 25

Problem 2:

f(x) = 7 + 2x

g(x) = 5x - 2

fg(7) = f(7) × g(7)

f(7) = 7 + 2(7)

f(7) = 7 + 14

f(7) = 21

g(x) = 5(7) - 2

g(x) = 35 - 2

g(x) = 33

Therefore:

fg(7) = 21 × 33

fg(7) = 693

8 0

3 years ago

Find the radius of the circle containing 60° arc of a circle whose length is 14 m.

Sauron [17]

Given:

It is given that

$\begin{gathered} \theta\text{ = 60}^0 \\ Arc\text{ length = 14}\pi \end{gathered}$

Required:

The radius of the circle

Explanation:

The length of an arc is given by the formula,

$\begin{gathered} Arc\text{ length = }\frac{\theta}{360}\text{ }\times\text{ 2}\pi r \\ \end{gathered}$

Substituting the values in the formula,

$\begin{gathered} 14\pi\text{ = }\frac{60}{360}\text{ }\times\text{ 2}\times\pi\times r \\ r\text{ = }\frac{14\times360}{60\times2} \\ r\text{ = }\frac{5040}{120} \\ r\text{ = 42} \end{gathered}$

Answer:

Thus the radius of the circle is 42 m.

5 0

2 years ago

From the first quiz to the second quiz, a score dropped eight points. The score on the second quiz was 39. What was the first qu

frutty [35]

Answer:47

Step-by-step explanation:

Because 39+8=47.

5 0

3 years ago

Read 2 more answers

3y = -x-3 <br> 2y-14 = 4x<br> 4x-3-y = 0<br> x-12 = -3

geniusboy [140]

x=-3 y=0 probably wrong but it’s the root of at least one of these

5 0

4 years ago

26. Define a relation ∼ ∼ on R 2 R2 by stating that ( a , b ) ∼ ( c , d ) (a,b)∼(c,d) if and only if a 2 + b 2 ≤ c 2 + d 2 . a2+

Tresset [83]

Answer:

~ is reflexive.

~ is asymmetric.

~ is transitive.

Step-by-step explanation:

~ is reflexive:

i.e., to prove $\forall (a, b) \in \mathbb{R}^2$ , $(a, b) R(a, b)$ .

That is, every element in the domain is related to itself.

The given relation is $\sim: (a,b) \sim (c, d) \iff a^2 + b^2 \leq c^2 + d^2$

Reflexive:

$(a, b) \sim (a, b)$ since $a^2 + b^2 = a^2 + b^2$

This is true for any pair of numbers in $\mathbb{R}^2$ . So, $\sim$ is reflexive.

Symmetry:

$\sim$ is symmetry iff whenever $(a, b) \sim (c, d)$ then $(c, d) \sim (a, b)$ .

Consider the following counter - example.

Let (a, b) = (2, 3) and (c, d) = (6, 3)

$a^2 + b^2 = 2^2 + 3^2 = 4 + 9 = 13$

$c^2 + d^2 = 6^2 + 3^2 = 36 + 9 = 42$

Hence, $(a, b) \sim (c, d)$ since $a^2 + b^2 \leq c^2 + d^2$

Note that $c^2 + d^2 \nleq a^2 + b^2$

Hence, the given relation is not symmetric.

Transitive:

$\sim$ is transitive iff whenever $(a, b) \sim (c, d) \hspace{2mm} \& \hspace{2mm} (c, d) \sim (e, f)$ then $(a, b) \sim (e, f)$

To prove transitivity let us assume $(a, b) \sim (c, d)$ and $(c, d) \sim (e, f)$ .

We have to show $(a, b) \sim (e, f)$

Since $(a, b) \sim (c, d)$ we have: $a^2 + b^2 \leq c^2 + d^2$

Since $(c, d) \sim (e, f)$ we have: $c^2 + d^2 \leq e^2 + f^2$

Combining both the inequalities we get:

$a^2 + b^2 \leq c^2 + d^2 \leq e^2 + f^2$

Therefore, we get: $a^2 + b^2 \leq e^2 + f^2$

Therefore, $\sim$ is transitive.

Hence, proved.

3 0

3 years ago