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2 years ago

The product of p and 6

Mathematics

1 answer:

inna [77]2 years ago

8 0

Answer:

plz tell the full question well i hope this will help.

Step-by-step explanation:

The key word to note is 'product'. A product is the answer to a multiplication operation.

The product of 3 and 4 means 3×4=12

12 is the product.

In this question we are working with algebra, but it means the same.

The product of pandq means p×q=pq

The product is pq. This is an algebraic term which cannot be simplified further and cannot be evaluated unless you know the values for pandq

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3(4-2) help please thank you

Talja [164]

Answer: 6

Explanation:
3(4-2) = 3(2) = 6

6 0

2 years ago

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For each problem below put the lettered measures in order for greatest to the least

Len [333]

Angles can be used to estimate the side lengths of a shape.

The letters in order of greatest to least are (a), (b) and (c)

The sum of angles at side (a) is:

$a = 30 + 50$

$a = 80$

The sum of angles at side (c) is:

$c = 50 + 90$

$c = 140$

The sum of angles at side (b) is:

$b = 40 + 90$

$b = 130$

So, we have:

$a = 80$

$b = 130$

$c = 140$

The smaller the base angles, the longer the side.

Hence, the letters in order of greatest to least are (a), (b) and (c)

Read more about angles at:

brainly.com/question/11585894

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2 years ago

What percent of 12 is 24

jeka57 [31]

Answer:

200%

Step-by-step explanation:

12 as a whole would be 100%. if you wanted to get to 24 (which would be double) you'd need to double the percentage! hope this helps

6 0

2 years ago

2: Which equation does not belong with the other three? Explain. Thank you so much in advance.

zmey [24]

Answer:

y=x² +6

Step-by-step explanation:

This is the only 2nd-degree equation in the group: y = x² + 6. The others are linear equations.

4 0

3 years ago

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the value of c guaranteed to exist by the Mean Value Theorem for V(x) = x² in the interval [0, 3] is...? a) 1 b) 2 c) 3/2 d) 1/2

lilavasa [31]

Answer: c) $\dfrac{3}{2}$ .

Step-by-step explanation:

Mean value theorem : If f(x) is defined and continuous on the interval [a,b] and differentiable on (a,b), then there is at least one number c in the interval (a,b) (that is a < c < b) such that

$\begin{displaymath}f'(c) = \frac{f(b) - f(a)}{b-a} \cdot\end{displaymath}$

Given function : $f(x) = x^2$

Interval : [0,3]

Then, by the mean value theorem, there is at least one number c in the interval (0,3) such that

$f'(c)=\dfrac{f(3)-f(0)}{3-0}\\\\=\dfrac{3^2-0^2}{3}=\dfrac{9}{3}\\\\=3$

$\Rightarrow\ f'(c)=3\ \ \ ...(i)$

Since $f'(x)=2x$

then, at x=c, $f'(c)=2c\ \ \ ...(ii)$

From (i) and (ii), we have

$2c=3\\\\\Rightarrow\ c=\dfrac{3}{2}$

Hence, the correct option is c) $\dfrac{3}{2}$ .

4 0

2 years ago