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jek_recluse [69]

3 years ago

14

Mrs.smith has 12 times as many markers as colored pencils.the total number of markers and colored pencils is 78.how many markers

does mrs.smith have?

1 answer:

cupoosta [38]3 years ago

6 0

Divide 78 and 12. You get 6

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Solve -4r + 18= 6 I need this equation

schepotkina [342]

The answer I got is R=3

3 0

3 years ago

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Factorising quadratics <br><br> All questions please to be completed. I’m stuck.

Alex777 [14]

Answer:

1. (x +1)(x - 1)

2. (x+5)(x-5)

3. (x+12)(x-12)

4. (x+14)(x-14)

5. (x+5)(x-7)

Step-by-step explanation:

The first 4 expressions can be factored with the difference of squares type of factoring.

For #5, the odd one out is (x+5)(x-7) because it isn't a factor of any of the standard form equations, while the others are

5 0

4 years ago

2+5 is going to be a long one ne

andrew-mc [135]

Answer: 7

Step-by-step explanation: Imagine having 5 chocolate bars and then adding 2, or you could try counting with your fingers. Another method is writing a number line and going plus 2 from five so you can visualize it more.

3 0

3 years ago

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Find the sum of the measures of the exterior angles on a 52-gon

juin [17]

Answer:

9000

Step-by-step explanation:

the equation to figure this out would be 180(n-2). N would equal the number of sides so you would plug it in 180(52-2). Then you end up with 180*50 which equals 9000

6 0

3 years ago

Which expression represents the number 2i4−5i3+3i2+−81‾‾‾‾√ rewritten in a+bi form?

vichka [17]

Answer:

The expression $-1+14i$ represents the number $2i^4-5i^3+3i^2+\sqrt{-81}$ rewritten in a+bi form.

Step-by-step explanation:

The value of $i$ is $i=\sqrt{-1}[tex] or [tex]i^{2}=-1[\tex].Now [tex]i^{4}$ in term of $i^{2}[\tex] can be written as, [tex]i^{4}=i^{2}\times i^{2}$

Substituting the value,

$i^{4}=\left(-1\right)\times \left(-1\right)$

Product of two negative numbers is always positive.

$\therefore i^{4}=1$

Now $i^{3}$ in term of $i^{2}[\tex] can be written as, [tex]i^{3}=i^{2}\times i$

Substituting the value,

$i^{3}=\left(-1\right)\times i$

Product of one negative and one positive numbers is always negative.

$\therefore i^{3}=-i$

Now $\sqrt{-81}$ can be written as follows,

$\sqrt{-81}=\sqrt{\left(81\right)\times\left(-1\right)}$

Applying radical multiplication rule,

$\sqrt{ab}={\sqrt{a}}\sqrt{b}$

$\sqrt{\left(81\right)\times\left(-1\right)}={\sqrt{81}}\sqrt{-1}$

Now, $\sqrt{\left(81\right)=9$ and $\sqrt{-1}}=i$

$\therefore \sqrt{\left(81\right)\times\left(-1\right)}=9i$

Now substituting the above values in given expression,

$2i^4-5i^3+3i^2+\sqrt{-81}=2\left(1\right)-5\left(-i\right)+3\left(-1\right)+9i$

Simplifying,

$2+5i-3+9i$

Collecting similar terms,

$2-3+5i+9i$

Combining similar terms,

$-1+14i$

The above expression is in the form of a+bi which is the required expression.

Hence, option number 4 is correct.

5 0

3 years ago