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luda_lava [24]
3 years ago
9

What is 5/12 × 6/35 please evaluate!

Mathematics
2 answers:
tensa zangetsu [6.8K]3 years ago
8 0

Answer:

\dfrac{1}{14}

Step-by-step explanation:

\dfrac{5}{12} \cdot  \dfrac{6}{35}

Once

\boxed{ \dfrac{a}{b} \cdot  \dfrac{c}{d} = \dfrac{ac}{bd} }

\dfrac{5}{12} \cdot  \dfrac{6}{35} = \dfrac{5\cdot  6}{12  \cdot  35} =  \dfrac{30}{420} = \dfrac{1}{14}

Andreas93 [3]3 years ago
5 0

Answer:

Step-by-step explanation:

You can do this without cancelation to start with.

5*6 / 12 * 35 = 30 / 420 = 1/14  

However you can cancel twice.

first cancel 5 into 35 that gives 7 where the 35 was

6 / (7 * 12)

Now you can cancel 6 into 12

1 / (7 *2)

1/14

Same answer.

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Evaluate each using the values given-6+4-y-z; use y= -4, and z = -1
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Answer:

Your answer is 15

Step-by-step explanation:

1) 6 + 4 = 10

2) -4 + -1 = -5

3) now add

4) 10 + -5 + 15

5) it's 15 because the bigger is positive.

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3 years ago
Sam ordered 2 tacos and 3 enchiladas for lunch at the restaurant. His bill came to $7.80. If enchiladas were $2 each, write and
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$0.90 for the tacos. 3 enchiladas multiples by $2.00 is $6.00. then $1.80 is left so then you divide $1.80 divided by 2 and you get 0.9 and add the $ and then the 0 at the end
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3 years ago
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4 years ago
A delivery truck drove 203 miles in 3.5 hours. After unloading and loading new cargo, it traveled on another 243 miles for 4.5 h
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4 years ago
What are the zeros of the quadratic function f(x) = 6x2 + 12x – 7? x = –1 – StartRoot StartFraction 13 Over 6 EndFraction EndRoo
ivanzaharov [21]

Answer:

x=-1+\sqrt{\frac{13}{6}}\\x=-1-\sqrt{\frac{13}{6}}

Step-by-step explanation:

In this problem, the function is

f(x)=6x^2+12x-7

The zeros of the function are the values of x for which the function is zero, so:

6x^2+12x-7=0

We start by dividing each term by 6:

x^2+2x-\frac{7}{6}=0

Then we add +1 and -1 to the equation:

x^2+2x+1-1-\frac{7}{6}=0

The first 3 terms x^2+2x+1 are the square of a binomial, (x+1), so this becomes

(x+1)^2-1-\frac{7}{6}=0

Which is equivalent to

(x+1)^2-\frac{13}{6}=0 (1)

Now we can use the following rule:

x^2-a^2 = (x+a)(x-a)

To rewrite (1) as:

(x+1)^2-\frac{13}{6}=\\=(x+1+\sqrt{\frac{13}{6}})(x+1-\sqrt{\frac{13}{6}})=0

This expression is equal to zero if either one of the two terms in the brackets is equal to zero. Therefore:

1)

x+1+\sqrt{\frac{13}{6}}=0 \rightarrow x=-1-\sqrt{\frac{13}{6}}

2)

x+1-\sqrt{\frac{13}{6}}=0 \rightarrow x=-1+\sqrt{\frac{13}{6}}

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3 years ago
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