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gayaneshka [121]
3 years ago
12

-(6)^-1 a.6 b.1/6 c.-1/6 d.-6 No idea what I'm doing...

Mathematics
1 answer:
Monica [59]3 years ago
7 0
-(6)^{-1} = ?
<span>Representing in fraction for the best understanding
</span>-(\frac{6}{1}) ^{-1} = ?<span>
When the exponent is negative, the inversion is made, the numerator becomes the denominator, and the denominator becomes the numerator, and the change of signal from the exponent, which was negative, becomes positive.

</span>-( \frac{1}{6} )^1 =
\boxed{\boxed{- \frac{1}{6}}}\end{array}}\qquad\quad\checkmark

Answer:
c.-1/6
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Answer:

$17.625

Step-by-step explanation:

7.5% = 0.075

235 x 0.075= 17.625

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AB has coordinates A(-5,9) and B(7,- 7). Points P, Q, and I are collinear
VMariaS [17]

Answer:

\overline{QT}

Step-by-step explanation:

We want to find the coordinates of a certain point C(x,y) such that C divides A(x_1,y_1) and B(x_2,y_2) in the ratio m:n=3:2

The x-coordinate is given by:

x=\frac{mx_2+nx_1}{m+n}

The y-coordinate is given by:

y=\frac{my_2+ny_1}{m+n}

AB has coordinates A(-5,9) and B(7,- 7)

We substitute the values to get:

x=\frac{3*7+2*-5}{3+2}

x=\frac{21-10}{5}

x=\frac{11}{5}

and

y=\frac{3*-7+2*9}{3+2}

y=\frac{-21+18}{5}

y=-\frac{3}{5}

Therefore C has coordinates  (\frac{11}{5},-\frac{3}{5})

The line segment that contains C is \overline{QT}

See attachment.

6 0
3 years ago
If an eight-inch square cake serves four people, how many twelve-inch square cakes are needed to provide epuivalent servings to
natta225 [31]
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7 0
4 years ago
From a box containing 10 cards numbered 1 to 10, four cards are drawn together. The probability that their sum is even is 21 21
ankoles [38]

Answer:

Step-by-step explanation:

We know that between 1 to 10 there are 5 even and 5 odd numbers.

We could get 4 even cards , 4 odd cards or 2 odd and 2 even cards

Let´s check all this combinations

Case 1: When all 4 numbers are even:  

We are going to take 4 of the 5 even numbers in the box so we have

5C4=5

Case 2: When all 4 numbers are odd:  

We are going to take 4 of the 5 odd numbers in the box, so we have

5C4=5

Case 3: When 2 are even and 2 are odd:

We are giong to take 2 from 5 even and odd cards in the box so we have

 

5C2 * 5C2

Remember that we obtain the probability from

\frac{Number-of-favourable-Outcome}{Total-number-of-outcomes}

So we have the number of favourable outcomes but we need the Total cases for drawing four cards, so we have that:  

We are taking 4 of the 10 cards:

10C_4=210

Hence we have that the probability that their sum is even

\frac{5+5+100}{210}=\frac{11}{21}

8 0
3 years ago
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