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Natali [406]
3 years ago
15

What times 1/11 equals 1?

Mathematics
1 answer:
valentinak56 [21]3 years ago
7 0

Answer:

11/1

Step-by-step explanation:

11/1 ×1/11=1

that fraction is the answer

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Which of the following pairs of numbers contains like fractions? A. 5⁄6 and 10⁄12 B. 3⁄2 and 2⁄3 C. 3 1⁄2 and 4 4⁄4 D. 6⁄7 and 1
ElenaW [278]
<h2>Hello!</h2>

The answers are:

A.

\frac{5}{6} and \frac{10}{12}

D.

\frac{6}{7} and 1\frac{5}{7}

<h2>Why?</h2>

To find which of the following pairs of numbers contains like fractions, we must remember that like fractions are the fractions that share the same denominator.

We are given two fractions that are like fractions. Those fractions are:

Option A.

\frac{5}{6} and \frac{10}{12}

We have that:

\frac{10}{12}=\frac{5}{6}

So, we have that the pairs of numbers

\frac{5}{6}

and

\frac{5}{6}

Share the same denominator, which is equal to 6, so, the pairs of numbers contains like fractions.

Option D.

\frac{6}{7} and 1\frac{5}{7}

We have that:

1\frac{5}{7}=1+\frac{5}{7}=\frac{7+5}{7}=\frac{12}{7}

So, we have that the pair of numbers

\frac{6}{7}

and

\frac{12}{7}

Share the same denominator, which is equal to 7, so, the pairs of numbers constains like fractions.

Also, we have that the other given options are not like fractions since both pairs of numbers do not share the same denominator.

The other options are:

\frac{3}{2},\frac{2}{3}

and

3\frac{1}{2},4\frac{4}{4}

We can see that both pairs of numbers do not share the same denominator so, they do not contain like fractions.

Hence, the answers are:

A.

\frac{5}{6} and \frac{10}{12}

D.

\frac{6}{7} and 1\frac{5}{7}

Have a nice day!

3 0
3 years ago
Three vertices of the trapezoid are A(4d,4e), B(4f,4e), and C(4g,0). The fourth vertex lies on the origin. Find the midpoint of
satela [25.4K]
We have that

point C and point D have y = 0-----------> (the bottom of the trapezoid).

point A and point B have y = 4e ---------- > (the top of the trapezoid)

the y component of  midpoint would be halfway between these lines
 y = (4e+ 0)/2 = 2e. 

<span>the x component of the midpoint of the midsegment would be halfway between the midpoint of AB and the midpoint of CD.

x component of midpoint of AB is (4d + 4f)/2.
x component of midpoint of CD is (4g + 0)/2 = 4g/2.
x component of a point between the two we just found is
[(4d + 4f)/2 + 4g/2]/2 = [(4d + 4f + 4g)/2]/2 = (4d + 4f + 4g)/4 = d + f + g. 

</span>therefore

the midpoint of the midsegment is (d + f + g, 2e)
8 0
3 years ago
PLEASE HELP. graph the set of points which model is most appropriate for the set (-6,4)(-4,0)(-3,-2)(-1,-7)
SOVA2 [1]
A linear equation would be the best fit, but the last point (-1,-7) kinda messes it up. If the -7 would have been a -6 the line y=-2x-8 would fit perfectly.
7 0
3 years ago
This is timed and I just need to see if i’m right! please check my work :)
Misha Larkins [42]
Hi! actually, the y intercept is 0, 8. hopefully this helps
4 0
3 years ago
Please help me with this geometry question <br><br> image attached
vodomira [7]

15/17. The value (ratio) of cos A is 15/17.

The trigonometric ratios of an acute angle are, basically, the sine, the cosine and the tangent. They are defined from an acute angle, α, of a right triangle, whose elements are the hypotenuse, the leg contiguous to the angle,  and the leg opposite the angle.

-The sine of the angle is the opposite leg divided by the hypotenuse.

-The cosine of the angle is the adjacent leg divided by the hypotenuse.

-The tangent of the angle is the opposite leg divided by the adjacent leg or, which is the same, the sine of the angle divided by the cosine of the angle.

cos A = adjacent leg/hypothenuse = BC/AC = 15/17

4 0
3 years ago
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