What is the value of 2 3/4 ➗ 3/4​

The answer is: (11/3) or (3 2/3) <-- (simplified)

Step-by-step explanation:

3 2/3 is the answer

Step-by-step explanation:

Conversion a mixed number 2 3/

4

to a improper fraction: 2 3/4 = 2 3/

4

= 2 · 4 + 3/

4

= 8 + 3/

4

= 11/

4

To find new numerator:

a) Multiply the whole number 2 by the denominator 4. Whole number 2 equally 2 * 4/

4

= 8/

4

b) Add the answer from previous step 8 to the numerator 3. New numerator is 8 + 3 = 11

c) Write a previous answer (new numerator 11) over the denominator 4.

Two and three quarters is eleven quarters

Divide: 11/

4

: 3/

4

= 11/

4

· 4/

3

= 11 · 4/

4 · 3

= 44/

12

= 4 · 11 /

4 · 3

= 11/

3

Dividing two fractions is the same as multiplying the first fraction by the reciprocal value of the second fraction. The first sub-step is to find the reciprocal (reverse the numerator and denominator, reciprocal of 3/

4

is 4/

3

) of the second fraction. Next, multiply the two numerators. Then, multiply the two denominators. In the next intermediate step , cancel by a common factor of 4 gives 11/

3

.

so you simplify to get 3 2/3

a. turk and mcalister

b. howard, mission, and market

c. yes. since streets end some do not cross yet are not parallel to others. example: mcalister and 5th

step-by-step explanation:

parallel lines are lines which do not intersect ever and look a lot like railroad tracks. turk, mcalister and golden gate are examples of these.

intersect are streets which cross each other and a car on one can reach the other. howard, mission, and market intersect with 6th.

skew lines are lines which neither intersect nor are parallel. these lines exist here because some streets end and never intersect another like mcalister and 5th.

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)