13) Tickets to a movie cost $7 for adults and $5 for students. A group of friends purchased Stickets for $52

An orchard has 651 orange trees. The number of rows exceeds the number of trees per row by 10. How many trees are there in a row

natima [27]

Answer:

The number of trees in each row is 21.

Step-by-step explanation:

Let us assume the number of trees in each row = m

So, the number of rows in the orchard = Number of trees in each row + 10

or, total number of trees in each row = (10 +m)

Now, Total number of Trees = Number of Rows x Number of trees in 1 row

⇒ 651 = m (m + 10)

$651 = m^2 + 10m\\\implies m^2 + 10m - 651 =0\\or, m^2 + 31m 21m - 651 =0\\or, m(m +31) -21(m+31) =0\\\implies (m +31) (m-21) = 0$

⇒ (m +31) = 0, or (m-21) = 0

⇒ m= -31, or m = 21

Now, as m = The number of trees in each row, so m ≠ -31

So, m = 21

Hence, the number of trees in each row is m = 21.

8 0

4 years ago

Help please someone math

NeX [460]

I think it is c, because the probability is the littlest thing and you add most of it, you are welcome my boy

3 0

3 years ago

Isosceles triangles have two equal sides. true or false

melomori [17]

True
isosceles triangles
have 2 equal sides

5 0

4 years ago

(cotx+cscx)/(sinx+tanx)

Butoxors [25]

Answer: $\bold{\dfrac{cot(x)}{sin(x)}}$

<u>Step-by-step explanation:</u>

Convert everything to "sin" and "cos" and then cancel out the common factors.

$\dfrac{cot(x)+csc(x)}{sin(x)+tan(x)}\\\\\\\bigg(\dfrac{cos(x)}{sin(x)}+\dfrac{1}{sin(x)}\bigg)\div\bigg(\dfrac{sin(x)}{1}+\dfrac{sin(x)}{cos(x)}\bigg)\\\\\\\bigg(\dfrac{cos(x)}{sin(x)}+\dfrac{1}{sin(x)}\bigg)\div\bigg[\dfrac{sin(x)}{1}\bigg(\dfrac{cos(x)}{cos(x)}\bigg)+\dfrac{sin(x)}{cos(x)}\bigg]\\\\\\\bigg(\dfrac{cos(x)}{sin(x)}+\dfrac{1}{sin(x)}\bigg)\div\bigg(\dfrac{sin(x)cos(x)}{cos(x)}+\dfrac{sin(x)}{cos(x)}\bigg)$

$\text{Simplify:}\\\\\bigg(\dfrac{cos(x)+1}{sin(x)}\bigg)\div\bigg(\dfrac{sin(x)cos(x)+sin(x)}{cos(x)}\bigg)\\\\\\\text{Multiply by the reciprocal (fraction rules)}:\\\\\bigg(\dfrac{cos(x)+1}{sin(x)}\bigg)\times\bigg(\dfrac{cos(x)}{sin(x)cos(x)+sin(x)}\bigg)\\\\\\\text{Factor out the common term on the right side denominator}:\\\\\bigg(\dfrac{cos(x)+1}{sin(x)}\bigg)\times\bigg(\dfrac{cos(x)}{sin(x)(cos(x)+1)}\bigg)$

$\text{Cross out the common factor of (cos(x) + 1) from the top and bottom}:\\\\\bigg(\dfrac{1}{sin(x)}\bigg)\times\bigg(\dfrac{cos(x)}{sin(x)}\bigg)\\\\\\\bigg(\dfrac{1}{sin(x)}\bigg)\times cot(x)}\qquad \rightarrow \qquad \dfrac{cot(x)}{sin(x)}$