What is the area of the triangle?

The sum of two numbers is 95, and their difference is 61. What are the two numbers?

BARSIC [14]

Let's call:

$first\#=x$

$second\#=y$

then

when we have a SUM, we have PLUS and when we have a DIFFERENCE, we have MINUS... Let's go then...

$\begin{Bmatrix}x+y&=&95\\x-y&=&61\end{matrix}$

now we can sum all the rows then we got it... (This is the other way to solved this question)

$x+y+(x-y)=95+61$

$x+y+x-y=156$

$2x=156$

$\boxed{x=78}$

now we can replace this value at first or at second row, you just need to pick up one...

I'll choose the second one

$x-y=61$

$78-y=61$

$y=78-61$

$\boxed{y=17}$

$\boxed{\boxed{\begin{Bmatrix}x&=&78\\y&=&17\end{matrix}}}$

8 0

4 years ago

Read 2 more answers

The ordered pair (a,b) satisfies the inequality y>x+3. Which statement is true ?

nordsb [41]

Answer:The correct option is(C). b is greater than 3.

Step-by-step explanation:

Given that the ordered pair (a, b) satisfied the following inequality :

We are to select the TRUE statement from the given options.

Since the ordered pair (a, b) satisfies inequality (i), so we have

Therefore, b is greater than 3.

Thus, the correct option is (C). b is greater than 3.

3 0

3 years ago

Which expression is equivalent to *picture attached*

DiKsa [7]

Answer:

The correct option is;

$4 \left (\dfrac{50 (50+1) (2\times 50+1)}{6} \right ) +3 \left (\dfrac{50(51) }{2} \right )$

Step-by-step explanation:

The given expression is presented as follows;

$\sum\limits _{n = 1}^{50}n\times \left (4\cdot n + 3 \right )$

Which can be expanded into the following form;

$\sum\limits _{n = 1}^{50} \left (4\cdot n^2 + 3 \cdot n\right ) = 4 \times \sum\limits _{n = 1}^{50} \left n^2 + 3 \times\sum\limits _{n = 1}^{50} n$

From which we have;

$\sum\limits _{k = 1}^{n} \left k^2 = \dfrac{n \times (n+1) \times(2n+1)}{6}$

$\sum\limits _{k = 1}^{n} \left k = \dfrac{n \times (n+1) }{2}$

Therefore, substituting the value of n = 50 we have;

$\sum\limits _{n = 1}^{50} \left k^2 = \dfrac{50 \times (50+1) \times(2\cdot 50+1)}{6}$

$\sum\limits _{k = 1}^{50} \left k = \dfrac{50 \times (50+1) }{2}$

Which gives;

$4 \times \sum\limits _{n = 1}^{50} \left n^2 = 4 \times \dfrac{n \times (n+1) \times(2n+1)}{6} = 4 \times \dfrac{50 \times (50+1) \times(2 \times 50+1)}{6}$

$3 \times\sum\limits _{n = 1}^{50} n = 3 \times \dfrac{n \times (n+1) }{2} = 3 \times \dfrac{50 \times (51) }{2}$

$\sum\limits _{n = 1}^{50}n\times \left (4\cdot n + 3 \right ) = 4 \times \dfrac{50 \times (50+1) \times(2\times 50+1)}{6} +3 \times \dfrac{50 \times (51) }{2}$

Therefore, we have;

$4 \left (\dfrac{50 (50+1) (2\times 50+1)}{6} \right ) +3 \left (\dfrac{50(51) }{2} \right )$ .

4 0

3 years ago

Help me figure out the answer

baherus [9]

Answer:

102

Step-by-step explanation:

lets say x=2

so,

2x+2 would be 4+2

x would be 2

and 3x+8 would be 6+8

add up the sums,

6+2+80+14

and you get 102

so your answer is 102

Hope this helps! :)

3 0

3 years ago

Read 2 more answers

PLEASEEEE HELP ASAP , due today

Levart [38]

Answer:

circumference=>2πr

circumference= 119.32 cm

π = 3.14

r =?

119.32 = 2 × 3.14 × r

r = 119.32/6.28

r = 19 cm

so area = πr²

area = 3.14 × (19)²

area = 3.14 × 361

area = 1,133.54 cm²

3 0

3 years ago