All categories

3 years ago

PLEASE HELP WILL MARK WILL MARK BRAINLIEST THANK YOU

Mathematics

1 answer:

Anika [276]3 years ago

6 0

Answer:

We are given that the frame is 2 units wide, which means that we subtract the length of the outer frame by 2 units for each side:

7 - 2 -2 = 3

9 - 2 -2 = 5

Now, we have the length and width for the smaller rectangle, so we use the area formula (area = length*width) to get:

3*5 = 15

Step-by-step explanation:

Please support my answer.

You might be interested in

The moon forms a right triangle with the Earth and the Sun during one of its phases, as shown below:

Readme [11.4K]

The scientist stayed awake in class, so knows that the cosine of angle x is the ratio of y to the sun-moon distance. The sun-moon distance can be found by dividing y by the cosine of x.

7 0

3 years ago

Is -2x=13 in standard form?

Lesechka [4]

I believe no, it would have to be x = -6.5

3 0

3 years ago

Describe the relationship between the two quantities. Distance traveled by car; Amount of gas in the car.

Alinara [238K]

Step-by-step explanation:

There are always two types of relationships between two quantities. Either the relationship is Directly proportional or it would be Indirectly proportional. Directly proportional is a relation in which one quantity increases with the increase of the other quantity and vice versa. Whereas in indirectly proportional relationship, one quantity decreases with the increase of the other quantity and vice versa.

Now in this question, the given quantities are Distance traveled by a car and the amount of gas in the car. These two quantities have indirect proportional relationship. When the Distance traveled by the car increases, the amount of gas in the car decreases and vice versa.

5 0

3 years ago

Connie is a nurse and works a 12- hour shift. If she has worked 4 hours, what fraction of her shift does she still have to work

vovikov84 [41]

The correct answer is B. 1/3

5 0

3 years ago

Read 2 more answers

I really stuck on trying to prove this

Alisiya [41]

$\displaystyle\sum_{r=1}^nr(r+1)\cdots(r+p-1)$

When $n=1$ ,

$\displaystyle\sum_{r=1}^1r(r+1)\cdots(r+p-1)=1(1+1)(1+2)\cdots(1+p-2)(1+p-1)=p!$

Meanwhile, you have on the right

$\dfrac{(1)(1+1)(1+2)\cdots(1+p-2)(1+p-1)(1+p)}{p+1}=(1)(1+1)(1+2)\cdots(p-1)(p)=p!$

so the equality holds for $n=1$ .

Assume it holds for $n=k$ , i.e. that

$\displaystyle\sum_{r=1}^kr(r+1)\cdots(r+p-1)=\frac{k(k+1)(k+2)\cdots(k+p-1)(k+p)}{p+1}$

Now for $n=k+1$ , you have

$\displaystyle\sum_{r=1}^{k+1}r(r+1)\cdots(r+p-1)=\sum_{r=1}^kr(r+1)\cdots(r+p-1)+(k+1)(k+2)\cdots(k+1+p-1)$
$=\displaystyle\frac{k(k+1)(k+2)\cdots(k+p-1)(k+p)}{p+1}+(k+1)(k+2)\cdots(k+1+p-2)(k+1+p-1)$
$=\displaystyle\frac{k(k+1)(k+2)\cdots(k+p-1)(k+p)}{p+1}+(k+1)(k+2)\cdots(k+p-1)(k+p)$
$=\left(\dfrac k{p+1}+1\right)(k+1)(k+2)\cdots(k+p-1)(k+p)$
$=\dfrac{k+p+1}{p+1}(k+1)(k+2)\cdots(k+p-1)(k+p)$
$=\dfrac{(k+1)(k+2)\cdots(k+p-1)(k+p)(k+p+1)}{p+1}$

as required.

3 0

3 years ago