All categories

3 years ago

Find the perimeter of a rhombus whose diagonals are

Mathematics

1 answer:

mojhsa [17]3 years ago

5 0

Answer:

Rhombus Side = sqr root [(34 / 2)^2 + (30 / 2)^2]

Rhombus Side = sqr root [17^2 + 15^2]

Rhombus Side = sqr root [289 + 225}

Rhombus Side = sqr root [514]

Rhombus Side = 22.6715680975

Rhombus Perimeter = side * 4 = 90.68627239 cm

Step-by-step explanation:

You might be interested in

What is half off 7 dollars and 50 cents (7.50)

Natali [406]

You have a starting value of 7.50. to cut something in half means you should divide that number by 2. So you would do:

7.50/2= $3.75

4 0

3 years ago

Read 2 more answers

Y=Mx+b what does x equal

Aleks [24]

Mx+b=y
subtract b from both sides
mx=y-b
divide m in both sides
x=(y-b)/m

3 0

4 years ago

the cooking club makes 3 chicken burritos for every 4 beef burritos they prepare. how many chicken burritos do they make if they

RideAnS [48]

They make 18 chicken burritos because when you do 24 beef burritos total divided by 4 beef burritos you get 6 beef then you muliply 6 by the number of chicken burrittos.

6 0

3 years ago

Read 2 more answers

What is the classification for this polynomial?<br> u^2 v^3 w

andrew11 [14]

Answer:

A...

Step-by-step explanation:

Not completely sure. :<

5 0

3 years ago

Before the industrial revolution in 1800 the concentration of carbon in Earth’s atmo- sphere was 280 ppm. The concentration in 2

slega [8]

Answer: There is increase of 4.255 in the amount of carbon in the atmosphere.

Step-by-step explanation:

Since we have given that

Concentration of carbon in Earth's atmosphere in 1800 = 280 ppm

Concentration of carbon in Earth's atmosphere in 2015 = 399 ppm

We need to find the percentage increase in the amount of carbon in the atmosphere.

So, Difference = 399-280 = 119 ppm

so, percentage increase in the amount of carbon is given by

$\dfrac{Difference}{Original}\times 100\\\\=\dfrac{119}{280}\times 100\\\\=\dfrac{11900}{280}\\\\=42.5\%$

Hence, there is increase of 4.255 in the amount of carbon in the atmosphere.

4 0

3 years ago