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1 year ago

Jason wants to plant 66 carrot seeds and 44 onion bulbs. He would like to plant them so

Mathematics

2 answers:

n200080 [17]1 year ago

8 0

Answer:

Step-by-step explanation:

Since we need an equal amount plants in each row, you have to find the greatest common factor (GCF) of 66 and 44.

The factors of 44 are: 1, 2, 4, 11, 22, 44

The factors of 66 are: 1, 2, 3, 6, 11, 22, 33, 66

Therefore, the greatest factor of 66 and 44 is 22. This means Jason will have 3 rows of carrot plants and 2 rows of onion plants. Each row will have 22 plants.

natta225 [31]1 year ago

3 0

Answer:

Step-by-step explanation:

see below

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What would be this answer for the algebra thing

skad [1K]

Answer:

c. 4x + 20 = 140

Step-by-step explanation:

Perimeter = 140 feet

Width = L + 10

P = 2w + 2L

P = 2(L + 10) + 2L

P = 2L + 20 + 2L

P = 4L + 20

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1 year ago

What is the perimeter, in centimeters, of the figure

xeze [42]

Answer:

30 cm

Step-by-step explanation:

add all sides and use 7 for the bottom line because 5+2=7 and use 8 for the side because 5+3=8.

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2 years ago

Hi again here's the picture

pshichka [43]

Answer:

Step-by-step explanation:

Given,

D = diameter of semicircular part = 42 m

L = Length of straight part = 110 m

Now, let's find the perimeter:

= 2 ( length of semi-circle + length of straight part )

$= 2( \frac{\pi \: d}{2} + l)$

Plug the values

$= 2( \frac{3.14 \times 42}{2} + 110)$

Calculate the product

$= 2( \frac{131.88}{2} + 110)$

Divide

$= 2 (65.94 + 110)$

Calculate the sum

$= 2 \times 175.94$

Calculate

$= 351.88 \: m$

Hope this helps..

Best regards!!

8 0

2 years ago

Read 2 more answers

PLEASE I need help on these 4 problems you don’t have to do them all but if you can at least do one of them TYSM

Brrunno [24]

Answer:

For right angle triangle,

we use Pythagoras theorem that is:

$c^{2} =a^{2} +b^{2}$

c = $\sqrt{a^{2} +b^{2} }$

For question 1:

c = ?

a = 40

b = 9

putting them in formula,

c = $\sqrt{40^{2} + 9^{2} }$

c = 41

For question 2:

c = ?

a = 12

b = 13

putting them in formula,

c = $\sqrt{12^{2} + 13^{2} }$

c = approximately 17.69181

For question 3:

c = 35

a = 20

b = ?

putting them in formula,

$c^{2} =a^{2} +b^{2}$

$35^{2} = 20^{2} + b^{2}$

1225 = 400 + $b^{2}$

$b^{2}$ = 1225 - 400

$b^{2}$ = 825

$\sqrt{b^{2} } = \sqrt{825}$

b = 5 $\sqrt{33}$

For question 4:

c = 37

a = 20

b = ?

putting them in formula,

$c^{2} =a^{2} +b^{2}$

$37^{2} = 20^{2} + b^{2}$

1369 = 400 + $b^{2}$

$b^{2}$ = 1369 - 400

$b^{2}$ = 969

Taking square root on both sides

b = 31.12

Hope it helps.

4 0

2 years ago

2p^2+10=0 <br><br> Please help!

ohaa [14]

$2p^2+10=0\ \ \ |-10\\\\2p^2=-10\ \ \ |:2\\\\p^2=-5$
No real solutions, because for all real numbers the square of number is greater or equal 0.

7 0

2 years ago