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Nana76 [90]
3 years ago
8

can someone help me. i am struggling to do this question. i need the other two boxes and thats it. please. who ever gets it righ

t gets the brainliest award.

Mathematics
1 answer:
Sedaia [141]3 years ago
3 0
The number of people who grow potatoes is 3 times the number that grow sugar beet.
 ⇒ 48 + x = 3( x + 12)

--------------------------------------------
Solve for x :
--------------------------------------------
48 + x = 3( x + 12)
48 + x = 3x + 36
2x = 12
x = 6

--------------------------------------------
People who do both grow either.
--------------------------------------------

Number of people who does not grow potato or sugar beets = 100 - 48 - 12 - 6 = 34

------------------------------------------------------------------------------
Answer: The answers are 6 and 34.
------------------------------------------------------------------------------
"6" is inbetween "48" and "12". 
"34" is outside the two circles.
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X +3/ 5 = 2 <br> i need help
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8 0
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What are the lower and upper quartiles of this data 122,164,71,98,84,147,114,111,98,85,104,71,77
RSB [31]
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7 0
3 years ago
What is the value of x?
Mariulka [41]

Answer:  x = 7

`Step-by-step explanation:

Because the figure shows two triangles that are similar. we can write and solve an equation of ratios:

 9 cm       72 cm

----------- = -----------

3x - 20      56 cm

Cross-multiplying, we get (3x - 20)(72 cm) = (9 cm)(56 cm) = 504 cm²

Dividing both sides by 72 cm, we get:

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4 0
3 years ago
Read 2 more answers
Find the angle between the given vectors. Round your answer, in degrees, to two decimal places. u=⟨2,−6⟩u=⟨2,−6⟩, v=⟨4,−7⟩
NISA [10]

Answer:

\theta = 108.29

Step-by-step explanation:

Given

u =

v =

Required:

Calculate the angle between u and v

The angle \theta is calculated as thus:

cos\theta = \frac{u.v}{|u|.|v|}

For a vector

A =

A = a * b

cos\theta = \frac{u.v}{|u|.|v|} becomes

cos\theta = \frac{.}{|u|.|v|}

cos\theta = \frac{2*6+4*-7}{|u|.|v|}

cos\theta = \frac{12-28}{|u|.|v|}

cos\theta = \frac{-16}{|u|.|v|}

For a vector

A =

|A| = \sqrt{a^2 + b^2}

So;

|u| = \sqrt{2^2 + 6^2}

|u| = \sqrt{4 + 36}

|u| = \sqrt{40}

|v| = \sqrt{4^2+(-7)^2}

|v| = \sqrt{16+49}

|v| = \sqrt{65}

So:

cos\theta = \frac{-16}{|u|.|v|}

cos\theta = \frac{-16}{\sqrt{40}*\sqrt{65}}

cos\theta = \frac{-16}{\sqrt{2600}}

cos\theta = \frac{-16}{\sqrt{100*26}}

cos\theta = \frac{-16}{10\sqrt{26}}

cos\theta = \frac{-8}{5\sqrt{26}}

Take arccos of both sides

\theta = cos^{-1}(\frac{-8}{5\sqrt{26}})

\theta = cos^{-1}(\frac{-8}{5 * 5.0990})

\theta = cos^{-1}(\frac{-8}{25.495})

\theta = cos^{-1}(-0.31378701706)

\theta = 108.288386087

<em></em>\theta = 108.29<em> (approximated)</em>

4 0
2 years ago
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