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____ [38]
3 years ago
11

Julio bought a total of 20 medium and large pumkins if he spent 53$ and bought 6 more large pumkins as medium pumkins how many l

arge pumkins did he buy large pumkin 3$ medium 2$ ans small 1$
Mathematics
1 answer:
aksik [14]3 years ago
3 0
L = large pumpkin

m = medium pumpkin


Write a system of equations: 

1)   L + m = 20
2)   m = L - 6
3)   $3L + $2m = $53

Substitute L - 6 for m in equation 1:

L + (L - 6) = 20

Add 6 to both sides:

2L = 26
(L + L = 2L)

Divide two on both sides:

L = 13

Julie bought 13 large pumpkins.

Check (use equation 3):

$3(13) + $2(7) = 53
(m = 7 because 20 - 13 = 7)

Multiply:

$39 + $14 = $53       :)
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Answer:

12

Step-by-step explanation:

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faust18 [17]

Answer:

B

Step-by-step explanation:

The domain is the set of x values for which the function is defined.

The range is the set of y values for which the function is defined.

Attached is the graph of the exponential function.

It is the basic graph of exponential function of y = 5^x which is shifted 6 units above (because of +6 at the end).

<em><u>Looking at the graph, the domain is the set of all x values.</u></em>

<em><u>The range is anything above 6.</u></em>

Correct answer is B.

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3 years ago
A chef prepared five chocolate tortes for a dinner party. The guests consumed 2 5/16 tortes. How many
Alenkinab [10]
If you would like to know how many tortes are left, you can calculate this using the following steps:

five chocolate tortes - 2 5/16 tortes = 5 - 2 5/16 = 5 - 37/16 =  80/16 - 37/16 = 43/16 = 2 11/16

The correct result would be D. 2 11/16.
3 0
3 years ago
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If sinA=√3-1/2√2,then prove that cos2A=√3/2 prove that
Ivan

Answer:

\boxed{\sf cos2A =\dfrac{\sqrt3}{2}}

Step-by-step explanation:

Here we are given that the value of sinA is √3-1/2√2 , and we need to prove that the value of cos2A is √3/2 .

<u>Given</u><u> </u><u>:</u><u>-</u>

• \sf\implies sinA =\dfrac{\sqrt3-1}{2\sqrt2}

<u>To</u><u> </u><u>Prove</u><u> </u><u>:</u><u>-</u><u> </u>

•\sf\implies cos2A =\dfrac{\sqrt3}{2}

<u>Proof </u><u>:</u><u>-</u><u> </u>

We know that ,

\sf\implies cos2A = 1 - 2sin^2A

Therefore , here substituting the value of sinA , we have ,

\sf\implies cos2A = 1 - 2\bigg( \dfrac{\sqrt3-1}{2\sqrt2}\bigg)^2

Simplify the whole square ,

\sf\implies cos2A = 1 -2\times \dfrac{ 3 +1-2\sqrt3}{8}

Add the numbers in numerator ,

\sf\implies cos2A =  1-2\times \dfrac{4-2\sqrt3}{8}

Multiply it by 2 ,

\sf\implies cos2A = 1 - \dfrac{ 4-2\sqrt3}{4}

Take out 2 common from the numerator ,

\sf\implies cos2A = 1-\dfrac{2(2-\sqrt3)}{4}

Simplify ,

\sf\implies cos2A =  1 -\dfrac{ 2-\sqrt3}{2}

Subtract the numbers ,

\sf\implies cos2A = \dfrac{ 2-2+\sqrt3}{2}

Simplify,

\sf\implies \boxed{\pink{\sf cos2A =\dfrac{\sqrt3}{2}} }

Hence Proved !

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Pani-rosa [81]

Answer:

LN = 18

Step-by-step explanation:

hope that's the answer..

6 0
3 years ago
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