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3 years ago

Please help me ASAP!

Mathematics

1 answer:

boyakko [2]3 years ago

7 0

Answer:

Step-by-step explanation:

6 is the smallest so the first one would be c because it is the smallest

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Can someone tell me the answer

astra-53 [7]

Answer:

-3

Step-by-step explanation:

(-6)/(+2)

-6/+2

-3

5 0

3 years ago

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There are 4 fewer cats than dogs.<br>There are 2 cats. How many dogs<br>are there?

allsm [11]

Answer:

there are 6 dogs

Step-by-step explanation:

x = dogs

x= 2 + 4

x = 6

8 0

3 years ago

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Write this ratio as a fraction in lowest terms $100 to $90

barxatty [35]

This ratio in fraction form is 10/9

7 0

4 years ago

Tyler went to the super market to buy food for a pantry. He has $36, and can carry up to 20pounds of food in his backpack. One p

AleksAgata [21]

This question is incomplete

Complete Question

Tyler went to the supermarket to buy food for a food pantry. He has $36, and can carry up to 20 pounds of food in his backpack. Pasta costs $1 for a 1-pound package. Pasta sauce costs $3 for a 1.5 pound jar. Let x = the number of packages of pasta and y = the number of jars of pasta sauce. One package of pasta is the right amount to go with one jar of pasta sauce. What is the best numbers of packages of pasta and jars of pasta sauce to buy for the food pantry? Explain your reasoning.

Answer:

Eight is the best numbers of packages of pasta and jars of pasta sauce to buy for the food pantry.

Step-by-step explanation:

Let

x = the number of packages of pasta

y = the number of jars of pasta sauce.

He has $36, and can carry up to 20 pounds of food in his backpack. Pasta costs $1 for a 1-pound package. Pasta sauce costs $3 for a 1.5 pound jar.

x + 1.5y ≤ 20....... Equation 1

x = 20 - 1.5y

x × $1 + y × $3 = $36

x + 3y ≤ 36..... Equation 2

20 - 1.5y + 3y = 36

-1.5y + 3y = 36 - 20

1.5y = 16

y = 16/1.5

y = 8

And x = 8

Therefore,

Eight is the best numbers of packages of pasta and jars of pasta sauce to buy for the food pantry.

6 0

3 years ago

Rationalise the denominator of:<br>1/(√3 + √5 - √2)

Paul [167]

Step-by-step explanation:

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm :\longmapsto\:\dfrac{1}{ \sqrt{3} + \sqrt{5} - \sqrt{2} }$

can be re-arranged as

$\rm :\longmapsto\:\dfrac{1}{ \sqrt{3} - \sqrt{2} + \sqrt{5} }$

$\rm \: = \: \dfrac{1}{( \sqrt{3} - \sqrt{2} ) + \sqrt{5} }$

On rationalizing the denominator, we get

$\rm \: = \: \dfrac{1}{( \sqrt{3} - \sqrt{2} ) + \sqrt{5} } \times \dfrac{( \sqrt{3} - \sqrt{2} ) - \sqrt{5} }{( \sqrt{3} - \sqrt{2} ) - \sqrt{5} }$

We know,

$\rm :\longmapsto\:\boxed{\tt{ (x + y)(x - y) = {x}^{2} - {y}^{2} \: }}$

So, using this, we get

$\rm \: = \: \dfrac{ \sqrt{3} - \sqrt{2} - \sqrt{5} }{ {( \sqrt{3} - \sqrt{2} )}^{2} - {( \sqrt{5}) }^{2} }$

$\rm \: = \: \dfrac{ \sqrt{3} - \sqrt{2} - \sqrt{5} }{3 + 2 - 2 \sqrt{6} - 5}$

$\rm \: = \: \dfrac{ \sqrt{3} - \sqrt{2} - \sqrt{5} }{5 - 2 \sqrt{6} - 5}$

$\rm \: = \: \dfrac{ \sqrt{3} - \sqrt{2} - \sqrt{5} }{ - 2 \sqrt{6}}$

$\rm \: = \: \dfrac{ - ( - \sqrt{3} + \sqrt{2} + \sqrt{5}) }{ - 2 \sqrt{6}}$

$\rm \: = \: \dfrac{- \sqrt{3} + \sqrt{2} + \sqrt{5}}{2 \sqrt{6}}$

On rationalizing the denominator, we get

$\rm \: = \: \dfrac{- \sqrt{3} + \sqrt{2} + \sqrt{5}}{2 \sqrt{6}} \times \dfrac{ \sqrt{6} }{ \sqrt{6} }$

$\rm \: = \: \dfrac{- \sqrt{18} + \sqrt{12} + \sqrt{30}}{2 \times 6}$

$\rm \: = \: \dfrac{- \sqrt{3 \times 3 \times 2} + \sqrt{2 \times 2 \times 3} + \sqrt{30}}{12}$

$\rm \: = \: \dfrac{- 3\sqrt{2} + 2 \sqrt{3} + \sqrt{30}}{12}$

Hence,

$\boxed{\tt{ \rm \dfrac{1}{ \sqrt{3} + \sqrt{5} - \sqrt{2} } =\dfrac{- \sqrt{3 \times 3 \times 2} + \sqrt{2 \times 2 \times 3} + \sqrt{30}}{12}}}$

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<h3><u>More Identities to </u><u>know:</u></h3>

$\purple{\boxed{\tt{ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2}}}}$

$\purple{\boxed{\tt{ {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2}}}}$

$\purple{\boxed{\tt{ {(x + y)}^{3} = {x}^{3} + 3xy(x + y) + {y}^{3}}}}$

$\purple{\boxed{\tt{ {(x - y)}^{3} = {x}^{3} - 3xy(x - y) - {y}^{3}}}}$

$\pink{\boxed{\tt{ {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2})}}}$

$\pink{\boxed{\tt{ {(x + y)}^{2} - {(x - y)}^{2} = 4xy}}}$

6 0

3 years ago