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Eva8 [605]
3 years ago
15

Help ASAP on #2 please

Mathematics
1 answer:
NikAS [45]3 years ago
5 0
Where is the picture
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Determine whether each pair of triangles is similar. Justify your answer.
USPshnik [31]
Add an attachment of the triangles maybe? :)! And i can help
4 0
3 years ago
Which of the following are not terms in the series below? Check all that apply.
Vlada [557]

Answer:

the answer is A and B on edg

Step-by-step explanation:

7 0
3 years ago
At a local movie theater sodas cost 4.50 and bags of popcorn cost 1.50. If Kirk buys three times as many bags as popcorn as soda
Arisa [49]

Answer:

12 bags of popcorn and 4 sodas

Step-by-step explanation:

Quantity: Popcorn =3x       Sodas=x

Value: Sodas - $4.50, Popcorn - $1.50

Total Value: $36

3x(1.5)+4.5x=36

4.5x+4.5x=36

9x=36

x=4

Therefore, 12 bags of popcorn and 4 sodas

3 0
3 years ago
The exprnditure of 15 days of a man is as follows. calculate this average expenditure. 30,25,35,24,40,35,30,25,
atroni [7]

Answer:

$30.53

Step-by-step explanation:

Find total sum

30+25+35=90

90+24=114

114+40=154

154+35=189

…

You will get 458

458/15=30.53333

7 0
3 years ago
What reasoning and explanations can be used when solving radical equations and how do extraneous solutions arise from radical eq
Mkey [24]

Answer:

We know that, a 'radical equation' is an equation that contains radical expressions, which further are the expressions containing radicals ( square roots and other roots of numbers ).

In order to solve radical equations, we use the rules of exponents and basic algebraic properties.

The common reasoning to use while solving a radical equation is:

1. Isolate the radical expression.

2. Square both sides of the equation to remove radical.

3. After removing the radical, solve the equation to find the unkown variable

4. Check the answer for the errors occurred by removing the radicals.

For e.g. \sqrt{x}-3=5

i.e. \sqrt{x}-3+3=5+3 ( Adding 3 on both sides )

i.e. \sqrt{x}=8

i.e. (\sqrt{x})^{2} =8^{2}

i.e. x=64.

So, the solution the the radical equation \sqrt{x}-3=5 is x = 64.

Further, we know that an 'extraneous solution' is that root of the radical equation which is not a root of the original equation and is excluded from the domain.

for e.g. Take \sqrt{x+4} =x-2

i.e. (\sqrt{x+4})^{2} =(x-2)^2

i.e. x+4=x^2+4-4x

i.e. 0=x^2-5x

i.e. 0=x(x-5)

i.e. x = 0 and x = 5.

Substituting x = 0 in \sqrt{x+4} =x-2, gives \sqrt{0+4} =0-2 i.e. \sqrt{4} =-2 i.e. 2=-2, which is not possible.

So, x = 0 is a solution that does not satisfy the equation.

Hence, x = 0 is an extraneous solution.

7 0
4 years ago
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