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4vir4ik [10]
2 years ago
11

Solve the systems of equations using substitution.

Mathematics
1 answer:
statuscvo [17]2 years ago
8 0

I'm sure on boxes 1 and 3 but not on the 2nd box

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Please help! Thanks! The midpoint between y and 33 is -7. Find y
Setler79 [48]

Answer: 32

Step-by-step explanation: hope this helps

4 0
2 years ago
Chad earned $30 for raking leaves. He has also earned money by doing chores for 6 weeks. Altogether, he has earned $120. How muc
galben [10]
You know that he makes $90 for doing chores, dividing that by 6 and you get that each week he earns $15 for chores.
6 0
2 years ago
Read 2 more answers
Mrs. Jackson put a total of $10,000 into two accounts one account earns 6% simple annual interest the other account earns 6.5% s
kozerog [31]

Answer:

The money invested in account at 6 % rate is $ 9230.77

The money invested in account at 6.5 % rate is $ 769.23

Step-by-step explanation:

Given as :

The total amount saves in two account = $ 10,000

The two rate of interest are 6 % and 6.5 % at simple interest

The interest earn at 6 % rate = $ 632

The interest earn at 6.5 % ate = $ 50

Now, Let The amount deposit at 6 % rate = $ x

And The amount deposit at 6.5 % rate = $ 10,000 - $ x

<u>From Simple Interest method</u>

Simple Interest = \dfrac{\textrm principal \times \textrm Rate\times \textrm time}{100}

or, At 6.5 % rate

$ 50 =  \dfrac{\textrm (10,000 - x ) \times \textrm 6.5\times \textrm 1}{100}

or, 10000 - x =  \frac{5000}{6.5}

Or, 10000 - x = 769.23

∴  x = 10000 - 769.23

I.e x = $ 9230.77

So , At 6.5 % the amount deposit = 10,000 - 9230.77 = $ 769.23

So, At 6 % The amount deposit  = $ 9230.77

Hence The money invested in account at 6 % rate is $ 9230.77

And  The money invested in account at 6.5 % rate is $ 769.23 Answer

7 0
3 years ago
in 2000 you weighed 175 pounds. in 2001 you weighed 62 pounds, and in 2002 you weighed 154 pounds. how many pounds did you lose
never [62]
You should get the answer 154 afte u subtract and at that school be your answer
6 0
3 years ago
Which of the following are identities? Check all that apply
Natasha2012 [34]

Answer:

A, C

Step-by-step explanation:

Actually, those questions require us to develop those equations to derive into trigonometrical equations so that we can unveil them or not. Doing it only two alternatives, the other ones will not result in Trigonometrical Identities.

Examining

A) True

\frac{1-tan^{2}x}{2tanx} =\frac{1}{tan2x} \\ \frac{1-tan^{2}x}{2tanx} =\frac{1}{\frac{2tanx}{1-tan^{2}x}}\\ tan2x=\frac{1-tan^{2}x}{2tanx}

Double angle tan2\alpha =\frac{1 -tan^{2}\alpha }{2tan\alpha}

B) False,

No further development towards a Trig Identity

C) True

Double Angle Sine Formula sin2\alpha =2sin\alpha *cos\alpha

sin(8x)=2sin(4x)cos(4x)\\2sin(4x)cos(4x)=2sin(4x)cos(4x)

D) False No further development towards a Trig Identity

[sin(x)-cos(x)]^{2} =1+sin(2x)\\ sin^{2} (x)-2sin(x)cos(x)+cos^{2}x=1+2sinxcosx\\ \\sin^{2} (x)+cos^{2}x=1+4sin(x)cos(x)

7 0
3 years ago
Read 2 more answers
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