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Molodets [167]
2 years ago
14

1. 1/8y < 34 Please solve and write the expression and steps down please

Mathematics
1 answer:
Sindrei [870]2 years ago
6 0

Answer:

y<272

Step-by-step explanation:

Step 1: Multiply both sides by 8.

8*(1/8y)<(8)*(34)

y<272

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150-5r&gt;87.5 Il mark brainliest
Mashutka [201]

Answer:

r < 12.5

Step-by-step explanation:

Given

150 - 5r > 87.5 ( subtract 150 from both sides )

- 5r > - 62.5

Divide both sides by - 5, reversing the symbol as a result of dividing by a negative quantity.

r < 12.5

8 0
3 years ago
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Evan paid $60.54 for 3 notebooks and a calculator. If the calculator costs $45.87, how much does each notebook cost? Write a two
Aleonysh [2.5K]

Answer:   4.89$

Step-by-step explanation:

60.54 is starting amount then subtract 45.87 left with 14.67 then divide by 3 get 4.89

60.54-45.87=14.67

14.67/3=4.89

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3 years ago
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Find the center, vertices, and foci of the ellipse with equation x squared divided by 400 plus y squared divided by 625 = 1
Inessa05 [86]

Answer:

x^2/400 + x^2/625

(x-0)^2/400) +(y-0^2/625)

x^2=400

X=sqrt. 400

x = 20

y^2=625

y = sqrt. 625

y= 25

a^2-c^2=b^2

sqrt 400-625 = c

20-25=c

The correct answer is c=-5

(-5,0)

(5,0)

Step-by-step explanation:

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2 years ago
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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)

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3 years ago
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Please help and I will give you a Cookie
Vesnalui [34]

Step-by-step explanation:

In triangle

x²=5²+12²

x²=25+144

x²=169

x= root of 169

x=13

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2 years ago
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