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natima [27]
3 years ago
13

Whas does 16g+3<-17+15g equal ​

Mathematics
1 answer:
ipn [44]3 years ago
4 0

Answer:

g∠-20

Step-by-step explanation:

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After 1 year, $800 deposited in a savings account with simple interest had earned $64 in
stepan [7]

Answer:

8%

Step-by-step explanation:

We will use the equation provided to solve this problem.

I=Prt

Lets plug in the values:

64=800x(1)     First, divide both sides by 800.

.08=x(1)         Multiply x by 1

x = .08

Now, we multiply .08 by 100 to get our interest rate:

(.08)100 = 8

The interest rate is 8%

7 0
3 years ago
Plzzzzzzzzzzzzzzzzz help quick
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Answer:

y=-2x-7 D

Step-by-step explanation:

3 0
3 years ago
Frank stands 450 feet from the base of the Statue of Liberty. If the Statue of Liberty is 306 feet tall, what is the angle of de
kari74 [83]
Let the angle of elevation be θ.


We have a right angled triangle with an opposite of 300.5 ft. (306 - 5.5) and an adjacent of 400 ft. Recalling SOH CAH TOA, tanθ = O/A.
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θ = tan^-1(300.5/400).
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8 0
3 years ago
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-3x+4y=-4 -2x+5y=9<br>solve the following system of equations
Vesnalui [34]

the answer is x= 8, and y=5

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3 years ago
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Help! If you know this can you tell me how to do it?
aleksandr82 [10.1K]

Answer:

c

Step-by-step explanation:

Here's how this works:

Get everything together into one fraction by finding the LCD and doing the math.  The LCD is sin(x) cos(x).  Multiplying that in to each term looks like this:

[sin(x)cos(x)]\frac{sin(x)}{cos(x)}+[sin(x)cos(x)]\frac{cos(x)}{sin(x)} =?

In the first term, the cos(x)'s cancel out, and in the second term the sin(x)'s cancel out, leaving:

\frac{sin^2(x)}{sin(x)cos(x)}+\frac{cos^2(x)}{sin(x)cos(x)}=?

Put everything over the common denominator now:

\frac{sin^2(x)+cos^2(x)}{sin(x)cos(x)}=?

Since sin^2(x)+cos^2(x)=1, we will make that substitution:

\frac{1}{sin(x)cos(x)}

We could separate that fraction into 2:

\frac{1}{sin(x)}×\frac{1}{cos(x)}

\frac{1}{sin(x)}=csc(x)  and  \frac{1}{cos(x)}=sec(x)

Therefore, the simplification is

sec(x)csc(x)

5 0
3 years ago
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