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Neko [114]
3 years ago
6

Helpppp!! Timed !!!!!

Mathematics
2 answers:
Hitman42 [59]3 years ago
5 0

Answer:

the last one

Step-by-step explanation:

Lapatulllka [165]3 years ago
4 0

Answer:

hi

Step-by-step explanation:

i think so,it is D

hope it helps

have a nice day

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X in 5xy = 10 to solve the system<br> below.<br> x = 25 - y<br> 5x + y =10
lora16 [44]

Answer:

Step-by-step explanation:

there are two ways to solve systems of equations,  elimination or substitution, this looks like elimination would work quickly

x= 25 -y

x + y = 25   (eq 1)

5x + y = 10  (eq 2)

subtract 2 from 1

x + y = 25  

-(5x + y = 10)

-4x + 0y = 15

-4x = 15

x = - 15/4

now plug in x into either starting equations and solve for y

-15/4 + y = 25

y = 25 + 15/4

y = 100/4 + 15 /4

y = 115 /4

y = 28 3/4

we have solved it :)

4 0
3 years ago
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guapka [62]

Answer: Mika has not described a proportional relationship. Although it is a linear function, it does not pass through the origin.

Step-by-step explanation: if you place those points on a graph, they do create a linear function; the line just doesn't pass through the origin.

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3 years ago
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Please do a step by step so that i can show work.
wariber [46]

Answer:

See this attachment...

7 0
3 years ago
Larry's father has 5 sons named... Ten, Twenty, Thirty and Forty... What would be the name if the fifth son???
ss7ja [257]
Larry because in the begging of  question it starts of with Larry's father has five son and then it names 4 of the sons and the fifth sons name is at the beginning<span />
4 0
4 years ago
PLEASE HELP IM DESPERATE precalculus. also 50 points!! please god help
kompoz [17]

Answer:

For B thru F these options will vary but here how you do it

B. Step 1 Draw the 4 Quadrants.

Then Draw the Triangle in the lower right quadrant which we call quadrant 4. Label the X axis as Adjacent and positive. Label the Y axis as Opposite and negative. Label the Slanted side as the hypotune and AS POSITVE SINCE HYPOTENUSE IS ALWAYS POSITIVE.

FOR C. IN QUADRANT 2, PLOT A POINT AT 0,12 AND AT (-5,0). CONNECT THE DOTS AND IT FORMS A TRIANGLE. Label the X axis as adjacent and negative and the y axis as positve and opposite and label the slanted side hypotunese and positive.

FOR D Draw a straight line along the x axis then draw a slanted line passing through (5,-1). In between them put the theta symbol in there.

The labeling is the same for C.

For E. Since tan must be positve and secant must be positve, our triangle must be in the 1st Quadrant. Draw any right triangle as long it is in the first quadrant

The x axis is adjacent and positve. The y axis is opposite and positve. The hypotenuse is the slanted side and it is positve.

For F. Since sin is negative and cos is positve the triangle is in the 4th quadrant. Draw any triangle in the 4th quadrant and the labeling is the same for Problem B.

2. We can find the sec of cos by flipping cosine.

\cos( \frac{x}{y} )  =  \sec( \frac{y}{x} )

\cos( \frac{1}{2} )  =  \sec(2 )

Sec is 2.

To find the cotangent, first let find the sin then tan.

We can use the identity

\cos( {theta}^{2} )  +  \sin( {theta}^{2} )  = 1

Let plug in the number

\cos( \frac{ {1}^{2} }{{2}^{2} } )  +  \sin(x {}^{2} )  = 1

\cos( \frac{1}{4} )  +  \sin(x {}^{2} )  = 1

\sin(x {}^{2} )  =   1 -  \frac{1}{4}

\sin(x {}^{2} )  =  \frac{3}{4}

\sin(x)  =  \frac{ \sqrt{3} }{ \sqrt{4} }

\sin(x)  =  \frac{ \sqrt{3} }{2}

Since sin is negative, sin x=

-  \frac{ \sqrt{3} }{2}

Now let apply the formula

\frac{ \sin(x) }{ \cos(x) }  =  \tan(x)

\frac{ \frac{ -  \sqrt{3} }{2} }{ \frac{1}{2} }  =  \tan(x)

-  \sqrt{3}

Now let find cotangent we can the reciprocal of

tan.

\tan=  -  \sqrt{3}

\cot =   - \frac{1}{ \sqrt{3} }

Rationalize denominator

\frac{ - 1}{ \sqrt{3} }  \times   \frac{ \sqrt{3} }{ \sqrt{3} }  =  \frac{ \sqrt -{ 3} }{3}

cotangent equal

-  \frac{ \sqrt{ 3} }{3}

4 0
3 years ago
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