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3 years ago

For the function y = 7-3x, what is ordered pair for x=5?

Mathematics

1 answer:

Strike441 [17]3 years ago

3 0

(5,0) if there is no y then it is a 0

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Zair has 48 marbles in his pocket. He has 3 more blues than the

d1i1m1o1n [39]

Answer:

# yellow = 20

# blue = 23

# red = 5

Step-by-step explanation:

let 'r' = # red marbles

let '4r' = # yellow marbles

let '3+4r' = # blue marbles

r + 4r + 3 + 4r = 48

9r + 3 = 48

9r = 45

r = 5

red = 5

yellow = 4(5) = 20

blue = 3+4(5) = 23

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3 years ago

Please help me taking a test

Paladinen [302]

Answer: the one that ends with 30. D the fourth one

Step-by-step explanation:

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3 years ago

Which equation(s) have x = –3 as the solution? log3(2x + 15) = 2 log5(8x + 9) = 2 log4(–20x + 4) = 3 logx81 = 4

baherus [9]

Answer:

log_3(2x+15)=2 and log_4(-20x+4)=3

Step-by-step explanation:

Plug it in and see!

$log_3(2x+15)=2\\log_3(2(-3)+15)=2\\log_3(-6+15)=2\\log_3(9)=2\\\text{ This is a true equation because } 3^2=9\\\\\\log_5(8(-3)+9)=2\\log_5(-24+9)=2\\log_5(-15)=2\\\\\text{ Not true because you cannot do log of a negative number }\\\\log_4(-20(-3)+4)=3\\log_4(64)=3\\\text{ this is true because } 4^3=64\\\\\\log_x (81)=4\\log_{-3}(81)=4\\\text{ the base cannot be negative }\\\\\\\\\text{ There is only two options here} \\\\log_3(2x+15)=2 \text{ and } log_4(-20x+4)=3$

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3 years ago

Read 2 more answers

Please help I am so lost!!!

ASHA 777 [7]

$\bf tan\left( \frac{x}{2} \right)+\cfrac{1}{tan\left( \frac{x}{2} \right)}\\\\
-----------------------------\\\\
tan\left(\cfrac{{{ \theta}}}{2}\right)=
\begin{cases}
\pm \sqrt{\cfrac{1-cos({{ \theta}})}{1+cos({{ \theta}})}}
\\ \quad \\

\cfrac{sin({{ \theta}})}{1+cos({{ \theta}})}
\\ \quad \\

\boxed{\cfrac{1-cos({{ \theta}})}{sin({{ \theta}})}}
\end{cases}\\\\$

$\bf -----------------------------\\\\
\cfrac{1-cos(x)}{sin(x)}+\cfrac{1}{\frac{1-cos(x)}{sin(x)}}\implies \cfrac{1-cos(x)}{sin(x)}+\cfrac{sin(x)}{1-cos(x)}
\\\\\\
\cfrac{[1-cos(x)]^2+sin^2(x)}{sin(x)[1-cos(x)]}\implies 
\cfrac{1-2cos(x)+\boxed{cos^2(x)+sin^2(x)}}{sin(x)[1-cos(x)]}
\\\\\\
\cfrac{1-2cos(x)+\boxed{1}}{sin(x)[1-cos(x)]}\implies \cfrac{2-2cos(x)}{sin(x)[1-cos(x)]}
\\\\\\
\cfrac{2[1-cos(x)]}{sin(x)[1-cos(x)]}\implies \cfrac{2}{sin(x)}\implies 2\cdot \cfrac{1}{sin(x)}\implies 2csc(x)$

4 0

3 years ago

Completing the ordered pair

Rainbow [258]

Answer:

(-4,-4)

Step-by-step explanation:

I checked the answer in desmos

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3 years ago