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3 years ago

What is constant o proportionality in the equation y=16.14z? PLEASE SHOW HIW TO DO THIS.

1 answer:

3 0

Y = 16.14z ; the constant of proportionality is 16.14. It is called constant of proportionality because it never changes. It remains constant or the same. Whatever the value of z is, y is directly proportional to z by the rate of the constant of proportionality.

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Help me Please! Find the volume and surface area for all.

HACTEHA [7]

Answer:

V =41.41³

A = 94.41²

----

V =225.16³

SA =283.25²

----

V = 64³

SA =113.32²

----

V =433.33³

SA = 378.57²

Step-by-step explanation:

Picture 2 = a = 1/2 base = 3.5 x 3.5 = 12.25 b= 5 x 5 = 25

c²= a² + b² = 3.5² + 5²

c ²= √12.25 + √25

c ²= √ 37.5 = 6.12372435696

c ² = 6.1237 missing side

Picture 1 + 2 formula SA = bh + (s1 + s2 + s3)H

V = V= 1/2 b x h h x SA

Picture 3 + 4 formula SA= a²+ 2a a² / 4 + h² V= a² h/3

5 0

3 years ago

Suppose you choose a marble from a bag containing 5 red marbles, 3 white marbles, and 6 blue marbles. You return the first marbl

Aneli [31]

$|\Omega|=14^2=196\\
|A|=5\cdot6\cdot2=60\\\\
P(A)=\dfrac{60}{196}=\dfrac{15}{49}$

5 0

3 years ago

Mr.chu is slicing 4 cantaloupes for his class.There are 24 students in class. If Me.Chu cut the cantaloupes to make an equal por

babunello [35]

1/6 because 4/24 = 1/6

8 0

3 years ago

3(4x-7)=27 What does x =? Please help asap im timed!!!

Scilla [17]

Answer:

x = 4

Step-by-step explanation:

3 x ( 4x -7 ) = 27

3 x 4x = 12x

3 x -7 = -21

12x -21 = 27

12x -21 + 21 = 27 + 21

12x = 48

12x/12 = 48/12

x = 4

3 0

3 years ago

I need help w/ this!! thank you

Fittoniya [83]

<h3>Answer:</h3>

$\boxed{\pink{\sf \leadsto Yes \ there \ is \ a \ solution \ of \ the \ given \ inequality .}}$

<h3>Step-by-step explanation:</h3>

A inequality is given to us and we need to convert it into standard form and see whether if it has a solution . So let's solve the inequality.

The inequality given to us is :-

$\bf\implies |2y + 3 | - 1 \leq 0 \\\\\bf\implies |2y+3|\leq 1 \\\\\bf\implies (|2y+3|)^2 \leq 1^2 \\\\\bf\implies (2y+3)^2 \leq 1 \\\\\bf\implies (2y)^2+3^2+2(2y)(3) \leq 1 \\\\\bf\implies 4y^2+9+12y - 1 \leq 0 \\\\\bf\implies 4y^2+12y+8 \leq 0 \\\\\bf\implies 4( y^2 + 3y + 2 ) \leq 0 \\\\\bf\implies y^2+3y +2 \leq 0 \:\:\bigg\lgroup \purple{\bf Standard \ form \ of \ inequality }\bigg\rgroup \\\\\bf\implies y^2y+2y+y+2 \leq 0 \\\\\bf\implies y(y+2)+1(y+2)\leq 0 \\\\\bf\implies ( y+2)(y+1)\leq 0 \\\\\bf\implies \boxed{\red{\bf y \leq (-2) , (-1) }}$

Let's plot a graph to see its interval . Graph attached in attachment .

Now we can see that the Interval notation of would be ,

$\boxed{\boxed{\orange \tt \purple{\leadsto}y \in [-2,-1] }}$

<h3>Hence the standard form of inequality is y²+3y +2 ≤ 0 and the Solution set of the inequality is [ -2 , -1 ] .</h3>

4 0

3 years ago