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3 years ago

Can i please get help with this question?

Mathematics

1 answer:

KATRIN_1 [288]3 years ago

4 0

Answer:

Not proportional.

Step-by-step explanation:

A proportional relationship is a relationship which crosses through the origin (0,0) and which has a proportional constant. We can determine this either by finding (0,0) where x=0 and y=0 in the table or by dividing y/x. None of the tables contain (0,0) so we will divide y by x. We are looking for a table which when each y is divided by its x we have the same constant appearing.

$\frac{2}{1} \neq \frac{3}{2} \neq \frac{4}{3}$

Though it has a point at the origin, these fractions are not equal. This is not proportional.

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What is 10(122(13(12)))

bearhunter [10]

10(122(13)(12))=10(122)(156)=(10)(19032)=190320
Hope You understand
This is just basic multiplication.

5 0

3 years ago

Read 2 more answers

Find all solutions to

BARSIC [14]

Answer:

x= 0 , $\frac{1}{14}$ , $\frac{-1}{12}$

Step-by-step explanation:

Given, equation is $\sqrt[3]{15x-1}$ + $\sqrt[3]{13x+1}$ = $4\sqrt[3]{x}$ . →→→ (1)

Now, by cubing the equation on both sides, we get

( $\sqrt[3]{15x-1}$ + $\sqrt[3]{13x+1}$ )³ = ( $4\sqrt[3]{x}$ )³

⇒ (15x-1) + (13x+1) + 3× $\sqrt[3]{15x-1}$ × $\sqrt[3]{13x+1}$ ( $\sqrt[3]{15x-1}$ + $\sqrt[3]{13x+1}$ ) = 64 x.

⇒ 28x + 3× $\sqrt[3]{15x-1}$ × $\sqrt[3]{13x+1}$ ( $4\sqrt[3]{x}$ ) = 64x.

(since from (1), $\sqrt[3]{15x-1}$ + $\sqrt[3]{13x+1}$ = $4\sqrt[3]{x}$ )

⇒ 12× $\sqrt[3]{15x-1}$ × $\sqrt[3]{13x+1}$ ( $\sqrt[3]{x}$ )= 36x.

⇒ 3x = $\sqrt[3]{(15x-1)(13+1)(x)}$ .

Now, once again cubing on both sides, we get

(3x)³ = ( $\sqrt[3]{(15x-1)(13+1)(x)}$ )³.

⇒ 27x³ = (15x-1)(13x+1)(x).

⇒ 27x³ = 195x³ + 2x² - x

⇒ 168x³ + 2x² - x = 0

⇒ x(168x² + 2x -1) = 0

⇒ by, solving the equation we get ,

x = 0 ; x = $\frac{1}{14}$ ; x = $\frac{-1}{12}$

therefore, solution is x= 0 , $\frac{1}{14}$ , $\frac{-1}{12}$

7 0

3 years ago

Destiny and Guadalupe are shopping. Destiny buys 2 pairs of pants and 6 bracelets and pays $178.Guadalupe buys 3 pairs of pants

ziro4ka [17]

For the entirety of this problem, p will represent a pair of pants and b will represent a bracelet.

Step 1) Set up equations for Destiny and Guadalupe

Destiny: 178 = 2p + 6b

Guadalupe: 155 = 3p + 2b

I will be using substitution to solve this problem, but elimination can also be used.

Step 2) Solve Destiny's equation for p

178 = 2p + 6b

178 - 6b = 2p

89 - 3b = p

Step 3) Substitute the found value of p from Destiny's equation into Guadalupe's equation and solve for b

155 = 3(89 - 3b) + 2b

155 = 267 - 9b + 2b

155 = 267 - 7b

-7b = -112

b = 16

Step 4) Use the value of b found in step 3, plug that back into our equation from step 2 and solve for p

89 - 3(16) = p

89 - 48 = p

p = 41

one pair of pants = $16

one bracelet = $41

Hope this helps!! :)

4 0

3 years ago

[20points][Precal] Find the cross product of -(3/4)V and (1/2)w if v = <-2, 12, -3> and w = <-7, 4, -6>

Lana71 [14]

$\mathbf v=\langle-2,12,-3\rangle=-2\,\vec i+12\,\vec j-3\,\vec k$
$-\dfrac34\mathbf v=\dfrac32\,\vec i-9\,\vec j+\dfrac94\,\vec k$

$\mathbf w=\langle-7,4,-6\rangle=-7\,\vec i+4\,\vec j-6\,\vec k$
$\dfrac12\mathbf w=-\dfrac72\,\vec i+2\,\vec j-3\,\vec k$

$-\dfrac34\mathbf v\times\dfrac12\mathbf w=\begin{vmatrix}\vec i&\vec j&\vec k\\\frac32&-9&\frac94\\-\frac72&2&-3\end{vmatrix}$
$=\left((-9)\times(-3)-\dfrac94\times2\right)\,\vec i-\left(\dfrac32\times(-3)-\dfrac94\times\left(-\dfrac72\right)\right)\,\vec j+\left(\dfrac32\times2-(-9)\times\left(-\dfrac72\right)\right)\,\vec k$
$=\dfrac{45}2\,\vec i-\dfrac{27}8\,\vec j-\dfrac{57}2\,\vec k=\left\langle\dfrac{45}2,-\dfrac{27}8,-\dfrac{57}2\right\rangle$

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2 years ago

Please give me an answer as fast as possible I need to finish this assignment.

TiliK225 [7]

Answer:

Step-by-step explanation:

I hope I'm right :)

5 0

3 years ago