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Gnoma [55]
3 years ago
15

Which representation does not show y as a function of x​

Mathematics
1 answer:
chubhunter [2.5K]3 years ago
8 0
J because the x value is used more then once.
You might be interested in
Y varies directly as the square of x. If y is 25 when x is 3 find y when x is 2
Vilka [71]

Answer:

y = \frac{100}{9}

Step-by-step explanation:

given y varies as the square of x then the equation relating them is

y = kx² ← k is the constant of variation

to find k use the given condition y = 25 when x = 3

k = \frac{y}{x^{2} } = \frac{25}{9}

⇒ y = \frac{25}{9} x²

when x = 2

y = \frac{25}{9} × 4 = \frac{100}{9}



8 0
3 years ago
Estimate 73% of 120 <br><br> Thanks if you help!
xxMikexx [17]

Answer:

73/100 (120)

answer:-87.6

3 0
3 years ago
Read 2 more answers
K=dF-DF solve for F<br><br> How do you solve for this??
aleksandrvk [35]

K=F/D it is very simple

8 0
3 years ago
Mason can walk 4 miles in 1 hour.how many miles can he walk in 3 hours
sdas [7]
Well if he can walk 4 miles in one hour, just multiply 4 by 3 to get 12. Hope this helps! Make sure to mark me brainlest(losing ranks like crazy for some reason)
3 0
3 years ago
Please help me!!!!!!!!!!!!!​
monitta

Answer:  see proof below

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

Use the following Half-Angle Identities:    tan (A/2) = (sinA)/(1 + cosA)

                                                                     cot (A/2) = (sinA)/(1 - cosA)

Use the Pythagorean Identity: cos²A + sin²B = 1

Use Unit Circle to evaluate: cos 45° = sin 45° = \frac{\sqrt2}{2}

<u>Proof LHS → RHS</u>

Given:                       cot\ (22\frac{1}{2})^o-tan\ (22\frac{1}{2})^o

Rewrite Fraction:     cot\ (\frac{45}{2})^o-tan\ (\frac{45}{2})^o

Half-Angle Identity:   \dfrac{sin(45)^o}{1-cos(45)^o}-\dfrac{sin(45)^o}{1+cos(45)^o}

Substitute:                  \dfrac{\frac{\sqrt2}{2}}{1-\frac{\sqrt2}{2}}-\dfrac{\frac{\sqrt2}{2}}{1+\frac{\sqrt2}{2}}

Simplify:                      \dfrac{\frac{\sqrt2}{2}}{\frac{2-\sqrt2}{2}}-\dfrac{\frac{\sqrt2}{2}}{\frac{2+\sqrt2}{2}}

                               =\dfrac{\sqrt2}{2-\sqrt2}-\dfrac{\sqrt2}{2+\sqrt2}

                               =\dfrac{\sqrt2}{2-\sqrt2}\bigg(\dfrac{2+\sqrt2}{2+\sqrt2}\bigg)-\dfrac{\sqrt2}{2+\sqrt2}\bigg(\dfrac{2-\sqrt2}{2-\sqrt2}\bigg)

                               =\dfrac{2\sqrt2+2}{4-2}-\dfrac{2\sqrt2-2}{4-2}

                               =\dfrac{4}{2}

                               = 2

LHS = RHS:  2 = 2  \checkmark

7 0
4 years ago
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