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3 years ago

Which equation represents a line that passes through (4, 1/3) and has a slope of 3/4

Mathematics

1 answer:

andrew-mc [135]3 years ago

8 0

Answer:

y = 3/4x - 8/3

Step-by-step explanation:

y = 3/4x + b

1/3 = 3/4(4) + b

1/3 = 3 + b

-8/3 = b

y = 3/4x - 8/3

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Pleaseeeeeeeee help!!!

JulijaS [17]

Answer:

The first one.

Step-by-step explanation:

The polynomial is made out of three terms so we would consider it a ”trinomial” and since it is a second degree polynomial (based on the highest degree) it would be “quadratic.”

3 0

3 years ago

Using the Factor Theorem, which of the polynomial functions has the zeros 4, <img src="https://tex.z-dn.net/?f=%5Csqrt7" id="Tex

Serjik [45]

Answer:

Step-by-step explanation:

According to the Factor Theorem, if (x - a) is a factor (where a is a zero) of the polynomial P(x), then P(a) must equal zero.

Our zeros are 4, √7, and - √7. Hence, when evaluating P(4), P(√7), and P(-√7), all must evaluate to zero.

Testing each choice, we can see that only choice B is true. That is:

$\displaystyle \displaystyle \begin{aligned} f(4)&= (4)^3 - 4(4)^2 - 7(4) + 28 \\ &= (64) - (64) - (28) + 28 \\ &= 0 \stackrel{\checkmark}{=}0 \\f(\sqrt{7}) &= (\sqrt{7})^3 - 4(\sqrt{7})^2 - 7(\sqrt{7}) + 28 \\ &= (7\sqrt{7}) - (28) - (7\sqrt{7}) + 28 \\ &= 0 \stackrel{\checkmark}{=} 0\\ f(-\sqrt{7}) &= (-\sqrt{7})^3 - 4(-\sqrt{7})^2 - 7(-\sqrt{7}) + 28 \\ &= (-7\sqrt{7}) - (28) + (7\sqrt{7}) + 28 \\ &= 0 \stackrel{\checkmark}{=} 0 \end{aligned}$

In conclusion, our answer is B.

8 0

3 years ago

Can someone help me with 21

victus00 [196]

1. 10.4
2. 9.3
3. 10.3

6 0

3 years ago

Whats the greatest ten you can multiply by 5 to get close to, but not go over, 335?

Amiraneli [1.4K]

65 is the answer because 65 x 5 =325 which is close but it’s 10 away.
I hope that helped

8 0

3 years ago

Read 2 more answers

What is the amplitude, period, and phase shift of y= 5cos( x/2 + 2pi/3)

nordsb [41]

Y = A cos(Bx + C)
A - amplitude
$\frac{2\pi}{B}$ - period
$- \frac{C}{B}$ - phase shift

$y=5\cos{( \frac{x}{2} + \frac{2\pi}{3} )}
\\
\\y=5\cos{( \frac{1}{2}x + \frac{2\pi}{3})}
\\
\\A=5,B= \frac{1}{2},C= \frac{2\pi}{3}$

Amplitude: 5
Period: $\frac{2\pi}{B}=\frac{2\pi}{\frac{1}{2}}=4\pi$
Phase shift: $- \frac{C}{B} =- \frac{\frac{2\pi}{3}}{\frac{1}{2}}= -\frac{4\pi}{3}$

3 0

3 years ago