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djyliett [7]
3 years ago
8

Daisy is trying to find the equation

Mathematics
1 answer:
podryga [215]3 years ago
5 0
The y-intercept is (0,12), or just 12, hope this helps
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compare to first factor to the value of the product how do i do that on module 4 lesson 21 exit ticket grade 5
Marina86 [1]
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5 0
3 years ago
Solve for qqq.
Anna11 [10]
3/5=q/10
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8 0
3 years ago
Read 2 more answers
Compare using <, >, = 5+(-4)____-4 + (-7)
nikitadnepr [17]

Good morning,

Answer:

<h2>5+(-4) > -4 + (-7)</h2>

Step-by-step explanation:

5+(-4) = 5 - 4 = 1

-4 + (-7) = -(4+7) = -11

since  1 > -11 then 5+(-4) > -4 + (-7)

:)

4 0
3 years ago
Convert y+7=4(x-3) standard form
True [87]

Answer:

-4x+y = -5

Step-by-step explanation:

Y+7=4(x-3)

Y+7=4x-12

+7. +7

Y=4x-5

-4x+y=-5

7 0
3 years ago
Math:
Dahasolnce [82]

Answer:

a) x_{1} = \frac{\pi}{3}\pm 2\pi \cdot i, \forall i \in \mathbb{N}_{O}, x_{2} = \frac{5\pi}{6}\pm 2\pi\cdot i, \forall i \in \mathbb{N}_{O}, b) x_{1} = \frac{\pi}{3}\pm 2\pi \cdot i, \forall i \in \mathbb{N}_{O}, x_{2} = \frac{5\pi}{3}\pm 2\pi\cdot i, \forall i \in \mathbb{N}_{O}

Step-by-step explanation:

a) The equation must be rearranged into a form with one fundamental trigonometric function first:

\sqrt{3}\cdot \csc x - 2 = 0

\sqrt{3} \cdot \left(\frac{1}{\sin x} \right) - 2 = 0

\sqrt{3} - 2\cdot \sin x = 0

\sin x = \frac{\sqrt{3}}{2}

x = \sin^{-1} \frac{\sqrt{3}}{2}

Value of x is contained in the following sets of solutions:

x_{1} = \frac{\pi}{3}\pm 2\pi \cdot i, \forall i \in \mathbb{N}_{O}

x_{2} = \frac{5\pi}{6}\pm 2\pi\cdot i, \forall i \in \mathbb{N}_{O}

b) The equation must be simplified first:

\cos x + 1 = - \cos x

2\cdot \cos x = -1

\cos x = -\frac{1}{2}

x = \cos^{-1} \left(-\frac{1}{2} \right)

Value of x is contained in the following sets of solutions:

x_{1} = \frac{\pi}{3}\pm 2\pi \cdot i, \forall i \in \mathbb{N}_{O}

x_{2} = \frac{5\pi}{3}\pm 2\pi\cdot i, \forall i \in \mathbb{N}_{O}

7 0
3 years ago
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