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4 years ago

Write each equation in standard form. 2y=–6+2x

Mathematics

2 answers:

jasenka [17]4 years ago

8 0

If I wrote it in standard from it would be

2x - 2y = 6

Hope that helps, Good luck! (:

Archy [21]4 years ago

6 0

Standard form is:

ax + by = c

So just arrange it like this:

-2x + 2y = -6

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PLEASE HELP >.<

belka [17]

Check the picture below.

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3 years ago

Read 2 more answers

Jodi poured herself a cold soda that had an initial temperature of 45degreesF and immediately went outside to sunbathe where the

arlik [135]

The initial temperature difference of 101-45 = 56 degees declined to 101-55 = 46 degrees in 8 minutes, We can write the exponential equation for the soda's temperature as

... T = 101 -56(46/56)^(t/8) . . . . where t is in minutes

After an additional 10 minutes, we have t=18, so the soda temperature will be

... T = 101 -56(46/56)^(18/8) ≈ 65.0 . . . degrees

6 0

3 years ago

Let h(x)equalsf(g(x)), where f and g are differentiable on their domains. If g(9)equalsnegative 6 and g prime(9)equals3,

artcher [175]

Answer:

Fghb

Step-by-step explanation:

was my time to

3 0

3 years ago

A=16,b=63<br> Find the missing angle

Julli [10]

180 - 16 - 63 = 101. The remaining angle is 101*

Hope I helped!

6 0

4 years ago

Read 2 more answers

List all possible rational roots. Then use synthetic division to confirm which rational roots exist:

Kisachek [45]

Answer:

$\boxed{(1) \, x = \, \pm \dfrac{1}{2}, \pm 1, \pm2, \pm \dfrac{5}{2}, \pm 5, \pm 10; (2) \, x = -2}$

Step-by-step explanation:

2x³+ 6x² - x - 10 = 0

(1) Possible roots

The Rational Roots Theorem states that, if a polynomial has any rational roots, they will have the form p/q, where p is a factor of the constant term and q is a factor of the leading coefficient.

$\text{Possible rational root} = \dfrac{ p }{ q } = \dfrac{\text{factor of constant term}}{\text{factor of leading coefficient}}$

In your function, the constant term is -10 and the leading coefficient is 2, so

$\text{Possible root} = \dfrac{\text{factor of 10}}{\text{factor of 2}}$

Factors of 10 = ±1, ±2, ±5, ±10

Factors of 2 = ±1, ±2

$\text{Possible roots are } \large \boxed{\mathbf{x = \pm \dfrac{1}{2}, \pm 1, \pm2, \pm \dfrac{5}{2}, \pm 5, \pm 10}}$

(2) Synthetic division

Rather than work through all 12 possibilities, I will do one that works.

$\begin{array}{r|rrrr}-2 & 2 & 6 & -1 & -10\\& & -4& -4 & 10\\& 2 & 2& -5 & 0\\\end{array}$

So, x = -2 is a root, and the quotient is 2x² + 2x - 5.

(3) Check for other rational roots

2x² + 2x - 5 = 0

D = b² - 4ac =2²- 4(2)(-5) = 4 + 40 = 44

√44 = 2√11, which is irrational.

Since irrational roots come in pairs, the cubic equation has two real, irrational roots and one rational root at x = -2.

6 0

4 years ago