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2 years ago

Help please (3x^2 + 5x - 9) + (-3x + 4)

Mathematics

2 answers:

mixas84 [53]2 years ago

4 0

Answer:

3x^2 + 2x - 5

Step-by-step explanation:

can i have a brainlest plz

emmainna [20.7K]2 years ago

4 0

Answer:

(3x^2+5x-9-3x+4)

or, 3x^2+2x-5

or, 3x^2+(5-3)x-5

or, 3x^2-3x+5x-5

or, 3x(x-1)+5(x-1)

or, (x-1) (3x+5)

hope u get it

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Find the slope of the line y = -7/6x - 8/3.

Nezavi [6.7K]

Hello!

This equation is written in slope-intercept form.

Slope-intercept form is y = mx + b, where m is the slope and y is the y-intercept.

Since the coefficient -7/6 is multiplied with x, the slope (or m) has to equal to -7/6.

Therefore, the slope of the line is -7/6.

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2 years ago

What is the answer to the equation: 8+10

alina1380 [7]

Answer:

18, the answer to 8+10 is 18, like you learn in 1st grade

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2 years ago

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In our Solar system six of the eight planets have moons what percentage of the planets have moons

Vladimir79 [104]

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2 years ago

Barry has 18 more marbles then Heidi. If Barry has 92 marbles, write and solve an equation to determine the number of marbles He

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3 years ago

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Newton's law of cooling is:

Mnenie [13.5K]

Answer:

$t = \frac{ln(\frac{21}{59})}{-0.15}=6.887 hr$

So it would takes approximately 6.9 hours to reach 32 F.

Step-by-step explanation:

For this case we have the following differential equationÑ

$\frac{du}{dt}= -k (u-T)$

We can reorder the expression like this:

$\frac{du}{u-T} = -k dt$

We can use the substitution $w = u-T$ and $dw =du$ so then we have:

$\frac{dw}{w} =-k dt$

IF we integrate both sides we got:

$ln |w| = -kt +C$

If we apply exponential in both sides we got:

$w = e^{-kt} *e^c$

And if we replace w = u-T we got:

$u(t)= T + C_1 e^{-kt}$

We can also express the solution in the following terms:

$u(t) = (T_i -T_{amb}) e^{kt} +T_{amb}$

For this case we know that $k =-0.15 hr$ since w ehave a cooloing, $T_{i}= 70 F, T_{amb}=11F$ , we have this model:

$u(t) = (70-11) e^{-0.15t} +11$

And if we want that the temperature would be 32F we can solve for t like this:

$32 = 59 e^{-0.15 t} +11$

$21=59 e^{-0.15 t}$

$\frac{21}{59} = e^{-0.15 t}$

If we apply natural logs on both sides we got:

$ln (\frac{21}{59}) =-0.15 t$

$t = \frac{ln(\frac{21}{59})}{-0.15}=6.887 hr$

So it would takes approximately 6.9 hours to reach 32 F.

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3 years ago