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3 years ago

10 Points Deals with Quadratic Functions

Mathematics

1 answer:

madam [21]3 years ago

3 0

Q(x) is just a fancy way of saying y, so when the graph intersects the x-axis, that will be your answer (which is A. -4)

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TRIANGLE CONGRUENCE! PICTURE ADDED. How is QTP congruent to SPT?

shtirl [24]

Answer:

The triangles are congruent because:

QP = ST
angle P = angle T
the side PT is in both triangles

We know when 2 sides of two triangles and the angle between them is congruent, the triangles are congruent.

5 0

3 years ago

Landon was taking a math test. He finished 1/2 of the test in 3/5 of an hour. How much of the test will he have done in one hour

timama [110]

Answer:

$\frac{5}{6}\ of\ the \ test$

Step-by-step explanation:

we know that

Landon finished 1/2 of the test in 3/5 of an hour

by proportion

Find out how much of the test will he have done in one hour

$\frac{(1/2)}{(3/5)}\frac{test}{hours} =\frac{x}{1}\frac{test}{hour}\\ \\x=(1/2)*(5/3)\\ \\x=\frac{5}{6}\ of\ the \ test$

7 0

3 years ago

What time is it 120 minutes after midday?

coldgirl [10]

Midday is 12pm so you add whatever hours to that. Which is 1:20pm after midday (12pm)

7 0

3 years ago

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.

7 0

4 years ago

What is the volume of the right triangular prism shown?

oksian1 [2.3K]

Answer:

area of the base triangle divided by two

Step-by-step explanation:

idk that's what Alexa said, sorry if it's wrong

5 0

3 years ago

***10 Points*** Deals with Quadratic Functions

10 Points Deals with Quadratic Functions