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2 years ago

5

Daniel wants to do 312 practice problems to prepare for an exam. if he wants to do approximately the same number of problems eac

h day for 36 days, which best describes the number of problems that he needs each to do each day?

1 answer:

alexandr1967 [171]2 years ago

6 0

Answer:

Step-by-step explanation:

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ABCD ~ EFGH<br> z=?<br> Please help me!!!

iragen [17]

Answer:

z = 110°

Step-by-step explanation:

Since the figures are similar, then corresponding angles are congruent, thus

∠H = ∠D, that is

z = 110°

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2 years ago

If f(t)= 3t^2-2, find f(k-2)

DedPeter [7]

By plugging (k-2) in t, we can write that
$f(k-2)=3(k-2)^{2}-2=3(k^{2}-4k+4)-2=3 k^{2}-12k+12-2=$ $3 k^{2}-12k+10$

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2 years ago

Calculate the value of x to one decimal place. inches

kakasveta [241]

Use the law of cosines: $c^2=a^2+b^2-2ab\cos C$

We have:

$c=x\\a=3in\\b=7in\\C=54^o$

substitute:

$\cos54^o\approx0.5878\\\\c^2=3^2+7^2-2\cdot3\cdot7\cdot0.5878\\\\c^2=9+49-24.6876\\\\c^2=33.3129\to c=\sqrt{33.3129}\\\\c\approx5.8$

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2 years ago

Emery looks out their apartment window to the building across the way. The building isknown to be 42 feet tall. The angle of dep

Amanda [17]

Answer:

$x=44.5ft$

Step-by-step explanation:

From the question we are told that:

Height $h=42ft$

Angle of depression $\theta=27\textdegree$

Angle of Elevation $\alpha=23 \textdegree$

Generally the equation for the vertical distance between Emery's distance x and the bottom of the building is mathematically given by

Since the angle of depression and elevation are given as

27 and 23 respectively

Therefore

Emery's view of the 42 ft building is

$\gamma=23+27$

$\gamma=50 \textdegree$

Therefore Emery's distance x to the base of the building h' is

$h'=\frac{27}{50}*42$

$h'=22.68ft$

Generally the Trigonometric equation for Emery's distance x is mathematically given by

$x=\frac{h'}{tan\theta}$

$x=\frac{22.68}{tan 27}$

$x=44.5ft$

6 0

1 year ago

An anthropologist finds that a prehistoric bone contains less than 8.1% of the amount of Carbon-14 the bones would have containe

stealth61 [152]

Answer:

20,944 years

Step-by-step explanation:

The formula you use for this type of decay problem is the one that uses the decay constant as opposed to the half life in years. We are given the k value of .00012. If we don't know how much carbon was in the bones when the person was alive, it would be safer to say that when he was alive he had 100% of his carbon. What's left then is 8.1%. Because the 8.1% is left over from 100% after t years, we don't need to worry about converting that percent into a decimal. We can use the 8.1. Here's the formula:

$N(t)=N_{0} e^{-kt}$

where N(t) is the amount left over after the decay occurs, $N_{0}$ is the initial amount, -k is the constant of decay (it's negative cuz decay is a taking away from as opposed to a giving to) and t is the time in years. Filling in accordingly,

$8.1=100e^{-.00012t}$

Begin by dividing the 100 on both sides to get

$.081=e^{-.00012t}$

Now take the natural log of both sides. Since the base of a natual log is e, natural logs and e "undo" each other, much like taking the square root of a squared number.

ln(.081)= -.00012t

Take the natual log of .081 on your calculator to get

-2.513306124 = -.00012t

Now divide both sides by -.00012 to get t = 20,944 years

4 0

2 years ago