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musickatia [10]

3 years ago

5

Can anyone help me please

2 answers:

Leno4ka [110]3 years ago

7 0

B is the answer. It demonstrates all of the terms placed correctly on the number line. (:

labwork [276]3 years ago

3 0

It’s B because -2 4/5 is near the -3 line, -1 is on -1, 0 is on 0, -2/5 is sorta near 0, and 5/6 is near 1

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slega [8]

Answer:

Step-by-step explanation:

If we think about a right triangle, we get that the hypotenuse is 24ft or the length of the ladder. The base of the ladder is 15ft away from the building which is the leg of the right triangle. The screenshot is the diagram (my drawing isn't that good lol:))

Thus we can use cosine function with the angle between the ladder and the ground since we need to find the angle and we get

$cos\alpha = \frac{15}{24}$

and $\alpha = 51.3 degrees$

The angle is 51.3 degrees hope that helps!

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2 years ago

Is -4^2<(-4)^2 a false statement

denis-greek [22]

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3 years ago

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A wedding planner purchased both small and large lanterns for a wedding reception. The planner purchased a total of 40 lanterns

Olegator [25]

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Step-by-step explanation:

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2 years ago

Leah's house has a media room with a large flat-screen television to watch movies. The media room television has a width of 8 fe

Naddik [55]

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8 0

3 years ago

A road perpendicular to a highway leads to a farmhouse located d miles away. An automobile traveling on this highway passes thro

pshichka [43]

Answer:

$\frac{dh}{dt}=\frac{30r}{\sqrt{d^{2}+900}}$

Step-by-step explanation:

A road is perpendicular to a highway leading to a farmhouse d miles away.

An automobile passes through the point of intersection with a constant speed $\frac{dx}{dt}$ = r mph

Let x be the distance of automobile from the point of intersection and distance between the automobile and farmhouse is 'h' miles.

Then by Pythagoras theorem,

h² = d² + x²

By taking derivative on both the sides of the equation,

$(2h)\frac{dh}{dt}=(2x)\frac{dx}{dt}$

$(h)\frac{dh}{dt}=(x)\frac{dx}{dt}$

$(h)\frac{dh}{dt}=rx$

$\frac{dh}{dt}=\frac{rx}{h}$

When automobile is 30 miles past the intersection,

For x = 30

$\frac{dh}{dt}=\frac{30r}{h}$

Since $h=\sqrt{d^{2}+(30)^{2}}$

Therefore,

$\frac{dh}{dt}=\frac{30r}{\sqrt{d^{2}+(30)^{2}}}$

$\frac{dh}{dt}=\frac{30r}{\sqrt{d^{2}+900}}$

3 0

3 years ago