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2 years ago

Can someone please help me ill mark you brainliest!!! !!!!

Mathematics

1 answer:

Aleksandr-060686 [28]2 years ago

5 0

Answer:

A and B

Step-by-step explanation:

It is an acute equilateral triangle

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Show all of your work. Simplify the expression: - 3a + 7 + 5a

Eva8 [605]

Answer:

2a + 7

Step-by-step explanation:

-3a + 7 +5a

<em>subtract the negative 3 from the 5</em>

2a + 7

4 0

2 years ago

Read 2 more answers

Write sin(19°) in terms of cosine.

Leto [7]

Answer:

cos(71)

Step-by-step explanation:

Since 19° is less than 90, we can express this in terms of confunction.

sin(θ) = cos(90-θ)

sin(19) = cos(90-19)

sin(19) = cos(71)

3 0

2 years ago

I have the question in a screengrab. Thank you, whoever helps me.

nata0808 [166]

To simplify

$\sqrt[4]{\dfrac{24x^6y}{128x^4y^5}}$

we need to use the fact that

$\sqrt[4]{x^4}=|x|$

Why the absolute value? It's because $(-x)^4=(-1)^4x^4=x^4$ .

We start by rewriting as

$\sqrt[4]{\dfrac{2^23x^6y}{2^6x^4y^5}}$

$\sqrt[4]{\dfrac{2^23x^4x^2y}{2^42^2x^4y^4y}}$

Since $x\neq0$ , we have $\dfrac xx=1$ , and the above reduces to

$\sqrt[4]{\dfrac{3x^2y}{2^4y^4y}}$

Then we pull out any 4th powers under the radical, and simplify everything we can:

$\dfrac1{\sqrt[4]{2^4y^4}}\sqrt[4]{\dfrac{3x^2y}{y}}$

$\dfrac1{|2y|}\sqrt[4]{3x^2}$

where $y>0$ allows us to write $\dfrac yy=1$ , and this also means that $|y|=y$ . So we end up with

$\dfrac{\sqrt[4]{3x^2}}{2y}$

making the last option the correct answer.

7 0

2 years ago

Five swimmers are entered into a competition. Four of the swimmers have had their

Mamont248 [21]

Answer:

Step-by-step explanation:

7 0

3 years ago

A sailboat is sitting at rest near its dock. A rope attached to the bow of the boat is drawn in over a pulley that stands on a p

irina [24]

Answer:

The boat is approaching the dock at rate of 2.14 ft/s

Step-by-step explanation:

The situation given in the question can be modeled as a triangle, please refer to the attached diagram.

A rope attached to the bow of the boat is drawn in over a pulley that stands on a post on the end of the dock that is 5 feet higher than the bow that means x = 5 ft.

The length of rope from bow to pulley is 13 feet that means y = 13 ft.

We know that Pythagorean theorem is given by

$x^{2} + y^{2} = z^{2}$

Differentiating the above equation with respect to time yields,

$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 2z\frac{dz}{dt}$

$x\frac{dx}{dt} + y\frac{dy}{dt} = z\frac{dz}{dt}$

dx/dt = 0 since dock height doesn't change

$y\frac{dy}{dt} = z\frac{dz}{dt}$

$\frac{dy}{dt} = \frac{z}{y} \frac{dz}{dt}$

The rope is being pulled in at a rate of 2 feet per second that is dz/dt = 2 ft/s

First we need to find z

z² = (5)² + (13)²

z² = 194

z = √194

z = 13.93 ft

So,

$\frac{dy}{dt} = \frac{z}{y} \frac{dz}{dt}$

$\frac{dy}{dt} = \frac{13.93}{13}(2)$

$\frac{dy}{dt} = 2.14$ $ft/s$

Therefore, the boat is approaching the dock at rate of 2.14 ft/s

8 0

2 years ago