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3 years ago

What do i do here please (just tell me what to do)

Mathematics

1 answer:

sweet-ann [11.9K]3 years ago

6 0

Construct a lunge segment from Side AB to side AC but make sure it is at a 90 degree angle or at least make it as close as possible.

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Solve for x: 1 < x + 3 < 4

Otrada [13]

Answer:

The answer is D

Step-by-step explanation:

$1 - 3 < x < 4 - 3 \\ - 2 < x < 1$

8 0

3 years ago

Find the value of x:

bixtya [17]

Answer:

Step-by-step explanation:

note: your question doesn't contain enough information

so I took it as midsegment

my answer is correct if it's midsegment

6 0

3 years ago

Read 2 more answers

The water from a fire hose follows a path described by y equals 2.0 plus 0.9 x minus 0.10 x squared (units are in meters). If

Alecsey [184]

Answer:

The resultant velocity is 12.21 m/s.

Step-by-step explanation:

We are given that the water from a fire hose follows a path described by y equals 2.0 plus 0.9 x minus 0.10 x squared (units are in meters).

Also, v Subscript x is constant at 10.0 m/s.

The water from a fire hose follows a path described by the following equation below;

$y=2.0 + 0.9x-0.10x^{2}$

The velocity of the $x$ component is constant at = $v_x=10.0 \text{ m/s}$

and the point at which resultant velocity has to be calculated is (9.0,2.0).

Let the velocity of x and y component be represented as;

$v_x=\frac{dx}{dt} \text{ and } v_y=\frac{dy}{dt}$

Now, differentiating the above equation with respect to t, we get;

$y=2.0 + 0.9x-0.10x^{2}$

$\frac{dy}{dt} =0 + 0.9\frac{dx}{dt} -(0.10\times 2)\frac{dx}{dt}$

$\frac{dy}{dt} = 0.9\frac{dx}{dt} -0.2\frac{dx}{dt}$

$v_y = 0.9v_x -0.2v_x$

$v_y = 0.7v_x$

Now, putting $v_x=10.0 \text{ m/s}$ in the above equation;

$v_y = 0.7 \times 10.0$ = 7 m/s

Now, the resultant velocity is given by = $v=\sqrt{v_x^{2}+v_y^{2} }$

$v=\sqrt{10^{2}+7^{2} }$

= $\sqrt{149}$ = 12.21 m/s

5 0

3 years ago

If lines l and m are parallel, then ∠2 and ∠3 are _______ angles.

Likurg_2 [28]

Answer=corresponding

3 0

3 years ago

6÷<img src="https://tex.z-dn.net/?f=%5Csqrt%7B9%7D" id="TexFormula1" title="\sqrt{9}" alt="\sqrt{9}" align="absmiddle" class="la

Ierofanga [76]

Step-by-step explanation:

6÷√9-2³×6

6÷3-8×6

2-48

-46

6 0

3 years ago