3. 5 points

Solve for x please help

nordsb [41]

Answer:

x = 87

Step-by-step explanation:

An exterior angle of a triangle is equal to the sum of the opposite interior angles.

Label the angles left and right of the right angle as "a" and "b". Then we have ...

a + b + 90 = 180

b = 90 - a . . . . subtract (90+a)

And we have a relation with the left exterior angle:

72 + a = 138

a = 138 -72 = 66 . . . . subtract 72

__

b = 90 -66 = 24 . . . . from above

And another relation with the right exterior angle:

(180 -x) +b = 117 . . . . . . the interior angle with "b" is the supplement of x

204 -x = 117 . . . . . . . . . substitute 24 for b

204 -117 = x = 87 . . . . add x-117

The measure of the angle marked x° is 87°.

4 0

3 years ago

PLEASEE HELPPPP!!!!! ASAP SEE ATTACHED!

stiks02 [169]

Short answer: C
Let's start with the pen.
The length is x
The width is 3 more.= x + 3
Perimeter of the pen = 2L + 2W
P = 2 L + 2W
P = 2*x + 2(x + 3)
P = 2x + 2x + 6
P = 4x + 6. Leave that for now.

Area of the pen without the doghouse = x(x + 3)
Area of the pen without the doghouse = x^2 + 3x
Area with the doghouse x^2 + 3x - y = 178 (that's his running around space.

Go to the dog house for a Moment
The perimeter of the doghouse and the area are equal and we are told to make them both y.
Area = y
Perimeter = y

Combined facts.
The perimeter of the Pen is three times larger than that of the dog house.
4x + 6 = 3y

The area of the Pen - The area of the doghouse = 178
x^2 + 3x - y = 178

Now what happens is that you have to juggle these two equations so they look like your answer
x^2 + 3x - y = 178
4x + 6 = 3y Subtract 3y from both sides.
4x - 3y + 6 = 0 Subtract 6 from both sides.
4x - 3y = - 6

Answer C

6 0

3 years ago

A large storage tank, open to the atmosphere at the top and filled with water, develops a small hole in its side at a point 10.4

vlabodo [156]

Answer:

Therefore the diameter of the hole is $1.94 \times 10^{-3}$ m.

Step-by-step explanation:

Bernoulli's equation,

$P_1+\frac12 \rho v^2_1+\rho g h_1= P_2+\frac12 \rho v^2_2+\rho g h_2$

P₁ = P₂= atmospheric presser

$\rho$ = density

$\frac12 \rho v^2_1+\rho g h_1= \frac12 \rho v^2_2+\rho g h_2$ [since P₁ = P₂]

$\Rightarrow\rho (\frac12 v^2_1+ g h_1)= \rho(\frac12 v^2_2+ g h_2)$

$\Rightarrow\frac12 v^2_1+ g h_1= \frac12 v^2_2+ g h_2$

$\Rightarrow\frac12 v^2_2-\frac12 v^2_1=g h_1- g h_2$

$\Rightarrow v^2_2- v^2_1=2g h$ [ $h_1-h_2=h$ ]

Here $v_1\approx 0$

$\Rightarrow v^2_2=2g h$

$\therefore v_2=\sqrt {2gh$

Here g= 9.8 m/s² , h = 10.4 m

The velocity of water that leaves from the hole $v_2$ = $\sqrt {2\times 9.8\times 10.4}$ m/s

=14.28 m/s.

Given, the rate of flow from the leak is $2.53\times 10^{-3}$ $m^3/min$

$=\frac{2.53\times 10^{-3}}{60}$ $m^3/s$

Let the diameter of the hole be d.

Then the cross section area of the hole is $=\pi (\frac d2)^2$

We know that,

The rate of flow = Cross section area × speed

$\Rightarrow \frac{2.53\times 10^{-3}}{60} =\pi (\frac d2)^2\times 14.28$

$\Rightarrow (\frac d2)^2=\frac{2.53\times 10^{-3}}{60\times 14.28\times \pi}$

$\Rightarrow d= 1.94 \times 10^{-3}$

Therefore the diameter of the hole is $1.94 \times 10^{-3}$ m.

4 0

4 years ago

What’s the answer?

zubka84 [21]

Answer:

1. Two real solutions

2. Two real solutions

3. One real solutions

4. Two real solutions

5. Two real solutions

5 0

3 years ago

Given cos theta=4/9 and csc theta 0 find sin theta and tan theta

Korvikt [17]

The correct question is
<span>Given cos theta=4/9 and csc theta < 0 find sin theta and tan theta
</span>
we know that
csc theta=1/sin theta
if csc theta < 0
then
sin theta < 0

we have that
<span>cos theta=4/9

we know that
sin</span>² theta+cos² theta=1
so
sin² theta=1-cos² theta-----> 1-(4/9)²----> 1-(16/81)----> 65/81
sin theta=-√(65/81)---->-√65/9

the answer Part a) is
sin theta=-√65/9

Part b) find tan theta

tan theta=sin theta/cos theta
tan theta=(-√65/9)/(4/9)-----> tan theta=-√65/4

the answer part b) is
tan theta=-√65/4

5 0

3 years ago

Read 2 more answers