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Semenov [28]
3 years ago
13

Triangle WXY is isosceles. ∠YWX and ∠YXW are the base angles. YZ bisects ∠WYX. m∠XYZ = (15x)°. m∠YXZ = (2x + 5)°. What is the me

asure of ∠WYX? 5° 15° 75° 150°
Mathematics
2 answers:
ozzi3 years ago
8 0

Answer:

150 degrees

Step-by-step explanation:

I just answered it on edginuty

kramer3 years ago
8 0

Answer:

D) 150 degrees

Step-by-step explanation:

on edge

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What is the 10th of a quarter of 1600
mihalych1998 [28]

Answer:

40

Step-by-step explanation:

A quarter of 1600 is 400. A tenth of that is 40.

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4 0
3 years ago
Read 2 more answers
The quotient of a number and 3 is 8
ahrayia [7]

Answer:

number = 24

Step-by-step explanation:

3 x 8 = 24

24 / 3 = 8

4 0
3 years ago
The total cost, in dollars, to order x units of a certain product is modeled by C(x)=5x²+320 . According to the model, for what
lora16 [44]

Answer:

The minimum cost per unit is obtained for an order of 8 units.

Step-by-step explanation:

Since the total cost is modeled by;

C(x)=5x²+320

Then;

1 unit costs; C(x)=5(1)²+320 = 325

cost per unit 325/1 = 325

8 units costs; C(x)=5(8)²+320 = 640

cost per unit = 640/8 = 80

80 units costs; C(x)=5(80)²+320 = 32320

Cost per unit = 32320/80 = 404

The minimum cost per unit is obtained for an order of 8 units.

7 0
3 years ago
Find the components of the vertical force Bold Upper FFequals=left angle 0 comma negative 10 right angle0,−10 in the directions
quester [9]

Solution :

Let $v_0$ be the unit vector in the direction parallel to the plane and let $F_1$ be the component of F in the direction of v_0 and F_2 be the component normal to v_0.

Since, |v_0| = 1,

$(v_0)_x=\cos 60^\circ= \frac{1}{2}$

$(v_0)_y=\sin 60^\circ= \frac{\sqrt 3}{2}$

Therefore, v_0 = \left

From figure,

|F_1|= |F| \cos 30^\circ = 10 \times \frac{\sqrt 3}{2} = 5 \sqrt3

We know that the direction of F_1 is opposite of the direction of v_0, so we have

$F_1 = -5\sqrt3 v_0$

    $=-5\sqrt3 \left$

    $= \left$

The unit vector in the direction normal to the plane, v_1 has components :

$(v_1)_x= \cos 30^\circ = \frac{\sqrt3}{2}$

$(v_1)_y= -\sin 30^\circ =- \frac{1}{2}$

Therefore, $v_1=\left< \frac{\sqrt3}{2}, -\frac{1}{2} \right>$

From figure,

|F_2 | = |F| \sin 30^\circ = 10 \times \frac{1}{2} = 5

∴  F_2 = 5v_1 = 5 \left< \frac{\sqrt3}{2}, - \frac{1}{2} \right>

                   $=\left$

Therefore,

$F_1+F_2 = \left< -\frac{5\sqrt3}{2}, -\frac{15}{2} \right> + \left< \frac{5 \sqrt3}{2}, -\frac{5}{2} \right>$

           $= = F$

3 0
3 years ago
What is the absolute value of -250?
Andrei [34K]

Answer:

250

Step-by-step explanation:

Pretty much every time it asks for the abasloute value it just asks how much to get the zero on the number line. It will ALWAYS be positive. For example -5 will be 5 to get to zero on the numberline. Same thing with a positive 5.

6 0
3 years ago
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