Ask question

Ask question

All categories

Artyom0805 [142]

2 years ago

7

Please help It due today MATH!!1

2 answers:

8_murik_8 [283]2 years ago

8 0

Answer:

Solid line y is less than or equal to -1+4

Dashed line y is greater than or equal to 1/3

Step-by-step explanation:

PLZ MARK BRAINLIEST

Klio2033 [76]2 years ago

4 0

Answer:

h bjwkjdfrekojfhrkeifcuhenjdc

Step-by-step explanation:

You might be interested in

Acone has a height of 2.5 in, and a radius of 5 in. What is the volume of the cone? (Use 3.14 for s. Round the answer to two

arlik [135]

Answer:

Step-by-step explanation:

8 0

3 years ago

345+ 223<br> Use rounding or compatible numbers to estimate the sum

Mars2501 [29]

I think it is 568 if I’m not mistaken

8 0

3 years ago

a waiter had nine tables he was waiting on, with seven women and three men at each table. how many customees total did the waite

Greeley [361]

Since each table had 7 women and 3 men at each table, that would mean each table has 10 people. Since there are 9 tables he was waiting on, and 10 people at each table, multiply 9 x 10, and you will find that he served 90 people in total.

8 0

3 years ago

Read 2 more answers

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

2 years ago

What is the length of side a?

Bad White [126]

$\huge \bf༆ Answer ༄$

According to given figure, The Triangle BDC is a right angled Triangle (Angle D = 90°)

So, we can apply Pythagoras theorem to find the Side a (Hypotenuse)

Now, let's solve ~

$\sf \: BD {}^{2} + CD {}^{2} = BC {}^{2}$

$\sf4 {}^{2} + {3}^{2} = {a}^{2}$

$\sf16 + 9 = {a}^{2}$

$\sf \: a {}^{2} = 25$

$\sf \: a = \sqrt{25}$

$\sf \: a = 5 \: units$

Therefore, Length of side a is 5 units ~

6 0

2 years ago