Ask question

Ask question

All categories

3 years ago

7

Need help with solving and understanding

1 answer:

prohojiy [21]3 years ago

5 0

Let start off with 5/8+7/8= 1 4/8
Now it's 13+1 4/8= 14 4/8 or 14 1/2 :)

You might be interested in

Helpppppppppppppppppp

Sever21 [200]

Answer:

-1

Step-by-step explanation:

Square both sides of the equation. (That is what is meant by "use the inverse relationship.")

$i^2=(\sqrt{-1})^2\\i^2=\boxed{-1} \qquad\text{simplify}$

8 0

3 years ago

Another question math

Genrish500 [490]

-6.7 and -8.8 and 11.(-2)

4 0

3 years ago

Read 2 more answers

Factor the trinomial 2x^2+15x+7

Mashutka [201]

Answer:

(2x+1)(x+7)

Step-by-step explanation:

2x2+15x+7

For a polynomial of the form ax2+bx+c

, rewrite the middle term as a sum of two terms whose product is a⋅c=2⋅7=14 and whose sum is b=15

.

Tap for more steps...

2x2+x+14x+7

Factor out the greatest common factor from each group.

Tap for more steps...

x(2x+1)+7(2x+1)

Factor the polynomial by factoring out the greatest common factor, 2x+1

.

(2x+1)(x+7)

3 0

2 years ago

Which number line shows the solution to the inequality -3x - 5 < -2 ?

Inessa05 [86]

The answer to his is A

3 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

2 years ago