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Nikolay [14]
2 years ago
5

17. What expression is equivalent to log(200) - log (2)? Calculate the answer.

Mathematics
1 answer:
IRINA_888 [86]2 years ago
3 0

Answer: 2

Step-by-step explanation:

Recall from the laws of Logarithms:

Log a - Log b = Log ( a/b )

That means

Log 200 - Log 2 = Log ( 200/2)

= Log 100 , which could be written as

Log 10^{2}

Recall from laws of Logarithms:

Log a^{b} = b Log a

Therefore:

Log10^{2} = 2 Log 10

Also from law of Logarithm

Log 10 = 1

Therefore 2 Log 10 = 2 x 1

= 2

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maxonik [38]

Answer:

(0,4) and (5,-3)

Step-by-step explanation:

7 0
2 years ago
Y= 1/2x-4 and y= -2x-9 what is the solution of equations
azamat

Answer:

(x,y)=(-2,-5)

Step-by-step explanation:

y=1/2x-4

y=-2x-9

(y=) 1/2x-4=-2x-9

1/2x +2x=-9+4

1/2x +4/2x=-5

5/2x=-5

5x=-5*2

5x=-10

x=-2

x=-2

y=-2x-9=-2*(-2)-9=4-9=-5

(x,y)=(-2,-5)

8 0
3 years ago
Someone please help me with this​
Sedbober [7]

Answer:

The area of the shaded figure is:

  • <u>20 units^2</u>

Step-by-step explanation:

To obtain the area of the shaded figure, first, you must calculate this as a rectangle, with the measurements: wide (4 units), and long (6 units):

  • Area of a rectangle = long * wide
  • Area of a rectangle = 6 * 4
  • Area of a rectangle = 24 units^2

How the figure isn't a rectangle, you must subtract the triangle on the top, so, now we calculate the area of that triangle with measurements: wide (4 units), and height (2 units):

  • Area of a triangle = \frac{wide*height}{2}
  • Area of a triangle = \frac{4*2}{2}
  • Area of a triangle =\frac{8}{2}
  • Area of a triangle = 4 units^2

In the end, you subtract the area of the triangle to the area of the rectangle, to obtain the area of the shaded figure:

  • Area of the shaded figure = Area of the rectangle - Area of the triangle
  • Area of the shaded figure = 24 units^2 - 4 units^2
  • <u>Area of the shaded figure = 20 units^2</u>

I use the name "units" because the exercise doesn't say if they are feet, inches, or another, but you can replace this in case you need it.

5 0
2 years ago
Someone please help me on this!
Alexxx [7]

Answer & Step-by-step explanation:

This can be proven with the SAS theorem (side-angle-side)

With a perpendicular bisector, the line it bisects is cut directly in half. This creates two equal sides:

GJ=IJ

and it creates two 90° angles:

∠GJH=∠IJH

And because of the reflexive property of congruence:

JH=JH

Side-Angle-Side.

:Done

6 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
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