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3 years ago

What statement complete the proof below given 3x+ 7=x-5

Mathematics

2 answers:

dusya [7]3 years ago

5 0

X = -2 ............................... .................... ( I had to add the dots so it would be 20 characters)

Advocard [28]3 years ago

4 0

Proof: x=-6

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Draw at least six different sized rectangles that have an area of 64 square units

olga_2 [115]

Let

x------> the length of the rectangle

y------> the width of the rectangle

we know that

The area of a rectangle is equal to

$A=x*y$

$A=64\ units^{2}$

$x*y=64$ --------> equation 1

let's assume different values of x to get the different values of y

case 1) For x=64 units

substitute in the equation 1

$x*y=64$

$y=64/x$

$y=64/64$

$y=1\ units$

the dimensions of the rectangle are 64 units x 1 unit

see the draw in the attached figure N 1

case 2) For x=32 units

substitute in the equation 1

$x*y=64$

$y=64/x$

$y=64/32$

$y=2\ units$

the dimensions of the rectangle are 32 units x 2 units

see the draw in the attached figure N 2

case 3) For x=16 units

substitute in the equation 1

$x*y=64$

$y=64/x$

$y=64/16$

$y=4\ units$

the dimensions of the rectangle are 16 units x 4 units

see the draw in the attached figure N 3

case 4) For x=30 units

substitute in the equation 1

$x*y=64$

$y=64/x$

$y=64/30$

$y=\frac{32}{15} =2\frac{2}{15} \ units$

the dimensions of the rectangle are 30 units x 2 (2/15) units

see the draw in the attached figure N 4

case 5) For x=40 units

substitute in the equation 1

$x*y=64$

$y=64/x$

$y=64/40$

$y=1.60\ units$

the dimensions of the rectangle are 40 units x 1.60 units

see the draw in the attached figure N 5

case 6) For x=60 units

substitute in the equation 1

$x*y=64$

$y=64/x$

$y=64/60$

$y=\frac{16}{15} =1\frac{1}{15} \ units$

the dimensions of the rectangle are 60 units x 1 (1/15) units

see the draw in the attached figure N 6

3 0

3 years ago

Sin(20)=x/670 ????????

mash [69]

Answer:

229.2

Step-by-step explanation:

Remark

There's a lot in this question. The x is not just any x. It is defined and limited by its relationship with the sine function.

The Sin is defined as Opposite over Hypotenuse in a right triangle. The 670 is the Hypotenuse. You know that by the definition of Sine.

So one thing you should always keep in mind: the x because of where it is in the definition, will always be smaller than the Hypotenuse. So you should get a value less than 670.

Formula

Just as you have it

Sin(20) = x / 670

Solution

Multiply both sides by 570

670 * sin(20) = x

sin(20) = 0.3420

x = 670 * 0.3420

x = 229.2

6 0

3 years ago

110 is 45% of what number

aev [14]

Answer:49.5

Step-by-step explanation:

To find the 45% of 110

45/100×110

=49.5

5 0

3 years ago

Read 2 more answers

What is the area of the polygon

Rina8888 [55]

The easiest way to do this is find the area of the large rectangle and subtract the two missing pieces. The length of the large rectangle is 24 and its height is 16. 24x16=384. The area is each of the two missing pieces is 2x2=4. Since there are two of them multiply by two 4x2=8. Then subtract eight from the first area calculated. 384-8=376 square meters

5 0

3 years ago

A. Evaluate ∫20 tan 2x sec^2 2x dx using the substitution u = tan 2x.

irakobra [83]

Answer:

The integral is equal to $5\sec^2(2x)+C$ for an arbitrary constant C.

Step-by-step explanation:

a) If $u=\tan(2x)$ then $du=2\sec^2(2x)dx$ so the integral becomes $\int 20\tan(2x)\sec^2(2x)dx=\int 10\tan(2x) (2\sec^2(2x))dx=\int 10udu=\frac{u^2}{2}+C=10(\int udu)=10(\frac{u^2}{2}+C)=5\tan^2(2x)+C$ . (the constant of integration is actually 5C, but this doesn't affect the result when taking derivatives, so we still denote it by C)

b) In this case $u=\sec(2x)$ hence $du=2\tan(2x)\sec(2x)dx$ . We rewrite the integral as $\int 20\tan(2x)\sec^2(2x)dx=\int 10\sec(2x) (2\tan(2x)\sec(2x))dx=\int 10udu=5\frac{u^2}{2}+C=5\sec^2(2x)+C$ .

c) We use the trigonometric identity $\tan(2x)^2+1=\sec(2x)^2$ is part b). The value of the integral is $5\sec^2(2x)+C=5(\tan^2(2x)+1)+C=5\tan^2(2x)+5+C=5\tan^2(2x)+C$ . which coincides with part a)

Note that we just replaced 5+C by C. This is because we are asked for an indefinite integral. Each value of C defines a unique antiderivative, but we are not interested in specific values of C as this integral is the family of all antiderivatives. Part a) and b) don't coincide for specific values of C (they would if we were working with a definite integral), but they do represent the same family of functions.

3 0

3 years ago