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3 years ago

Addition. 9.69 + 1.4

Mathematics

2 answers:

zzz [600]3 years ago

8 0

The answer is 11.09. you add .4 to .6 and get 1. then you add 1 to 9 and get 10. 10 + 1 = 11 + .09 and you get 11.09

Allushta [10]3 years ago

5 0

Your answer is 11.09. Since the 4 in 1.4 is in the same spot the 6 is in 9.69 you add the 4 and 6 which makes 10. 10 is a whole so you would move the one up to 9 +1 =10 then +1which would be 11. Now since there is no number behind the 4 in 1.4 the last 9 in 9.69 will be the last number so it would be 11.09.

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-3^(-x)-6=-3^x+10<br><br>Solve the equation below for x by graphing plz

vitfil [10]

To graph it, just graph $y=-3^{-x}-6$ and $y=-3^x+10$ and see where they intersect

I would like to solve it by using math and not graphing
if you don't want to see the math, just don't scroll down
the graphical meathod is above, first line, just read it

hmm
multiply both sides by -1
$3^{-x}+6=3^x-10$
multiply both sides by $3^x$
$3^0+6(3^x)=3^{2x}-10(3^x)$
$1+6(3^x)=3^{2x}-10(3^x)$
minus 1 from both sides and minus 6(3^x) from both sides
$0=3^{2x}-16(3^x)-1$
use u subsitution
$u=3^x$
we can rewrite it as
$0=u^2-16u-1$
now factor
I mean use quadratic formula
for $ax^2+bx+c=0$ $x=\frac{-b+/-\sqrt{b^2-4ac}}{2a}$
so for 0=u^2-16u-1, a=1, b=-16, c=-1
$u=\frac{-(-16)+/-\sqrt{(-16)^2-4(1)(-1)}{2(1)}$
$u=8+/-\sqrt{65}$
remember that u=3^x so u>0
if we have u=8+√65, it's fine, but u=8-√65 is negative and not allowed
so therfor
$u=8+\sqrt{65}=3^x$
$8+\sqrt{65}=3^x$
if you take the log base 3 of both sides you get
$log_3(8+\sqrt{65})=x$
if you use ln then
$ln(8+\sqrt{65})=xln(3)$
then
$\frac{ln(8+\sqrt{65})}{ln(3)}=x$

8 0

3 years ago

Read 2 more answers

The ratio of the surface areas of two similar solids is 4:49. What is the ratio of a volumes?

Fynjy0 [20]

Answer: 8/343

Step-by-Step:

If two similar solids have a scale factor of a:b then,

The ratio of their surface areas (2D) is- a^2:b^2

The ratio of their volume (3D) is- a^3:b^3

First you want to square root the scale factor numbers to find the original value. Then you cube the original numbers to find the answer.

6 0

4 years ago

Consider the following differential equation to be solved by undetermined coefficients. y(4) − 2y''' + y'' = ex + 1 Write the gi

kompoz [17]

Answer:

The general solution is

$y= (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x +\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))$

+ $\frac{x^2}{2}$

Step-by-step explanation:

Step :1:-

Given differential equation y(4) − 2y''' + y'' = e^x + 1

The differential operator form of the given differential equation

$(D^4 -2D^3+D^2)y = e^x+1$

comparing f(D)y = e^ x+1

The auxiliary equation (A.E) f(m) = 0

$m^4 -2m^3+m^2 = 0$

$m^2(m^2 -2m+1) = 0$

$(m^2 -2m+1) this is the expansion of (a-b)^2$

$m^2 =0 and (m-1)^2 =0$

The roots are m=0,0 and m =1,1

complementary function is $y_{c} = (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x$

<u>Step 2</u>:-

The particular equation is $\frac{1}{f(D)} Q$

P.I = $\frac{1}{D^2(D-1)^2} e^x+1$

P.I = $\frac{1}{D^2(D-1)^2} e^x+\frac{1}{D^2(D-1)^2}e^{0x}$

P.I = $I_{1} +I_{2}$

$\frac{1}{D^2} (\frac{x^2}{2!} )e^x + \frac{1}{D^{2} } e^{0x}$

$\frac{1}{D} means integration$

$\frac{1}{D^2} (\frac{x^2}{2!} )e^x = \frac{1}{2D} \int\limits {x^2e^x} \, dx$

applying in integration u v formula

$\int\limits {uv} \, dx = u\int\limits {v} \, dx - \int\limits ({u^{l}\int\limits{v} \, dx } )\, dx$

$I_{1}$ = $\frac{1}{D^2(D-1)^2} e^x$

$\frac{1}{2D} (e^x(x^2)-e^x(2x)+e^x(2))$

$\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))$

$I_{2}$ $= \frac{1}{D^2(D-1)^2}e^{0x}$

$\frac{1}{D} \int\limits {1} \, dx= \frac{1}{D} x$

again integration $\frac{1}{D} x = \frac{x^2}{2!}$

The general solution is $y = y_{C} +y_{P}$

$y= (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x +\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))$

+ $\frac{x^2}{2!}$

3 0

3 years ago

will the contents of a cylindrical can of soup with a 1.5-inch radius and a 5-inch height fit into a pan with a 4-inch radius an

Anna007 [38]

Answer:

Answer: The volume of the soup can is less that that of the pan, so the soup will fit in the pan.

Step-by-step explanation:

a. V = s3

V = (7 km)3

V = 7 km × 7 km × 7 km

V = 343 km3

b. V = s3

V = (11 yd.)3

V = 11 yd. × 11 yd. × 11 yd.

V = 1,331 km cu. yd.

a. V = l × w × h

V = 6 m × 2 m × 3 m

V = 36 m3

b. V = l × w × h

V = 5 ft. × 10 ft. × 8 ft.

V = 400 cu. ft.

a. V = πr2h

V = 3.14 × (3 in.)2 × 3 in.

V = 3.14 × 9 sq. in. × 3 in.

V = 84.78 cu. in.

b. V = πr2h

V = 3.14 × (1 m)2 × 4 m

V = 3.14 × 1 m2 × 4 m

V = 12.56 m3

V = s3

V = (2 ft.)3

V = 2 ft. × 2 ft. × 2 ft.

V = 8 cu. ft.

Answer: The capacity of the ice chest is 8 cubic feet.

V = πr2h

V = 3.14 × (1 ft.)2 × 3 ft.

V = 3.14 × 1 sq. ft. × 3 ft.

V = 9.42 cu. ft.

Answer: The volume of the trash can is 9.42 cubic feet.

Closet #1

V = l × w × h

V = 4 ft. × 5 ft. × 8 ft.

V = 160 cu. ft.

Closet #2

V = l × w × h

V = 6 ft. × 3 ft. × 8 ft.

V = 144 cu. ft.

Compare the size of the closets

160 cu. ft. > 144 cu. ft.

Answer: The first closet is larger.

Both the can and the pan are cylindrical shapes. Compare the volumes of the two objects.

Volume of Can

V = πr2h

V = 3.14 × (1.5 in.)2 × 5 in.

V = 3.14 × 2.25 sq. in. × 5 in.

V = 35.325 cu. in.

Volume of Pan

V = πr2h

V = 3.14 × (4 in.)2 × 3 in.

V = 3.14 × 16 sq. in.× 3 in.

V = 150.72 cu. in.

35.325 cu. in. soup can < 150.72 cu. in pan

Answer: The volume of the soup can is less that that of the pan, so the soup will fit in the pan.

3 0

3 years ago

What’s the Answer to this I got 24 somehow

Bess [88]

Answer:

Step-by-step explanation:

So we have the expression:

$\frac{2x+y}{xy-2x}$

And we want to evaluate it for x=-4 and y=8.

So, substitute:

$\frac{2(-4)+(8)}{(-4)(8)-2(-4)}$

Multiply:

$=\frac{-8+8}{-32+8}$

Add:

$=\frac{0}{-24}$

Evaluate:

$=0$

So, our answer is 0.

And we're done!

6 0

4 years ago