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3 years ago

I need help with this! I'll give you a scobby snack if you explain it and get the right answer!

Mathematics

2 answers:

zubka84 [21]3 years ago

8 0

The basic idea is that multiplying by 10 just moves the decimal point one place to the right. It makes the number bigger.

9.3 x 10 = 93

9.3 x 10 x 10 = 93 x 10 = 930

Each time you multiply by another 10, the decimal point moves one more time.

10^4 says you're going to multiply by 10 four times in a row. So that will move the decimal point 4 places to the right.

So what do you get if you move the decimal in 9.3 four places to the right?

dem82 [27]3 years ago

7 0

Answer:

9300

See the attachment for explanaton

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A bottle of shampoo costs $4 at a store. It is on sale for 25% off, and the sales tax in 5.5%. What is the sales price of the sh

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1.22 is the price of the shampoo after sales tax

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3 years ago

5 ^ 3 + 1/2 ^ 2. I need help.

sineoko [7]

Answer:

125.25

Step-by-step explanation:

5 cube =125

1/2=0.5

0.5 squared =0.25

125+0.25=125.25

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4 years ago

X ^ (2) y '' - 7xy '+ 16y = 0, y1 = x ^ 4

nignag [31]

Standard reduction of order procedure: suppose there is a second solution of the form $y_2(x)=v(x)y_1(x)$ , which has derivatives

$y_2=vx^4$
${y_2}'=v'x^4+4vx^3$
${y_2}''=v''x^4+8v'x^3+12vx^2$

Substitute these terms into the ODE:

$x^2(v''x^4+8v'x^3+12vx^2)-7x(v'x^4+4vx^3)+16vx^4=0$
$v''x^6+8v'x^5+12vx^4-7v'x^5-28vx^4+16vx^4=0$
$v''x^6+v'x^5=0$

and replacing $v'=w$ , we have an ODE linear in $w$ :

$w'x^6+wx^5=0$

Divide both sides by $x^5$ , giving

$w'x+w=0$

and noting that the left hand side is a derivative of a product, namely

$\dfrac{\mathrm d}{\mathrm dx}[wx]=0$

we can then integrate both sides to obtain

$wx=C_1$
$w=\dfrac{C_1}x$

Solve for $v$ :

$v'=\dfrac{C_1}x$
$v=C_1\ln|x|+C_2$

Now

$y=C_1x^4\ln|x|+C_2x^4$

where the second term is already accounted for by $y_1$ , which means $y_2=x^4\ln x$ , and the above is the general solution for the ODE.

4 0

4 years ago

EX2. Find the solution of the following matrix using

stealth61 [152]

Answer:

please post a more cohesive question. The values are very confusing

7 0

3 years ago

Sand is being dumped from a conveyor belt and forms a conical pile. Assuming that the height of this cone is always exactly 3 ti

viva [34]

Answer:

dh/dt = 0.4 m/min

Step-by-step explanation:

The volume of the cone is:

V(c) = (1/3)*r² *h if always h = 3r then r = h/3

The volume of the cone as a function of h will be:

V(h) = (1/3)* (h/3)²*h

V(h) = (1/27)*h³

The increasing rate of the volume is equal to the rate of sand added the:

D(V)/dt = (1/27)*3*h²*dh/dt

D(v) / dt = 10 m³/min

h = 15 m and dh/dt is the rate of increasing of the height

By substitution

10 m³/min = ( 1/9)* 225 * dh/dt (m²)

dh/dt = 90 / 225 m/min

dh/dt = 0.4 m/min

3 0

3 years ago