Solve for m -3/7m=12 A: M= -28 B: M= -26 C: M= 26 D: M= 28

The earth has a mass of approximately 6\cdot 10^{24}6⋅10 24 6, dot, 10, start superscript, 24, end superscript kilograms (\text{

Alex_Xolod [135]

Answer:

0.02

Step-by-step explanation:

The volume of the earth's oceans is approximately 1.34\cdot 10^{9}1.34⋅10

9

1, point, 34, dot, 10, start superscript, 9, end superscript cubic kilometers (\text{km}^3)(km

3

)left parenthesis, start text, k, m, end text, cubed, right parenthesis, and ocean water has a mass of about 1.03 \cdot 10^{12}\,\dfrac{\text{kg}}{\text{km}^3}1.03⋅10

12

km

3

kg

1, point, 03, dot, 10, start superscript, 12, end superscript, start fraction, start text, k, g, end text, divided by, start text, k, m, end text, cubed, end fraction .

To simplify, we will use the product of powers property of exponents that says that x^a\cdot x^b = x^{a+b}x

a

⋅x

b

=x

a+b

x, start superscript, a, end superscript, dot, x, start superscript, b, end superscript, equals, x, start superscript, a, plus, b, end superscript.

\qquad 1.34\cdot 10^{9}\,\cancel{\text{km}^3} \cdot 1.03 \cdot 10^{12}\,\dfrac{\text{kg}}{\cancel{\text{km}^3}} = 1.3802 \cdot 10^{21}\,\text{kg}1.34⋅10

9

km

3

⋅1.03⋅10

12

km

3

kg

=1.3802⋅10

21

kg1, point, 34, dot, 10, start superscript, 9, end superscript, start cancel, start text, k, m, end text, cubed, end cancel, dot, 1, point, 03, dot, 10, start superscript, 12, end superscript, start fraction, start text, k, g, end text, divided by, start cancel, start text, k, m, end text, cubed, end cancel, end fraction, equals, 1, point, 3802, dot, 10, start superscript, 21, end superscript, start text, k, g, end text

Hint #2

Next we want to know what portion of the earth's mass this represents. We have:

\qquad \begin{aligned} \dfrac{\text{mass of the oceans}}{\text{total mass of the earth}} &= \dfrac{1.3802 \cdot 10^{21}\,\text{kg}}{6\cdot 10^{24}\,\text{kg}} \\\\ &= \dfrac{1.3802}{6\cdot 10^{3}} \\\\ &= \dfrac{1.3802}{6000} \\\\ &= 0.0002300\overline{3} \end{aligned}

total mass of the earth

mass of the oceans

=

6⋅10

24

kg

1.3802⋅10

21

kg

=

6⋅10

3

1.3802

=

6000

1.3802

=0.0002300

3

To convert this to a percent, we multiply by 100100100, so the oceans represent 0.02300\overline{3}\%0.02300

3

%0, point, 02300, start overline, 3, end overline, percent of the earth's total mass, according to these figures.

Hint #3

To the nearest hundredth of a percent, 0.020.020, point, 02 percent of the earth's mass is from oceans.

6 0

2 years ago

Help me with this lol value of( 2^3)^2

ss7ja [257]

Answer:

64

Step-by-step explanation:

first, we'll start with the value inside the parenthesis, which is 2³

2³ means to multiply 2 by itself 3 times

so it would be

2×2×2=8

which then means that the expression is now (8)²

now we need to do (8)², which is to multiply 8 by itself twice

8*8=64

hope this helps!

6 0

2 years ago

What are the excluded values of x for x+4/-3x^2+12x+36

LekaFEV [45]

I will assume that you meant:

(x+4)/(-3x^2+12x+36) factor the denominator...

(x+4)/(-3(x^2-4x-12))

To factor a quadratic of the form ax^2+bx+c you need to find two values, j and k, which satisfy two conditions...

jk=ac=-12 and j+k=b=-4 so j and k must be 2 and -6 and the factors are then:

(x+2)(x-6) so we are now left with:

(x+4)/(-3(x+2)(x-6))

The only restriction on this function is that division by zero is undefined, so x cannot equal -2 or 6

3 0

3 years ago

Read 2 more answers

“An architect is planning to incorporate several stone spheres of different sizes into the landscaping of a public park, and wor

Artist 52 [7]

Answer:

Part 1) The value that is closest to the cost of finishing a sphere with a 5.50-meter circumference is $900

Part 2) The value that is closest to the cost of finishing a sphere with a 7.85-meter circumference is $1,800

Step-by-step explanation:

Step 1

Find the radius of each sphere

we know that

The circumference of a circle is equal to

$C=2\pi r$

Find the radius of the sphere with a 5.50-meter circumference

For $C=5.50\ m$

assume

$\pi =3.14$

substitute and solve for r

$5.50=2(3.14)r$

$r=5.50/[2(3.14)]=0.88\ m$

Find the radius of the sphere with a 7.85-meter circumference

For $C=7.85\ m$

assume

$\pi =3.14$

substitute and solve for r

$7.85=2(3.14)r$

$r=7.85/[2(3.14)]=1.25\ m$

step 2

Find the surface area of each sphere

The surface area of sphere is equal to

$SA=4\pi r^{2}$

Find the surface area of sphere with a 5.50-meter circumference

For $r=0.88\ m$

assume

$\pi =3.14$

substitute

$SA=4(3.14)(0.88)^{2}$

$SA=9.73\ m^{2}$

Find the surface area of sphere with a 7.85-meter circumference

For $r=1.25\ m$

assume

$\pi =3.14$

substitute

$SA=4(3.14)(1.25)^{2}$

$SA=19.63\ m^{2}$

step 3

Find the cost of finishing each sphere

we know that

To find out the cost , multiply the surface area by $92 per square meter

Find the cost of sphere with a 5.50-meter circumference

$9.73*(92)=\$895.16$

therefore

The value that is closest to the cost of finishing a sphere with a 5.50-meter circumference is $900

Find the cost of sphere with a 7.85-meter circumference

$19.63*(92)=\$1,805.96$

therefore

The value that is closest to the cost of finishing a sphere with a 7.85-meter circumference is $1,800

6 0

3 years ago

Assume the hold time of callers to a cable company is normally distributed with a mean of 5.5 minutes and a standard deviation o

Ierofanga [76]

Answer:

The percent of callers are 37.21 who are on hold.

Step-by-step explanation:

Given:

A normally distributed data.

Mean of the data, $\mu$ = 5.5 mins

Standard deviation, $\sigma$ = 0.4 mins

We have to find the callers percentage who are on hold between 5.4 and 5.8 mins.

Lets find z-score on each raw score.

⇒ $z_1=\frac{x_1-\mu}{\sigma}$ ...raw score, $x_1$ = $5.4$

⇒ Plugging the values.

⇒ $z_1=\frac{5.4-5.5}{0.4}$

⇒ $z_1=-0.25$

For raw score 5.5 the z score is.

⇒ $z_2=\frac{5.8-5.5}{0.4}$

⇒ $z_2=0.75$

Now we have to look upon the values from Z score table and arrange them in probability terms then convert it into percentages.

We have to work with P(5.4<z<5.8).

⇒ $P(5.4$

⇒ $P(-0.25$

⇒ $P(z$

⇒ $z(1.5)=0.7734$ and $z(-0.25)=0.4013$ ...from z -score table.

⇒ $0.7734-0.4013$

⇒ $0.3721$

To find the percentage we have to multiply with 100.

⇒ $0.3721\times 100$

⇒ $37.21$ %

The percent of callers who are on hold between 5.4 minutes to 5.8 minutes is 37.21

4 0

2 years ago