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Bad White [126]

2 years ago

13

Which is the solution to the equation -6x-15=-1.5

2 answers:

sergey [27]2 years ago

7 0

Answer:

x = -2.25

Step-by-step explanation:

cupoosta [38]2 years ago

5 0

Answer:

x = -2.25

Step-by-step explanation:

Add 15 to both sides of the equation

-6x - 15 = -1.5

-6x-15+15=-1.5+15

Simplify

-6x = 13.5

Divide both sides of the equation by the same term

-6x/6 = 13.5/6

Simplify

x = -2.25

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A circular pool is 3.5 meters deep and has a radius of 9 meters. What is the volume

mote1985 [20]

Answer:

$890.75m^3$

Step-by-step explanation:

Step one:

given data

heigth/depth = 3.5m

radius= 9m

Step two:

The shape is that of a cylinder

Hence the volume of the pool is

$V= \pi r^2h$

substitute

$V= 3.142*9^2*3.5\\\\V=890.75 m^3$

7 0

3 years ago

$10,788 per dozen, find for EACH computer (a) dollar markup, (b) percent markup on cost, and (c) percent markup on selling price

swat32

It’s A dollar markup

5 0

3 years ago

Two streams flow into a reservoir. Let X and Y be two continuous random variables representing the flow of each stream with join

zlopas [31]

Answer:

c = 0.165

Step-by-step explanation:

Given:

f(x, y) = cx y(1 + y) for 0 ≤ x ≤ 3 and 0 ≤ y ≤ 3,

f(x, y) = 0 otherwise.

Required:

The value of c

To find the value of c, we make use of the property of a joint probability distribution function which states that

$\int\limits^a_b \int\limits^a_b {f(x,y)} \, dy \, dx = 1$

where a and b represent -infinity to +infinity (in other words, the bound of the distribution)

By substituting cx y(1 + y) for f(x, y) and replacing a and b with their respective values, we have

$\int\limits^3_0 \int\limits^3_0 {cxy(1+y)} \, dy \, dx = 1$

Since c is a constant, we can bring it out of the integral sign; to give us

$c\int\limits^3_0 \int\limits^3_0 {xy(1+y)} \, dy \, dx = 1$

Open the bracket

$c\int\limits^3_0 \int\limits^3_0 {xy+xy^{2} } \, dy \, dx = 1$

Integrate with respect to y

$c\int\limits^3_0 {\frac{xy^{2}}{2} +\frac{xy^{3}}{3} } \, dx (0,3}) = 1$

Substitute 0 and 3 for y

$c\int\limits^3_0 {(\frac{x* 3^{2}}{2} +\frac{x * 3^{3}}{3} ) - (\frac{x* 0^{2}}{2} +\frac{x * 0^{3}}{3})} \, dx = 1$

$c\int\limits^3_0 {(\frac{x* 9}{2} +\frac{x * 27}{3} ) - (0 +0) \, dx = 1$

$c\int\limits^3_0 {(\frac{9x}{2} +\frac{27x}{3} ) \, dx = 1$

Add fraction

$c\int\limits^3_0 {(\frac{27x + 54x}{6}) \, dx = 1$

$c\int\limits^3_0 {\frac{81x}{6} \, dx = 1$

Rewrite;

$c\int\limits^3_0 (81x * \frac{1}{6}) \, dx = 1$

The $\frac{1}{6}$ is a constant, so it can be removed from the integral sign to give

$c * \frac{1}{6}\int\limits^3_0 (81x ) \, dx = 1$

$\frac{c}{6}\int\limits^3_0 (81x ) \, dx = 1$

Integrate with respect to x

$\frac{c}{6} * \frac{81x^{2}}{2} (0,3) = 1$

Substitute 0 and 3 for x

$\frac{c}{6} * \frac{81 * 3^{2} - 81 * 0^{2}}{2} = 1$

$\frac{c}{6} * \frac{81 * 9 - 0}{2} = 1$

$\frac{c}{6} * \frac{729}{2} = 1$

$\frac{729c}{12} = 1$

Multiply both sides by $\frac{12}{729}$

$c = \frac{12}{729}$

$c = 0.0165 (Approximately)$

8 0

3 years ago

A fastball is hit straight up over home plate. The ball's height, h (in feet), from the ground is modeled by ℎ = −16푡 ଶ+103푡+5 w

Liono4ka [1.6K]

Question:

A fastball is hit straight up over home plate. The ball's height, h (in feet), from the ground is modeled by h(t)=-16t^2+80t+5, where t is measured in seconds. Write an equation to determine how long it will take for the ball to reach the ground.

Answer:

$t = 5.0625$

Step-by-step explanation:

Given

$h(t)=-16t^2+80t+5$

Required

Find t when the ball hits the ground

This implies that h(t) = 0

So, we have:

$0=-16t^2+80t+5$

Reorder

$-16t^2+80t+5 = 0$

Using quadratic formula, we have:

$t = \frac{-b\±\sqrt{b^2 - 4ac}}{2a}$

Where

$a = -16$ $b =80$ $c = 5$

So, we have:

$t = \frac{-80\±\sqrt{80^2 - 4*-16*5}}{2*-16}$

$t = \frac{-80\±\sqrt{6400 +320}}{-32}$

$t = \frac{-80\±\sqrt{6720}}{-32}$

$t = \frac{-80\±82.0}{-32}$

This gives:

$t = \frac{-80+82.0}{-32}$ or $t = \frac{-80-82.0}{-32}$

$t = \frac{2}{-32}$ or $t = \frac{-162}{-32}$

$t = -\frac{2}{32}$ or $t = \frac{162}{32}$

But time can not be negative.

So, we have:

$t = \frac{162}{32}$

$t = 5.0625$

<em>Hence, time to hit the ground is 5.0625 seconds</em>

3 0

3 years ago

5x=2x+24 what the anwser

d1i1m1o1n [39]

5x=2x+24
5x-2x=24
3x=24
x=24/3
x=8

Now you can verify by putting 8 in the x :)

5 0

3 years ago

Read 2 more answers