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3 years ago

Solve for x. 2(3 + x) = 5x + 54A. 17B. 16C. -16D. -17

Mathematics

2 answers:

nataly862011 [7]3 years ago

5 0

Answer:

X= -12

Step-by-step explanation:

Hope this helps please make me the brainliest!

ddd [48]3 years ago

4 0

Answer: x = -16; correct option is C

Step-by-step explanation:

From the question, 2(3 + x) = 5x + 54

First open the bracket

6 + 2x = 5x +54

Combine like terms

6 - 54 = 5x - 2x

-48 = 3x

Divide both sides by 3

x = -48/3

x = -16

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This is slopes is this supposed to be negative or positive????

Shtirlitz [24]

This slope will be zero

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3 years ago

A radar is designed to report the track of an aircraft every 2 seconds. If this radar reports 15 reports 15 tracks in one minute

meriva

Answer:

Step-by-step explanation:

Since radar reports 30 tracks in one minute in which 60 seconds makes 1 minute

Let calculate using this formula

Using this formula

Percentage of time=Tracks reported in minutes/60 seconds*100

Let plug in the formula

Percentage of time=30 tracks/60 seconds*100

Percentage of time=0.5*100

Percentage of time=50%

Inconclusion The percentage of the time that radar track the aircraft is 50%

5 0

3 years ago

Read 2 more answers

60>5(6-2x) ok help me so I can get an answer

Dima020 [189]

Inequalities are used to represent unequal expressions

The solution to the inequality is: $x > -3$

<h3>How to determine the inequality solution</h3>

The inequality is given as:

60>5(6-2x)

Rewrite the inequality properly as:

$60>5(6-2x)$

Divide both sides of the inequality by 5

$12>6-2x$

Subtract 6 from both sides of the inequality

$6>-2x$

Divide both sides of the inequality by -2

$-3Rewrite the inequality as[tex]x > -3$

Hence, the solution to the inequality is: $x > -3$

Read more about inequalities at:

brainly.com/question/11234618

5 0

2 years ago

A sphere of radius r is cut by a plane h units above the equator, where

Anika [276]

Consider the top half of a sphere centered at the origin with radius $r$ , which can be described by the equation

$z=\sqrt{r^2-x^2-y^2}$

and consider a plane

$z=h$

with $0$ . Call the region between the two surfaces $R$ . The volume of $R$ is given by the triple integral

$\displaystyle\iiint_R\mathrm dV=\int_{-\sqrt{r^2-h^2}}^{\sqrt{r^2-h^2}}\int_{-\sqrt{r^2-h^2-x^2}}^{\sqrt{r^2-h^2-x^2}}\int_h^{\sqrt{r^2-x^2-y^2}}\mathrm dz\,\mathrm dy\,\mathrm dx$

Converting to polar coordinates will help make this computation easier. Set

$\begin{cases}x=\rho\cos\theta\sin\varphi\\y=\rho\sin\theta\sin\varphi\\z=\rho\cos\var\phi\end{cases}\implies\mathrm dx\,\mathrm dy\,\mathrm dz=\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi$

Now, the volume can be computed with the integral

$\displaystyle\iiint_R\mathrm dV=\int_0^{2\pi}\int_0^{\arctan\frac{\sqrt{r^2-h^2}}h}\int_{h\sec\varphi}^r\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\varphi\,\mathrm d\theta$

You should get

$\dfrac{2\pi}3\left(r^3\arctan\dfrac{\sqrt{r^2-h^2}}h-\dfrac{h^3}2\left(\dfrac{r\sqrt{r^2-h^2}}{h^2}+\ln\dfrac{r+\sqrt{r^2-h^2}}h\right)\right)$

5 0

3 years ago

Factoring Trinomials

alexandr402 [8]

25 a² + 20 ab + 4 b²

for formula $a^x+bx^ky^k+cy ^{2k }$ find u,v such that: u.v = a

and u+v = b then group according to $(ax ^{2k} +ux^ky^k)+(vx^ky^k+cy ^{2k} )$

= $(25a^2+10ab)+(10ab+4b^2)$

factor out 5a from : $25a^2+10ab : 5a(2b+5a)$

factor out 2b from : $10ab+4b^2: 2b(2b+5a)$

= $5a(2b+5a) + 2b(2b+5a)$

factor out : $(5a+2b)$

= $(5a+2b)(5a+2b)$

Refine:

$(2b+5a)^2$

hope this helps!.

3 0

4 years ago

Solve for x. 2(3 + x) = 5x + 54A. 17B. 16C. -16D. -17​

Solve for x. 2(3 + x) = 5x + 54A. 17B. 16C. -16D. -17