All categories

3 years ago

1/3 (6x - 15) = 1/2 (10x - 4 ) a. -10 b. 1 c. 1/10 d. -1

Mathematics

1 answer:

Ulleksa [173]3 years ago

5 0

Answer:

d)-1

Step-by-step explanation:

1/3 (6x-15)

6 times -1 equals -6

-6 minus 15 equals -21

1/3 times -21 equals -7

1/2 (10x-4)

10 times -1 equals -10

-10 minus 4 equals -14

1/2 times -14 equals -7

You might be interested in

For each function, what is the output of the given input? for f(x) = 4x 7, find f(4)

shusha [124]

I hope this helps you

x =4

f (4)= 4.4+7 = 16+7=23

f (4)= 4.4-7=16-7=9

6 0

3 years ago

Which figure has two bases and one lateral face that is rectangular?

vodomira [7]

Rectangular pyramid

4 0

3 years ago

Read 2 more answers

Which graph shows a negative rate of change for the interval 0 to 2 on the x-axis?

valentina_108 [34]

It seems that the four graphs are the same and they do not have a negative change rate in the interval 0 to 2 in the x-axis.

A negative change rate means that when x increases the value of the function (y) decreases; this is, the function is decreasing in the interval being estudied, which is the same that going downward.

So, you must look for in your graphs where the equation is going downward.

For example, in the graph attached, that happens in any interval from negative infitity to 1.5.

The vertex will help you to identify it.

Given that the graph goes downward from negative infinity to the vertex, any interval that includes that range will have negative change.

You must look for a parabola that opens upward and whose vertex is in x = 2.

Read more on Brainly.com - brainly.com/question/3774202#readmore

5 0

4 years ago

Read 2 more answers

Can anyone help me integrate :

worty [1.4K]

Rewrite the second factor in the numerator as

$2x^2+6x+1=2(x+2)^2-2(x+2)-3$

Then in the entire integrand, set $x+2=\sqrt3\sec t$ , so that $\mathrm dx=\sqrt3\sec t\tan t\,\mathrm dt$ . The integral is then equivalent to

$\displaystyle\int\frac{(\sqrt3\sec t-2)(6\sec^2t-2\sqrt3\sec t-3)}{\sqrt{(\sqrt3\sec t)^2-3}}(\sqrt3\sec t)\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\sec^2t-1}}\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\tan^2t}}\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{|\tan t|}\,\mathrm dt$

Note that by letting $x+2=\sqrt3\sec t$ , we are enforcing an invertible substitution which would make it so that $t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3}$ requires $0\le t$ or $\dfrac\pi2$ . However, $\tan t$ is positive over this first interval and negative over the second, so we can't ignore the absolute value.

So let's just assume the integral is being taken over a domain on which $\tan t>0$ so that $|\tan t|=\tan t$ . This allows us to write

$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\tan t}\,\mathrm dt$
$=\displaystyle\int(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\csc t\,\mathrm dt$

We can show pretty easily that

$\displaystyle\int\csc t\,\mathrm dt=-\ln|\csc t+\cot t|+C$
$\displaystyle\int\sec t\csc t\,\mathrm dt=-\ln|\csc2t+\cot2t|+C$
$\displaystyle\int\sec^2t\csc t\,\mathrm dt=\sec t-\ln|\csc t+\cot t|+C$
$\displaystyle\int\sec^3t\csc t\,\mathrm dt=\frac12\sec^2t+\ln|\tan t|+C$

which means the integral above becomes

$=3\sqrt3\sec^2t+6\sqrt3\ln|\tan t|-18\sec t+18\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|-6\ln|\csc t+\cot t|+C$
$=3\sqrt3\sec^2t-18\sec t+6\sqrt3\ln|\tan t|+12\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|+C$

Back-substituting to get this in terms of $x$ is a bit of a nightmare, but you'll find that, since $t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3}$ , we get

$\sec t=\dfrac{x+2}{\sqrt3}$
$\sec^2t=\dfrac{(x+2)^2}3$
$\tan t=\sqrt{\dfrac{x^2+4x+1}3}$
$\cot t=\sqrt{\dfrac3{x^2+4x+1}}$
$\csc t=\dfrac{x+2}{\sqrt{x^2+4x+1}}$
$\csc2t=\dfrac{(x+2)^2}{2\sqrt3\sqrt{x^2+4x+1}}$

etc.

3 0

3 years ago

A boy band has released two albums. Their first album sold 5.9×10^5 copies. Their second album sold 1.3×10^6 copies. How many to

prisoha [69]

The band has sold 1,890,000 copies or 1.89 x 10^6

3 0

3 years ago

Read 2 more answers