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3 years ago

Pleaseeeee helppp!!!!!!!!!!!

Mathematics

1 answer:

Mrrafil [7]3 years ago

5 0

$4^{2x}=7^{x-1}\\
16^x=7^x\cdot7^{-1}\\
\dfrac{16^x}{7^x}=\dfrac{1}{7}\\
\left(\dfrac{16}{7}\right)^x=\dfrac{1}{7}\\
x=\log_{\tfrac{16}{7}}\dfrac{1}{7}\approx-2.35389$

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PLEASE HELP WILL GIVE BRAINLIEST.

kari74 [83]

Answer:

Choice D

Step-by-step explanation:

For this one I would find if the point lands on the line.

Choice A:

What we have to do is to plug in -4 for x and 4 for y.

$y=2x+10\\(4)=2(-4)+10\\4=-8+10\\4\ne2$

The point is not on this line so this cannot be it.

Choice B:

We pug what we know again.

$y=\frac{1}{3}x+1\\(4)=\frac{1}{3}(-4)+1\\(4)=-\frac{4}{3}+1\\4\ne-\frac{1}{3}$

The point is not on this line so it can't be it.

Choice C:

We pug in what we know again.

$y=3x-9\\(4)=3(-4)-9\\(4)=-12-9\\4\ne-21$

The point is not on this line so it can't be it.

The next one has to be it, but we'll check it just in case.

Choice D:

We plug in what we know again.

$y= \frac{1}{2}x+6\\(4)=\frac{1}{2}(-4)+6\\(4)=-2+6\\4=4$

The point is on this line so this is the line.

3 0

2 years ago

Help ppppppppppppllllllllllllllllsssssssss

iVinArrow [24]

Answer:

The correct option is D ....

Step-by-step explanation:

Angle is ending at the point = (-3, -4)

Signs of both x and y are negative therefore it is in the third quadrant.

To find the the tangent of an angle we have to find the ratio of the length of the opposite side to the length of the adjacent side.

In this question the opposite side of the angle is of 4 units and the adjacent side of the angle is 3 units.

tanθ = y/x

tanθ = -4/-3

tanθ = 4/3

Thus the correct option is D ....

5 0

3 years ago

What is the mean, medium, mode of 54,220

irina [24]

Answer:

Mean: 16.75

Median: 15.5

Mode: 13

Range: 16

Minimum: 9

Maximum: 25

Count n: 16

Sum: 268

Step-by-step explanation:

hope this helps!

7 0

2 years ago

Find the integral using substitution or a formula.

Nadusha1986 [10]

$\rm \int \dfrac{x^2+7}{x^2+2x+5}~dx$

Derivative of the denominator:
$\rm (x^2+2x+5)'=2x+2$

Hmm our numerator is 2x+7. Ok this let's us know that a simple u-substitution is NOT going to work. But let's apply some clever Algebra to the numerator splitting it up into two separate fractions. Split the +7 into +2 and +5.

$\rm \int \dfrac{x^2+2+5}{x^2+2x+5}~dx$

and then split the fraction,

$\rm \int \dfrac{x^2+2}{x^2+2x+5}~dx+\int\dfrac{5}{x^2+2x+5}~dx$

Based on our previous test, we know that a simple substitution will work for the first integral: $\rm \quad u=x^2+2x+5\qquad\to\qquad du=2x+2~dx$

So the first integral changes,

$\rm \int \dfrac{1}{u}~du+\int\dfrac{5}{x^2+2x+5}~dx$

integrating to a log,

$\rm ln|x^2+2x+5|+\int\dfrac{5}{x^2+2x+5}~dx$

Other one is a little tricky. We'll need to complete the square on the denominator. After that it will look very similar to our arctangent integral so perhaps we can just match it up to the identity.

$\rm x^2+2x+5=(x^2+2x+1)+4=(x+1)^2+2^2$

So we have this going on,

$\rm ln|x^2+2x+5|+\int\dfrac{5}{(x+1)^2+2^2}~dx$

Let's factor the 5 out of the intergral,
and the 4 from the denominator,

$\rm ln|x^2+2x+5|+\frac54\int\dfrac{1}{\frac{(x+1)^2}{2^2}+1}~dx$

Bringing all that stuff together as a single square,

$\rm ln|x^2+2x+5|+\frac54\int\dfrac{1}{\left(\dfrac{x+1}{2}\right)^2+1}~dx$

Making the substitution: $\rm \quad u=\dfrac{x+1}{2}\qquad\to\qquad 2du=dx$

giving us,

$\rm ln|x^2+2x+5|+\frac54\int\dfrac{1}{\left(u\right)^2+1}~2du$

simplying a lil bit,

$\rm ln|x^2+2x+5|+\frac52\int\dfrac{1}{u^2+1}~du$

and hopefully from this point you recognize your arctangent integral,

$\rm ln|x^2+2x+5|+\frac52arctan(u)$

undo your substitution as a final step,
and include a constant of integration,

$\rm ln|x^2+2x+5|+\frac52arctan\left(\frac{x+1}{2}\right)+c$

Hope that helps!
Lemme know if any steps were too confusing.

8 0

3 years ago

Suppose the lengths of two sides of a right triangle are represented by 2x and 3 (x + 1), and the longest side is 17 units. Find

densk [106]

Answer:

x=4

Step-by-step explanation:

Step 1:-

given the lengths of two sides of a right angle are represented by 2x and 3(x+1) and longest side is 17 units.

AB = 2x and BC = 3(x+1) and longest side AC= 17

by using Pythagoras theorem

$AC^2 = AB^2 + BC^2$