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2 years ago

Plz help me solve this small equation plz show working Thank you so much

Mathematics

1 answer:

neonofarm [45]2 years ago

5 0

Answer:

can you type it out??

Step-by-step explanation:

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What is the domain of this

Kaylis [27]

Answer:

$\boxed{\sf B. \ All \ real \ numbers}$

Step-by-step explanation:

The domain is all possible values for x.

$f(x)=(\frac{1}{4} )^x$

There are no restrictions on x.

The domain is all real numbers.

3 0

2 years ago

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Enter the value of p so the expression 5/6-1/3nis equivalent to p(5-2n).

SIZIF [17.4K]

Take LCD of 6 and 3 = 6 is the lcm
5/6 -2/6n
(5-2n)/6.

Now, what number can be multiplied to the right side of the equal sign to get the value of P

The answer is 1/6 as the left side is divided by 6, so multiply the right side by 1/6 to make it equivalent.

6 0

3 years ago

If the Leaning Tower of Pisa (which is roughly cylindrical) has a height of 55 meters, radius of 8 meters, what is the volume?

valentinak56 [21]

Answer:

Option C. $V=3,520\pi\ m^3$

Step-by-step explanation:

we know that

The volume of a cylinder is equal to

$V=\pi r^{2} h$

we have

$r=8\ m$

$h=55/ m$

substitute

$V=\pi (8)^{2} (55)$

$V=3,520\pi\ m^3$

8 0

2 years ago

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The sum of first three terms of a finite geometric series is -7/10 and their product is -1/125. [Hint: Use a/r, a, and ar to rep

Alchen [17]

Ooh, fun

geometric sequences can be represented as
$a_n=a(r)^{n-1}$
so the first 3 terms are
$a_1=a$
$a_2=a(r)$
$a_2=a(r)^2$

the sum is -7/10
$\frac{-7}{10}=a+ar+ar^2$
and their product is -1/125
$\frac{-1}{125}=(a)(ar)(ar^2)=a^3r^3=(ar)^3$

from the 2nd equation we can take the cube root of both sides to get
$\frac{-1}{5}=ar$
note that a=ar/r and ar²=(ar)r
so now rewrite 1st equation as
$\frac{-7}{10}=\frac{ar}{r}+ar+(ar)r$
subsituting -1/5 for ar
$\frac{-7}{10}=\frac{\frac{-1}{5}}{r}+\frac{-1}{5}+(\frac{-1}{5})r$
which simplifies to
$\frac{-7}{10}=\frac{-1}{5r}+\frac{-1}{5}+\frac{-r}{5}$
multiply both sides by 10r
-7r=-2-2r-2r²
add (2r²+2r+2) to both sides
2r²-5r+2=0
solve using quadratic formula
for $ax^2+bx+c=0$
$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$
so
for 2r²-5r+2=0
a=2
b=-5
c=2

$r=\frac{-(-5) \pm \sqrt{(-5)^2-4(2)(2)}}{2(2)}$
$r=\frac{5 \pm \sqrt{25-16}}{4}$
$r=\frac{5 \pm \sqrt{9}}{4}$
$r=\frac{5 \pm 3}{4}$
so
$r=\frac{5+3}{4}=\frac{8}{4}=2$ or $r=\frac{5-3}{4}=\frac{2}{4}=\frac{1}{2}$

use them to solve for the value of a
$\frac{-1}{5}=ar$
$\frac{-1}{5r}=a$
try for r=2 and 1/2
$a=\frac{-1}{10}$ or $a=\frac{-2}{5}$

test each
for a=-1/10 and r=2
a+ar+ar²= $\frac{-1}{10}+\frac{-2}{10}+\frac{-4}{10}=\frac{-7}{10}$
it works

for a=-2/5 and r=1/2
a+ar+ar²= $\frac{-2}{5}+\frac{-1}{5}+\frac{-1}{10}=\frac{-7}{10}$
it works

both have the same terms but one is simplified

the 3 numbers are $\frac{-2}{5}$ , $\frac{-1}{5}$ , and $\frac{-1}{10}$

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3 years ago

Determine whether the graphs of y=-2x-7 and -y=2x+13 are parallel, perpendicular, coincident, or none of these.

Elden [556K]

It's parallel. Definitely true

3 0

2 years ago

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