The perimeter of a rectangular field is 342 yards. If the width of the field is 72 yards, what is its length

Whats the difference between "6 + A" and "6 x A"

Nata [24]

Answer:

To determine what is the difference between "6 + A" and "6 x A", the logic of the proposed mathematical operations must be explained:

In "6 + A", the value A is added to the initial value 6. Thus, for example, if A were worth 10, to the initial value 6 10 units are added, with which the final value is 16.

In contrast, in "6 x A", the initial value 6 is multiplied by as many times as the value A indicates. Therefore, continuing with the value of A as 10, in this case 6 would be multiplied by 10 times, giving a final value of 60.

6 0

3 years ago

Please tell me what question b and c is ASAP

artcher [175]

Answer:

5432gffhjkkkolllmnnn

7 0

3 years ago

Read 2 more answers

(PLEASE ANSWER + BRAINLIEST!!)

Licemer1 [7]

The answer is x=20. Hope this helps.

8 0

3 years ago

The sum of two numbers is 64. The difference between the numbers is 14. Write a system

Anastasy [175]

Answer:

They say that the sum of the two numbers (x and y) is 64, so we can write our first equation by adding them and setting that sum equal to 64:

x + y = 64

The question also tells us that their difference is 14. Similarly to before, we'll just subtract the two numbers and set that difference equal to 14:

x - y = 14

Now from here, you know how to continue with substitution to find the values for x and y. Just remember when you get a word problem to break it down and look for key words like sum, difference, or product, and from there you'll be able to build your system of equations.

8 0

2 years ago

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 $\frac{-1}{10}$

6 0

3 years ago