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3 years ago

2/3 of the product of 3/8 and 16

1 answer:

7 0

First setting up the first step of the equation
3/8*16=?
Does 16/1=16? Yes it does! So you can put 16/1 in place of the 16!
So, 3/8*16/1= 48/8
And 48/8=6
<span>The solution is 6! Do you have further questions?</span>

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So lost. sum one pls help

Brums [2.3K]

Answer: 3x-y=10 , 6x-2y=5. 3x−y=10 3 x - y = 10 , 6x−2y=5 6 x - 2 y = 5. Solve for y y in the first equation. Tap for more steps... Subtract 3x 3 x from

Graph y=6x-10 |

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3 years ago

-6=-8+<br><img src="https://tex.z-dn.net/?f=%20%5Cfrac%7Bp%7D%7B11%7D%20" id="TexFormula1" title=" \frac{p}{11} " alt=" \frac{p}

vekshin1

P=22 rewrite the question as -8+p/11=-6 then move all terms not containing p to the right side p/11=2 multiply both sides of the given equation by 11, the reciprocal, 11*p/11=11*2

8 0

3 years ago

64÷54 is the number of.... whar

ankoles [38]

Your answer is 1.18518519

8 0

3 years ago

Read 2 more answers

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

given that BD is the median of ABD and that ABD is isosceles congruence postulate SSS can be used to prove which of the followin

Alja [10]

It is given in the question that

BD is the median. So it divides the opposite sides in two equal parts .

Therefore in triangles BAD and BCD,

AB and AC are congruent because of isosceles triangle.

AD and CD are congruent because of the median BD.

And BD and BD are congruent .

So the two triangles are congruent by SSS and the correct option is the first option .

3 0

3 years ago

Read 2 more answers