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3 years ago

A triangle has sides with lengths of 21 meters, 70 meters, and 73 meters. Is it a right triangle?

Mathematics

1 answer:

Elena-2011 [213]3 years ago

8 0

9514 1404 393

Answer:

Step-by-step explanation:

It is close, but not exactly a right triangle.

If it were a right triangle, we would have ...

21² +70² = 73²

We have instead ...

441 +4900 ≠ 5329

5341 ≠ 5329 . . . . . . . . not a right triangle

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Need help dont know how to round round this number to the nearest thousand? 0.7966347977

tatyana61 [14]

Answer:

.797

Step-by-step explanation:

If your talking about a decimal number, you would say the thousandth. I know, weird, but don't blame me.

To round up, any number above 5 rounds up to the next value place, and any number below 5 rounds down to the last value place.

Ex: 1.3483838383

Since 8 is greater than 5, we could get rid of all the 83s and just round up to 1.35.

So the thousandths place in a decimal would be 0.79666357977, so we just ask if the next number after the thousandths place is greater than 5. The place number after .796 is 6, so since it is greater than 5, we would round up to .797.

Our answer is .797.

Hope this helps!

3 0

3 years ago

Read 2 more answers

6570 ÷ 16 410 410 1/16 410 3/8 410 5/8

UNO [17]

Answer:

Exact Form:

3285

Decimal Form:

410.625

Mixed Number Form:

410

thats what i got i hope this helps

Step-by-step explanation:

4 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

Help pls

gogolik [260]

Answer:

I think the answer is 3x

Step-by-step explanation:

If i am wrong please correct me

4 0

2 years ago

The sum of one-third of a number and three-fourths of the number exceeds that number by one. Which equation could be used to fin

Marrrta [24]