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2 years ago

Please Answer These questions (on the picture) 20 Points

Mathematics

1 answer:

telo118 [61]2 years ago

8 0

#13 is 17
#14 is 42
#15 is 15
#16 is 7
#17 is 3
#18 is 4

only giving you those fam good luck homie

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What is the measure ?

Ivenika [448]

Check the picture below.

4 0

3 years ago

The areas of two similar triangles are 72dm2 and 50dm2. The sum of their perimeters is 226dm. What is the perimeter of each of t

astraxan [27]

Answer:

Hence, the perimeter of the triangles are:

P=123.2727 dm

P'=102.7272 dm

Step-by-step explanation:

In two similar triangles:

The ratio of the areas of two triangle is equal to the square of their perimeters.

Let A and A' represents the area of two triangles and P and P' represents their perimeter.

Then they are related as:

$\dfrac{A}{A'}=\dfrac{P^2}{P'^2}$

We are given:

A=72 dm^2 , A'=50 dm^2

and P+P'=226 dm.-----------(1)

i.e. $\dfrac{72}{50}=\dfrac{P^2}{P'^2}\\\\\dfrac{36}{25}=\dfrac{P^2}{P'^2}$

on taking square root on both the side we get:

$\dfrac{P}{P'}=\dfrac{6}{5}\\\\P=\dfrac{6}{5}P'$

Now putting the value of P in equation (1) we obtain:

$\dfrac{6}{5}P'+P'=226\\\\\dfrac{6P'+5\times P'}{5}=226\\\\\dfrac{6P'+5P'}{5}=226\\\\11P'=226\times 5\\\\11P'=1130\\\\P'=\dfrac{1130}{11}=102.7272$

Hence,

P=226-102.7272=123.2727

Hence, the perimeter of the triangles are:

P=123.2727 dm

P'=102.7272 dm

6 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

ms. Omar runs the school tennis club. She has a bin of tennis balls and rackets. For every 5 tennis balls in the bin, there are

irina1246 [14]

Answer:

$a=\frac{5}{3}b\\\\or\\\\3a=5b$