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2 years ago

I need help on understanding this concept

Mathematics

1 answer:

Sedbober [7]2 years ago

3 0

So the slope formula is m= (y2-y1) /
(x2-x1). You can label -2 as x1 and 11 as y1. And the other point is x2 and y2. They you just plug it in the m= (y2-y1) /
(x2-x1) formula and solve. To get a better understanding look up, how to find the slope of a line with two points. Hope it helps

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271,403 to the nearest hundred thousand

11Alexandr11 [23.1K]

Answer: Your answer would be 300,000

Step-by-step explanation: Hope this helped

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3 years ago

A recipe requires

Alla [95]

Answer:

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6 0

2 years ago

A small explanation would be really appreciated

shutvik [7]

Step-by-step explanation:

the 2 angles marked with orange and green are equal becos the question said that the indicated angles are congruent meaning same deg. so 180 - that given angle = the 2 marked angles (based on the rule known as angles on a str line are supplementary)

since the 2 marked angles are same, you can say that d and k are parallel due to the rule of alternate angles

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7 0

3 years ago

Determine the value of x in the diagram where lines a and b are parallel

Yakvenalex [24]

If lines a and b are parallel, the two angles given would be the same.
2x - 5 = x + 20
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6 0

3 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