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3 years ago

A point at (-5,7) in the standard (x,y) coordinate

1 answer:

dsp733 years ago

7 0

The correct answer is actually (2,2)

What you do is you add 7 to the x coordinate since it is 7 to the right, -5+7=2 so the new x coordinate is 2

Then you subtract 5 from the y coordinate since it is 5 down, 7-5=2 so the new y coordinate is 2

So the answer is (2,2)

You might be interested in

Find the value of z.

AVprozaik [17]

Answer:

z = 110°

Step-by-step explanation:

Angles z and the one marked 70° form an linear pair, hence are supplementary.

z = 180° -70° = 110°

<h3>Angles where chords cross</h3>

The angle made by two chords is half the sum of the arcs intercepted by those chords. Here, that means ...

70° = 1/2(60° +x) ⇒ x = 140° -60° = 80°

The arc w completes the circle of 360°, so we have ...

w +x +79° +60° = 360° ⇒ w = 360° -219° = 141°

Finally, z is the average of w and 79°:

z = (w +79°)/2 = (141° +79°)/2 = 110° . . . as above

8 0

2 years ago

Solve for j.<br>-13j - 20 = -8j + 20

aleksley [76]

Answer:

j = -8

Step-by-step explanation:

-13j - 20 = -8j + 20

Add 13 j to each side

-13j+13j - 20 = -8j+13j + 20

-20 = 5j+20

subtract 20 from each side

-20-20 = 5j +20-20

-40 = 5j

Divide by 5

-40/5 = =5j/5

-8 =j

8 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

Which is the correct comparison of solutions for 2x + 6 < 10 and -2x + 22 < 18?

Ronch [10]

IDK. I looked up the answer but i couldn't find it. I'm on UsaTestPrep and they gave me the same question.

5 0

3 years ago

I REALLY NEED HELP WITH THIS

gtnhenbr [62]

Answer:

x=24

Step-by-step explanation:

6 0

3 years ago