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ikadub [295]
3 years ago
7

Slope = 3; (-2, -2) plzz help

Mathematics
1 answer:
vladimir2022 [97]3 years ago
6 0

Answer:

y = 3x +4

Step-by-step explanation:

Given

m = 3 --- slope

(x,y) = (-2,-2)

Required

Determine the equation and plot the graph

The formula for the equation is:

y = m(x - x_1) + y_1

This gives:

y = 3(x - (-2)) - 2

y = 3(x +2) - 2

Open bracket

y = 3x +6 - 2

y = 3x +4

<em>See attachment for graph</em>

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Claire is going to invest in an account paying an interest rate of 5.4% compounded daily. How much would Claire need to invest,
Delicious77 [7]

Answer:

P ≈ 100

Step-by-step explanation:

260 = P(1+0.054/365)^365(14)

260 = P(1.000147945)^5110

260= P(2.129621112)

P= 122.087444818

P ≈ 100

4 0
3 years ago
Will mark brainliest! Questions are in the picture!
serg [7]
Option a, c,d are correct.
step-by-step explanation:
from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have
tz=sz(given)
bz=zb(common)
therefore, by rhs rule,δtzb ≅δszb
by cpctc, sz≅tz
also, from δctz and δasz,
tz=sz(given)
∠tcz=∠saz(90°)
by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz
also,from δasz and δzsb,
zs=sz(common)
∠zbs=∠saz=90°
by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb
hence, option a, c,d are correct.
5 0
2 years ago
32. Parallelogram ABCD has vertices A(0,0), B(2,4), and C(10,4). Find the coordinates of D.
goblinko [34]

Answer:

The fourth vertex is D(8, 0)

Step-by-step explanation:

Let A(0, 0), B(2, 4), and C(10, 4) be the three vertices of a parallelogram ABCD and let its fourth vertex be D(a, b).

Join AC and BD. Let AC and BD intersect at point O.

  • We have known that the diagonals of a parallelogram bisect each other.  

So, O would be the midpoint of AC as well as that of BD.

A\left(0,\:0\right)=\left(x_1,\:x_2\:\right)

C\left(10,\:4\right)=\left(x_2,\:x_2\:\right)

The midpoint of AC is:

\mathrm{Midpoint\:of\:}\left(x_1,\:y_1\right),\:\left(x_2,\:y_2\right):\quad \left(\frac{x_2+x_1}{2},\:\:\frac{y_2+y_1}{2}\right)

\left(x_1,\:y_1\right)=\left(0,\:0\right),\:\left(x_2,\:y_2\right)=\left(10,\:4\right)

=\left(\frac{10+0}{2},\:\frac{4+0}{2}\right)

=\left(5,\:2\right)

The midpoint of BD is:

=\left(\frac{a+2}{2},\:\frac{b+4}{2}\right)

so

∵  \frac{a+2}{2}=5

a+2=10

a=8

∵ \frac{b+4}{2}=2

b+4=4

b=0

Hence, the fourth vertex is D(8, 0)

4 0
3 years ago
Which values for e have the same reference angles?
Harman [31]

Hey there! :)

Answer:

Choice 4: π/4, 3π4, 7π/4

Step-by-step explanation:

For the values of θ to have the same reference angles, the denominators of the radians must be the same. Therefore:

Choice 1: π/6, π/3, 5π/6 - incorrect. π/3 will have a different reference angle as it has a denominator of 3.

Choice 2: π/3, 5π/6, 4π/3 - incorrect. 5π/6 will have a different reference angle as it has a denominator of 6.

Choice 3: π/2, 5π/4, 7π/4 - incorrect. π/2 will have a different reference angle.

Choice 4: π/4, 3π4, 7π/4 - correct. All of these radians contain the same denominator, and will each have the same reference angles of π/4.

7 0
3 years ago
Please help????<br> ?????
3241004551 [841]
The answer is Angle BCA. I’m pretty sure
5 0
4 years ago
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