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antoniya [11.8K]

3 years ago

12

an electronics store sells large flat screen tvs for 1699 last month the store sold 8 of this tvs anout how much did the tvs sel

l for

1 answer:

rusak2 [61]3 years ago

4 0

The answer is $212.37 for each large flatscreen tv

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I need help 25 points plz

Dmitriy789 [7]

Answer:

-1

Step-by-step explanation:

6 0

4 years ago

Find c such that (−3,−8), (−7,−6), and (c,4) lie on a line.

bearhunter [10]

If the 3 points are collinear, then the slopes of all line segments connecting the points are the same.

Thus,

-6 - (-8) 2
m = ------------- = -------- = -1/2
-7 - (-3) -4

Then the following must be true:

4-(-6) 10
-1/2 = ------------ = ---------
c - (-7) c + 7

Cross multiplying, -(c+7) = 20, and c+7 = -20, so that c = -27

7 0

3 years ago

1. Quadrilateral ABCD has vertices A(-1, 1), B(2, 3), C(6, 0) and D(3, -2). Determine using coordinate geometry whether or not t

deff fn [24]

Answer:

The Conclusion is

Diagonals AC and BD,

a. Bisect each other

b. Not Congruent

c. Not Perpendicular

Step-by-step explanation:

Given:

[]ABCD is Quadrilateral having Vertices as

A(-1, 1),

B(2, 3),

C(6, 0) and

D(3, -2).

So the Diagonal are AC and BD

To Check

The diagonals AC and BD

a. Bisect each other. B. Are congruent. C. Are perpendicular.

Solution:

For a. Bisect each other

We will use Mid Point Formula,

If The mid point of diagonals AC and BD are Same Then

Diagonal, Bisect each other,

For mid point of AC

$Mid\ point(AC)=(\dfrac{x_{1}+x_{2} }{2},\dfrac{y_{1}+y_{2} }{2})$

Substituting the coordinates of A and C we get

$Mid\ point(AC)=(\dfrac{-1+6}{2},\dfrac{1+0}{2})=(\dfrac{5}{2},\dfrac{1}{2})$

Similarly, For mid point of BD

Substituting the coordinates of B and D we get

$Mid\ point(BD)=(\dfrac{2+3}{2},\dfrac{3-2}{2})=(\dfrac{5}{2},\dfrac{1}{2})$

Therefore The Mid point of diagonals AC and BD are Same

Hence Diagonals,

a. Bisect each other

B. Are congruent

For Diagonals to be Congruent We use Distance Formula

For Diagonal AC

$l(AC) = \sqrt{((x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2} )}$

Substituting A and C we get

$l(AC) = \sqrt{((6-(-1))^{2}+(0-1)^{2} )}=\sqrt{(49+1)}=\sqrt{50}$

Similarly ,For Diagonal BD

Substituting Band D we get

$l(BD) = \sqrt{((3-2))^{2}+(-2-3)^{2} )}=\sqrt{(1+25)}=\sqrt{26}$

Therefore Diagonals Not Congruent

For C. Are perpendicular.

For Diagonals to be perpendicular we need to have the Product of slopes must be - 1

For Slope we have

$Slope(AC)=\dfrac{y_{2}-y_{1} }{x_{2}-x_{1} }$

Substituting A and C we get

$Slope(AC)=\dfrac{0-1}{6--1}\\\\Slope(AC)=\dfrac{-1}{7}$

Similarly, for BD we have

$Slope(BD)=\dfrac{-2-3}{3-2}\\\\Slope(BD)=\dfrac{-5}{1}$

The Product of slope is not -1

Hence Diagonals are Not Perpendicular.

6 0

4 years ago

Please helpppppp me

alex41 [277]

Answer:

B. $4 \times 10^{12}$

Step-by-step explanation:

$(6 \times 10^{8}) \div (1.5 \times 10^{-4}) =$

$= \dfrac{6 \times 10^{8}}{1.5 \times 10^{-4}}$

$= \dfrac{6}{1.5} \times \dfrac{10^{8}}{10^{-4}}$

$= 4 \times 10^{8 - (-4)}$

$= 4 \times 10^{12}$

7 0

3 years ago

Help me out hear ples

tatyana61 [14]

Answer:

tony:25,000 explmation 2,500x10 shoe:25% discount:22 because 120 +10 is 22

Step-by-step explanation:

give brainliest and thank ty

8 0

3 years ago