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3 years ago

Helppppp meeee pleaseeee

Mathematics

1 answer:

ss7ja [257]3 years ago

4 0

B is 0 hope thsi helps and to die lol

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The product of (n2 – 6n + 3) and -4n is________ When this product is multiplied by -n, the result is________

krek1111 [17]

-4n^3+24n^2-12n, 4n^4-24n^3+12n^2

3 0

2 years ago

Read 2 more answers

Based on the attendance at past picnics, Luis expects 95% of the students in his class to attend this years picnic. If 120 stude

Sever21 [200]

Answer:

95 / 100 * 120 = 114 students

Thank you and please rate me as brainliest as it will help me to level up

6 0

3 years ago

If the nth term of a sequence is 3n+2 them T³= T⁵=T¹⁵=

Katena32 [7]

Answer:

T^75

Step-by-step explanation:

3 times 5 = 15

5 times 15 = 75

7 0

3 years ago

Given the set of vertices, determine whether parallelogram ABCD is a rhombus, rectangle or square. List all that apply. A(7,-4),

Sloan [31]

Given:

Vertices of a parallelogram ABCD are A(7,-4), B(-1,-4), C(-1,-12), D(7, -12).

To find:

Whether the parallelogram ABCD is a rhombus, rectangle or square.

Solution:

Distance formula:

$D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Using distance formula, we get

$AB=\sqrt{(-4-(-4))^2+(-1-7)^2}$

$AB=\sqrt{(-4+4)^2+(-8)^2}$

$AB=\sqrt{0+64}$

$AB=8$

Similarly,

$BC=\sqrt{(-1-(-1))^2+(12-(-4))^2}=8$

$CD=\sqrt{(7-(-1))^2+(-12-(-12))^2}=8$

$AD=\sqrt{(7-7)^2+(-12-(-4))^2}=8$

All sides of parallelogram are equal.

$AC=\sqrt{(-1-7)^2+(-12-(-4))^2}=8\sqrt{2}$

$BD=\sqrt{(7-(-1))^2+(-12-(-4))^2}=8\sqrt{2}$

Both diagonals are equal.

Since, all sides are equal and both diagonals are equal, therefore, the parallelogram ABCD is a square.

We know that, a square is special case of rectangles and rhombus.

So, parallelogram ABCD is a rhombus, rectangle or square. Therefore, the correct option is c.

7 0

2 years ago

Solve.<br> 10x^2 - 6 = 9x

grin007 [14]

Answer:

Step-by-step explanation:

use quadratic formula: (-b ±√b²-4ac) ÷ 2a

a = 10, b = -9, c = -6

7 0

2 years ago

Helppppp meeee pleaseeee​

Helppppp meeee pleaseeee