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3 years ago

Least to greatest -55,143,18,-79,44,101

Mathematics

2 answers:

Dahasolnce [82]3 years ago

8 0

Answer:

Step-by-step explanation:

-79 , -55 , 18 , 44 , 101 , 143

katovenus [111]3 years ago

3 0

Answer:

-79, -55, 18, 44, 101, 143

Step-by-step explanation:

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Describe the translation 7 units to the left, 12 units up using a vector.

Lelu [443]

The magnitude of the vector is √(7^2+12^2)=√(49+144)=√193

The direction is α=arctan(y/x)=arctan(12/-7)≈-59.74 however this is relative to the negative x-axis so we add 180° to get the standard angle...180-59.74=120.26°

So the vector is √193 @ 120.26°

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3 years ago

Find the common difference if the 8th term is 55 and the first term is 13

Vlad [161]

Answer:

Step-by-step explanation:

Equation

L = a + (n - 1)*d

Givens

L = 55

a = 13

n =8

Solution

55 = 13 + (8 - 1)*d Combine

55 = 13 + 7d Subtract 13 from both sides

55 - 13 = 7d

42 = 7d Divide by 7

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2 years ago

Directions:Simplify the following monomials.SHOW ALL STEPS!

Mandarinka [93]

Answer:

7. Ans;

$\frac{6 {a}^{5} {b}^{7} }{ - 2 {a}^{3} {b}^{7} } = \frac{2 {a}^{3} {b}^{7}(3 {a}^{2} ) }{ - 2 {a}^{3} {b}^{7} } = - 3 {a}^{2}$

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8. Ans;

$\frac{ - 20 {x}^{3} {y}^{2} }{ - 5 {x}^{3}y } = \frac{ - 5 {x}^{3}y(4y) }{ - 5 {x}^{3}y } = 4y$

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9. Ans;

$\frac{ - 16 a {b}^{4} }{4 {b}^{3} } = \frac{4 {b}^{3}( - 4ab) }{4 {b}^{3} } = - 4ab$

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10. Ans;

$\frac{21 {m}^{8} {n}^{5} }{27 {m}^{5} {n}^{4} } = \frac{3 {m}^{5} {n}^{4}(7 {m}^{3} n) }{3 {m}^{5} {n}^{4}(9) } = \frac{7 {m}^{3}n }{9}$

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11. Ans;

$\frac{ - 15 {x}^{5} {y}^{4} }{45x {y}^{3} } = \frac{ 15x {y}^{3} ( - {x}^{4} y)}{15x {y}^{3}(3) } = - \frac{ {x}^{4}y }{3}$

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12.Ans;

$\frac{7 {p}^{2} {q}^{2} }{14 {p}^{2} {q}^{2} } = \frac{7 {p}^{2} {q}^{2} }{7 {p}^{2} {q}^{2} (2) } = \frac{1}{2} = 0.5$

I hope I helped you^_^

7 0

2 years ago

Read 2 more answers

What is the answer to this question

Sonja [21]

First we need to understand that the sum of adjacent angles on a straight line is 180°,meaning that when all the angles are added up the answer would be 180°

Here's how we can set up the equation:

2x+100°=180°
2x=80°
x=40°

Thus the answer is 40°

Again,hope it helps!

6 0

3 years ago

The polynomial P(x) = 2x^3 + mx^2-5 leaves the same remainder when divided by (x-1) or (2x + 3). Find the value of m and the rem

Zigmanuir [339]

Answer:

m=7

Remainder =4

If q=1 then r=3 or r=-1.

If q=2 then r=3.

They are probably looking for q=1 and r=3 because the other combinations were used earlier in the problem.

Step-by-step explanation:

Let's assume the remainders left when doing P divided by (x-1) and P divided by (2x+3) is R.

By remainder theorem we have that:

P(1)=R

P(-3/2)=R

$P(1)=2(1)^3+m(1)^2-5$

$=2+m-5=m-3$

$P(\frac{-3}{2})=2(\frac{-3}{2})^3+m(\frac{-3}{2})^2-5$

$=2(\frac{-27}{8})+m(\frac{9}{4})-5$

$=-\frac{27}{4}+\frac{9m}{4}-5$

$=\frac{-27+9m-20}{4}$

$=\frac{9m-47}{4}$

Both of these are equal to R.

$m-3=R$

$\frac{9m-47}{4}=R$

I'm going to substitute second R which is (9m-47)/4 in place of first R.

$m-3=\frac{9m-47}{4}$

Multiply both sides by 4:

$4(m-3)=9m-47$

Distribute:

$4m-12=9m-47$

Subtract 4m on both sides:

$-12=5m-47$

Add 47 on both sides:

$-12+47=5m$

Simplify left hand side:

$35=5m$

Divide both sides by 5:

$\frac{35}{5}=m$

$7=m$

So the value for m is 7.

$P(x)=2x^3+7x^2-5$

What is the remainder when dividing P by (x-1) or (2x+3)?

Well recall that we said m-3=R which means r=m-3=7-3=4.

So the remainder is 4 when dividing P by (x-1) or (2x+3).

Now P divided by (qx+r) will also give the same remainder R=4.

So by remainder theorem we have that P(-r/q)=4.

Let's plug this in:

$P(\frac{-r}{q})=2(\frac{-r}{q})^3+m(\frac{-r}{q})^2-5$

Let x=-r/q

This is equal to 4 so we have this equation:

$2u^3+7u^2-5=4$

Subtract 4 on both sides:

$2u^3+7u^2-9=0$

I see one obvious solution of 1.

I seen this because I see 2+7-9 is 0.

u=1 would do that.

Let's see if we can find any other real solutions.

Dividing:

1 | 2 7 0 -9

| 2 9 9

-----------------------

2 9 9 0

This gives us the quadratic equation to solve:

$2x^2+9x+9=0$

Compare this to $ax^2+bx+c=0$

$a=2$

$b=9$

$c=9$

Since the coefficient of $x^2$ is not 1, we have to find two numbers that multiply to be $ac$ and add up to be $b$ .

Those numbers are 6 and 3 because $6(3)=18=ac$ while $6+3=9=b$ .

So we are going to replace $bx$ or $9x$ with $6x+3x$ then factor by grouping:

$2x^2+6x+3x+9=0$

$(2x^2+6x)+(3x+9)=0$

$2x(x+3)+3(x+3)=0$

$(x+3)(2x+3)=0$

This means x+3=0 or 2x+3=0.

We need to solve both of these:

x+3=0

Subtract 3 on both sides:

x=-3

----

2x+3=0

Subtract 3 on both sides:

2x=-3

Divide both sides by 2:

x=-3/2

So the solutions to P(x)=4:

$x \in \{-3,\frac{-3}{2},1\}$

If x=-3 is a solution then (x+3) is a factor that you can divide P by to get remainder 4.

If x=-3/2 is a solution then (2x+3) is a factor that you can divide P by to get remainder 4.

If x=1 is a solution then (x-1) is a factor that you can divide P by to get remainder 4.

Compare (qx+r) to (x+3); we see one possibility for (q,r)=(1,3).

Compare (qx+r) to (2x+3); we see another possibility is (q,r)=(2,3).

Compare (qx+r) to (x-1); we see another possibility is (q,r)=(1,-1).

6 0

2 years ago