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3 years ago

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Mathematics

2 answers:

Alenkinab [10]3 years ago

8 0

Answer:

Step-by-step explanation:

To solve this, you can add 2/5 b to both sides of the equation. You now have 3 + 5/5 (or 1) b = 11. => 3 + b = 11 => Now, subtract 3 from both sides, and you have b = 11 - 3, and then b = 8.

taurus [48]3 years ago

6 0

Step-by-step explanation:

3 + 2/5 b =11 - 2/5 b (this is the given question is it?)

2/5 b + 2/5 b= 11 - 3 (like terms together, you transpose -2/5b to the other side as shown.)

2b+2b = 8 ( at that stage you would find t

5 hat 5 is the common value to go into 5 itself, as shown.

4b = 8 ( you just add 2b + 2b to have 4b)

4b = 8 ( at that point you cross multiply

5 1 as shown)

4b = 8 × 5 ( simple math as shown)

4b = 40 ( you multiply 8 × 5 to obtain 40)

b =10 you can prove that. Thank you.

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Convert 75 degrees to radican measure

NeX [460]

Answer:5π/12

Step-by-step explanation:

180°=π

75°=x

x=75π ➗ 180

x=5π/12

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3 years ago

WILL DO BRAINLIEST!!!! Suppose the store wants to earn a daily profit of $150 from the sale of soccer balls. To earn this profit

lara [203]

Think about it this way, what is 5x3??? 15 right. Ok then what is 5x30???? 150. So the store should sell each soccer ball for $5 each:)

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4 years ago

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List all possible rational roots. Then use synthetic division to confirm which rational roots exist:

Kisachek [45]

Answer:

$\boxed{(1) \, x = \, \pm \dfrac{1}{2}, \pm 1, \pm2, \pm \dfrac{5}{2}, \pm 5, \pm 10; (2) \, x = -2}$

Step-by-step explanation:

2x³+ 6x² - x - 10 = 0

(1) Possible roots

The Rational Roots Theorem states that, if a polynomial has any rational roots, they will have the form p/q, where p is a factor of the constant term and q is a factor of the leading coefficient.

$\text{Possible rational root} = \dfrac{ p }{ q } = \dfrac{\text{factor of constant term}}{\text{factor of leading coefficient}}$

In your function, the constant term is -10 and the leading coefficient is 2, so

$\text{Possible root} = \dfrac{\text{factor of 10}}{\text{factor of 2}}$

Factors of 10 = ±1, ±2, ±5, ±10

Factors of 2 = ±1, ±2

$\text{Possible roots are } \large \boxed{\mathbf{x = \pm \dfrac{1}{2}, \pm 1, \pm2, \pm \dfrac{5}{2}, \pm 5, \pm 10}}$

(2) Synthetic division

Rather than work through all 12 possibilities, I will do one that works.

$\begin{array}{r|rrrr}-2 & 2 & 6 & -1 & -10\\& & -4& -4 & 10\\& 2 & 2& -5 & 0\\\end{array}$

So, x = -2 is a root, and the quotient is 2x² + 2x - 5.

(3) Check for other rational roots

2x² + 2x - 5 = 0

D = b² - 4ac =2²- 4(2)(-5) = 4 + 40 = 44

√44 = 2√11, which is irrational.

Since irrational roots come in pairs, the cubic equation has two real, irrational roots and one rational root at x = -2.

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3 years ago

Write in slope intercept form an equation of the line that passes through the given points. (-1,7) , (3,-1)

Tcecarenko [31]

Answer! :

y = -2x + 5

Step by Step! :

slope intercept form:

y = mx + b

m being slope, b being the y intercept.

To find the slope, use this equation:

y^2 - y^1 / x^2 - x^1

Plug in your points.

(-1) - (7) / (3) - (-1)

-8 / 4

-2 is your slope! (y = -2x + b)

To solve for b, plug in any one of your points and solve for b. Let's use (3, -1)

-1 = -2(3) + b

-1 = -6 + b

5 = b

Your new equation is...

y = -2x + 5

6 0

2 years ago

Which ordered pair makes both inequalities true? y < 3x – 1 y > –x + 4 On a coordinate plane, 2 straight lines are shown.

Katen [24]

Answer:

(4,0)

Step-by-step explanation:

we have

$y< 3x-1$ ----> inequality A

$y \geq -x+4$ ----> inequality B

we know that

If a ordered pair is a solution of the system of inequalities, then the ordered pair must satisfy both inequalities (makes true both inequalities)

Verify each ordered pair

case 1) (4,0)

Inequality A

$0< 3(4)-1$

$0< 11$ ----> is true

Inequality B

$0 \geq -(4)+4$

$0 \geq 0$ ----> is true

the ordered pair makes both inequalities true

case 2) (1,2)

Inequality A

$2< 3(1)-1$

$2< 2$ ----> is not true

the ordered pair not makes both inequalities true

case 3) (0,4)

Inequality A

$4< 3(0)-1$

$4< -1$ ----> is not true

the ordered pair not makes both inequalities true

case 4) (2,1)

Inequality A

$1< 3(2)-1$

$1< 5$ ----> is true

Inequality B

$1 \geq -(2)+4$

$1 \geq 2$ ----> is not true

the ordered pair not makes both inequalities true

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3 years ago

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