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3 years ago

What is the answer for the three questions please help

Mathematics

1 answer:

ycow [4]3 years ago

7 0

The driver can drive around the track 54 times for $12. Thus for one dollar, he can drive around the track 54/12 times, or 4.5 times. If for each dollar he drives around 4.5 times, for 8 dollars, he can drive around 8*4.5 or 36 times.

If there are 207 students in 9 classes, in 1 class, there are 207/9 students, or 23 students per class. Since there are 23 students per class and there are classes, there are a total of 23*6 or 138 students in total.

Liam practices 24 times in 4 days. Thus he practices 24/4 or 6 times per day. His brother practices 12 times in 3 days. Thus he practices 12/3 or 4 times a day. Thus their rates are not equivalent.

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What is the value of x? 20 35 60 70

baherus [9]

Answer:

20°

Step-by-step explanation:

Step 1:

x + 40° = 3x Vertical ∠'s

Step 2:

40° = 2x Subtract x on both sides

Step 3:

x = 40° ÷ 2 Divide

Answer:

x = 20°

Hope This Helps :)

8 0

3 years ago

Provide a statement and reason for each of the following using the indicated

grin007 [14]

2x + 5 = 11

(-5)

2x = 6

(/2)

x = 3

5 0

2 years ago

What is the standard form equation for y=2x-4?

Oksanka [162]

Isn't it already in standard form?

4 0

3 years ago

(a) Determine a cubic polynomial with integer coefficients which has $\sqrt[3]{2} + \sqrt[3]{4}$ as a root.

Keith_Richards [23]

Answer:

(a) $x\³ - 6x - 6$

(b) Proved

Step-by-step explanation:

Given

$r = $\sqrt[3]{2} + \sqrt[3]{4}$$ --- the root

Solving (a): The polynomial

A cubic function is represented as:

$f = (a + b)^3$

Expand

$f = a^3 + 3a^2b + 3ab^2 + b^3$

Rewrite as:

$f = a^3 + 3ab(a + b) + b^3$

The root is represented as:

$r=a+b$

By comparison:

$a = $\sqrt[3]{2}$

$b = \sqrt[3]{4}$$

So, we have:

$f = ($\sqrt[3]{2})^3 + 3*$\sqrt[3]{2}*\sqrt[3]{4}$*($\sqrt[3]{2} + \sqrt[3]{4}$) + (\sqrt[3]{4}$)^3$

Expand

$f = 2 + 3*$\sqrt[3]{2*4}*($\sqrt[3]{2} + \sqrt[3]{4}$) + 4$

$f = 2 + 3*$\sqrt[3]{8}*($\sqrt[3]{2} + \sqrt[3]{4}$) + 4$

$f = 2 + 3*2*($\sqrt[3]{2} + \sqrt[3]{4}$) + 4$

$f = 2 + 6($\sqrt[3]{2} + \sqrt[3]{4}$) + 4$

Evaluate like terms

$f = 6 + 6($\sqrt[3]{2} + \sqrt[3]{4}$)$

Recall that: $r = $\sqrt[3]{2} + \sqrt[3]{4}$$

So, we have:

$f = 6 + 6r$

Equate to 0

$f - 6 - 6r = 0$

Rewrite as:

$f - 6r - 6 = 0$

Express as a cubic function

$x^3 - 6x - 6 = 0$

Hence, the cubic polynomial is:

$f(x) = x^3 - 6x - 6$

Solving (b): Prove that r is irrational

The constant term of $x^3 - 6x - 6 = 0$ is -6

The divisors of -6 are: -6,-3,-2,-1,1,2,3,6

Calculate f(x) for each of the above values to calculate the remainder when f(x) is divided by any of the above values

$f(-6) = (-6)^3 - 6*-6 - 6 = -186$

$f(-3) = (-3)^3 - 6*-3 - 6 = -15$

$f(-2) = (-2)^3 - 6*-2 - 6 = -2$

$f(-1) = (-1)^3 - 6*-1 - 6 = -1$

$f(1) = (1)^3 - 6*1 - 6 = -11$

$f(2) = (2)^3 - 6*2 - 6 = -10$

$f(3) = (3)^3 - 6*3 - 6 = 3$

$f(6) = (6)^3 - 6*6 - 6 = 174$

For r to be rational;

The divisors of -6 must divide f(x) without remainder

i.e. Any of the above values must equal 0

Since none equals 0, then r is irrational

3 0

2 years ago

What is the product of 14.5×100

liberstina [14]

Answer:

1450

Step-by-step explanation:

Any number times 10 goes up one value.

ones become tens, and tens become hundreds etc..

87x10=870

And a number times 100 goes up two values.

23.5x100= 2350

hope this helped ;)

3 0

3 years ago

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