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andrew-mc [135]
3 years ago
5

How do you solve this problem?

Mathematics
2 answers:
myrzilka [38]3 years ago
6 0
You would need a protractor to find the problem
Gnom [1K]3 years ago
6 0
M<BCD = 180 - <DCE - <ECG
             = 180 -  31 - 24
             = 125
             
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1,236,840 in scientific notation.
Leni [432]
The answer is 123684 x 10^1. Hope i helped.(whatever comes after ^ is an exponent)
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3 years ago
Find the angle between the vectors u=i-9j and v=8i+5j
Katyanochek1 [597]

Answer: \cos^{-1} \left(\frac{-37}{\sqrt{7298}} \right) \approx 115.665^{\circ}

Step-by-step explanation:

u \cdot v=(1)(8)+(-9)(5)=-37\\\\|u|=\sqrt{1^{2}+(-9)^{2}}=\sqrt{82}\\\\|v|=\sqrt{8^{2}+5^{2}}=\sqrt{89}\\\\\cos \theta=\frac{-37}{\sqrt{82}\sqrt{89}}\\\\\theta=\cos^{-1} \left(\frac{-37}{\sqrt{82}\sqrt{89}} \right)\\\\\theta=\boxed{\cos^{-1} \left(\frac{-37}{\sqrt{7298}} \right) \approx 115.665^{\circ}}

7 0
2 years ago
Read the sentence below and answer the following question: Roy was very careful to cleave the apple into even portions. What doe
irga5000 [103]
To separate into parts
5 0
3 years ago
Anderson is a plumber and is paid a fixed amount just to make a house visit. The remainder of his fee, y, is determined by the n
Serhud [2]

Answer:

y=10x+50

Explanation:

The line starts at point (0,50) which is the y intercept (how much he gets paid to visit house).The line crosses points (10,150) and (20,250) and the slope of these points are 100/10 which when simplified equals 10x which is how much money he gets paid by the hour(x=hours) so y=10x plus the y intercept y=10x+50

7 0
3 years ago
Read 2 more answers
(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

<em>Since none equals 0, then r is irrational</em>

3 0
3 years ago
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